Slide 1

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 2

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 3

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 4

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 5

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 6

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 7

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 8

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 9

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 10

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 11

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 12

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 13

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 14

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 15

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 16

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 17

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 18

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 19

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 20

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 21

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 22

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 23

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 24

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 25

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 26

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 27

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 28

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 29

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 30

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 31

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 32

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 33

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 34

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 35

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 36

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 37

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 38

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 39

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 40

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 41

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 42

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 43

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 44

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 45

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 46

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 47

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 48

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 49

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations


Slide 50

Electronics
Principles & Applications
Sixth Edition

Charles A. Schuler
Chapter 6

Introduction to
Small-Signal Amplifiers
(student version)
©2003

Glencoe/McGraw-Hill

INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations

Dear Student:
This presentation is arranged in segments.
Each segment is preceded by a Concept
Preview slide and is followed by a Concept
Review slide. When you reach a Concept
Review slide, you can return to the
beginning of that segment by clicking on the
Repeat Segment button. This will allow you
to view that segment again, if you want to.

Concept Preview
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.

• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.
• System gain or loss is found by adding dB stage
gains or losses.

1.5 V

In

5V

Amplifier

Out

The units cancel
Gain =

5V
Out
InV
1.5

= 3.33

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2

10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 3981

The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

The dB unit is based on a power ratio.

P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
The dB unit can be adapted to a voltage ratio.

dB = 20 x log

VOUT
VIN

This equation assumes VOUT and VIN
are measured across equal impedances.

+10 dB

-6 dB

+30 dB

-8 dB

+20 dB

dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB

Gain Quiz
Amplifier output is equal to the input
________ by the gain.

multiplied

Common logarithms are ________ of
the number 10.

exponents

Doubling a log is the same as _________
the number it represents.

squaring

System performance is found by
________ dB stage gains and losses.

adding

Logs of numbers smaller than one
are ____________.

negative

Concept Review
• Voltage gain is the ratio of Vout to Vin.
• Power gain is the ratio of Pout to Pin.
• Common logarithms are exponents of 10.
• Gain or loss in decibels is equal to 10 times the
log of the power ratio or 20 times the log of the
voltage ratio.
• dB voltage gain equals dB power gain when the
input impedance equals the output impedance.

• System gain or loss is found by adding dB stage
gains or losses.
Repeat Segment

Concept Preview
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.
• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.
• If a signal drives the amplifier beyond either or
both limits the output will be clipped.

• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.

A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.

RB

RL
C

CC

B

E

VCC

The output
is phase inverted.

RB

RL
C

CC

B

E

VCC

When the input signal goes positive:
The base current increases.
The collector current increases b times.

So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.

RB

RL
C

CC

B

E

VCC

When the input signal goes negative:
The base current decreases.
The collector current decreases b times.

So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.

RB

RL
C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.

14 V
IC(MAX) =
1 kW

350 kW

1 kW
C

CC

B

E

14 V

The load line connects the limits.
This end is called
saturation.

The linear region is between the limits.
100 mA

14
12
10
IC in mA 8
6
4
2

SAT.

LINEAR

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
This end is called cutoff.

CUTOFF

0 mA

Use Ohm’s Law to determine the base current:

14 V
IB =
350 kW

350 kW

1 kW
C

CC

= 40 mA

B

E

14 V

An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.

100 mA

14
12
10
IC in mA 8
6
4
2

Q

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
Q = the quiescent point

0 mA

The input signal varies the base
current above and below the Q point.

100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 mA

Overdriving the amplifier causes clipping.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
The output is non-linear.

0 mA

What’s wrong with this Q point?
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18

VCE in Volts
How about this one?

0 mA

14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V

VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 150

14 V
= 40 mA (IB is not affected)
IB =
350 kW
IC = b x IB = 350 x 40 mA = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW = 14 V (VR is higher)
L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification.
350 kW

1 kW
C

CC

B

E

14 V
b = 350
b is higher

The higher b causes
saturation.
100 mA

14
12
10
IC in mA 8
6
4
2

80 mA
60 mA
40 mA
20 mA

0 2 4 6

8 10 12 14 16 18
VCE in Volts

The output is non-linear.

0 mA

This common-emitter amplifier is not practical.

It’s b dependent!
It’s also temperature dependent.
RB

RL
C

CC

B

E

VCC

Basic C-E Amplifier Quiz
The input and output signals in C-E
are phase ______________.

inverted

The limits of an amplifier’s load line
are saturation and _________.

cutoff

Linear amplifiers are normally operated
near the _________ of the load line.
center

The operating point of an amplifier is
also called the ________ point.

quiescent

Single resistor base bias is not practical
since it’s _________ dependent.

b

Concept Review
• In a common emitter amplifier (C-E), the base
is the input and the collector is the output.

• C-E amplifiers produce a phase inversion.
• The circuit limits are called saturation and
cutoff.

• If a signal drives the amplifier beyond either or
both limits the output will be clipped.
• The operating point (Q-point) should be
centered between saturation and cutoff.
• Beta-dependent amplifiers are not practical.
Repeat Segment

Concept Preview
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.

• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.

• Emitter bypassing increases the voltage gain.

This common-emitter amplifier is practical.
RB1

RL
C

CC
B
RB2

VCC

E
RE

It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

+VCC

Voltage divider bias
RB1

RL

RB1 and RB2 form
a voltage divider
RB2

RE

+VCC

Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

B
RB2 2.7 kW

VB =

RB2
RB1 + RB2

x VCC

RL= 2.2 kW
C

2.7 kW
x 12 V
VB =
2.7 kW + 22 kW

E

VB = 1.31 V

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E

RE = 220 W

Solving the practical circuit for its dc conditions:

VCC = 12 V
RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

IE =

IE =

VE

0.61 V
220 W

RE

= 2.77 mA

E

RE = 220 W

IC @ IE

Solving the practical circuit for its dc conditions:
VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kW

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

VRL = 6.09 V
VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 W

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:
1. Calculate the base
voltage using the
voltage divider
equation.
2. Subtract 0.7 V to get
the emitter voltage.

3. Divide by emitter
resistance to get the
emitter current.

4. Determine the drop
across the collector
resistor.

5. Calculate the collector
to emitter voltage
using KVL.
6. Decide if the Q-point
is linear.
7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V
The ac emitter resistance is rE:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

RE = 220 W

25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V
The voltage gain from base to collector:

RB1 22 kW

RL= 2.2 kW
C

B
RB2 2.7 kW

E

AV =

RE = 220 W

AV =

RL
RE + rE

2.2 kW
220 W + 9.03 W

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V
RB1 22 kW

An emitter bypass capacitor
can be used to increase AV:

RL= 2.2 kW

AV =

C

B
RB2 2.7 kW

E
RE

AV =

CE

2.2 kW
9.03 W

RL
rE
= 244

Practical C-E Amplifier Quiz
b-dependency is reduced with emitter
feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is
subtracted from ____________.

VB

To find VCE, VRL and VE are
subtracted from _________.

VCC

Voltage gain is equal to the collector
resistance _______ by the emitter resistance. divided
Voltage gain can be increased by
________ the emitter resistor.

bypassing

Concept Review
• C-E amplifiers can be stabilized by using
voltage divider base bias and emitter feedback.
• The base current can be ignored when
analyzing the divider for the base voltage (VB).
• Subtract VBE to find VE.
• Use Ohm’s Law to find IE and VRL.
• Use KVL to find VCE.
• IE determines the ac emitter resistance (rE).
• RL, RE and rE determine the voltage gain.
• Emitter bypassing increases the voltage gain.
Repeat Segment

Concept Preview
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF
applications.
• The analysis procedure for PNP amplifiers is
the same as for NPN.

The common-emitter configuration is used most often.
It has the best power gain.

VCC
Medium output Z

RB1

RL
C

B
Medium input Z

RB2

E

RE

CE

The common-collector configuration is shown below.
Its input impedance and current gain are both high.

VCC
RB1

It’s often called an
emitter-follower.

RC
C

B
RB2

E

RL

In-phase
output
Low output Z

The common-base configuration is shown below.
It’s used most
at RF.

Its voltage gain is high.

VCC
RB1

RL
C

B
RB2
Low input Z

E

RE

In-phase
output

PNP C-E Amplifier
VB = - 3.75 V
1.5 kW
22 kW

12 V

+

VE = - 3.05 V
IE = 2.913 mA rE = 8.58 W

10 kW

47 W

1 kW

VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V

AV = 27

Amplifier Configuration Quiz
In a C-E amplifier, the base is the input
and the __________ is the output.

collector

In an emitter-follower, the base is the
input and the ______ is the output.

emitter

The only configuration that phaseinverts is the ________.

C-E

The configuration with the best power
gain is the ________.

C-E

In the common-base configuration,
the ________ is the input terminal.

emitter

Concept Review
• C-E amplifiers are the most widely applied.
• C-E amplifiers are the only ones that provide a
phase inversion.
• C-C amplifiers are also called emitter followers.
• C-C amplifiers are noted for their low output
impedance.
• C-B amplifiers are noted for their low input
impedance.
• C-B amplifiers are used mostly in RF applications.
• The analysis procedure for PNP amplifiers is the
same as for NPN.
Repeat Segment

REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations