Transcript Lecture 1 - Ilam university

BJT Amplifiers (cont’d)
Common-base topology
OUTLINE
• Common-base topology
– CB core
– CB stage with source resistance
– Impact of base resistance
• CB stage with biasing
– Emitter follower (Common-collector amplifier)
– Analysis of emitter follower core
– Impact of source resistance
– Impact of Early effect
• Emitter follower with biasing
Reading: Chapter 5.3.2, 5.3.3-5.4
Common Base (CB) Amplifier

The base terminal is biased at a fixed voltage; the input
signal is applied to the emitter, and the output signal
sensed at the collector.
Small-Signal Analysis of CB Core

The voltage gain of a CB stage is gmRC, which is
identical to that of a CE stage in magnitude and
opposite in phase.
Av  g m RC
Tradeoff between Gain and Headroom

To ensure that the BJT operates in active mode, the
voltage drop across RC cannot exceed VCC-VBE.
IC
VCC  VBE
Av 
RC 
VT
VT
Simple CB Stage Example
VCC = 1.8V
IC = 0.2mA
IS = 5x10-17 A
b = 100
Vb  1.354V 
R2
VCC if I1  I B
R1  R2
Choose I1  10I B  20A 
1
Av  g m RC 
 2230  17 .2
130
VCC
R1  R2
 R1  22.3k, R2  67.7k
Input Impedance of a CB Stage

The input impedance of a CB stage is much smaller
than that of a CE stage.
1
Rin 
if VA  
gm
CB Stage with Source Resistance

With the inclusion of a source resistance, the input
signal is attenuated before it reaches the emitter of
the amplifier; therefore, the voltage gain is lowered.

This effect is similar to CE stage emitter degeneration.
Av 
RC
1
 RS
gm
Practical Example of a CB Stage

An antenna usually has low output impedance;
therefore, a correspondingly low input impedance is
required for the following stage.
Output Impedance: CE vs. CB Stages

The output impedances of emitter-degenerated CE
and CB stages are the same. This is because the
circuits for small-signal analysis are the same when
the input port is grounded.
Output Impedance of a CB Stage

The output impedance of a CB stage is equal to RC in
parallel with the impedance looking into the collector.
Rout1  1  g m ( RE || r )rO  RE || r 
Rout 2  RC || Rout1
Av of CB Stage with Base Resistance (VA = ∞)

With base resistance, the voltage gain degrades.
vout
vout   g m v RC  v  
g m RC
v
vout
r  RB 
vP   r  RB  
r
r g m RC
vP 
vout
r  RB 
bRC
vout
r  RB   vin




v
v v
v
bR
1
KCL at node P :   g mv  P in    g m   out   C
r
RE
RE
 r
 g m RC 
vout
bRC
RC


1
RB
vin r  b  1RE  RB
 RE 
gm
b 1
Voltage Gain: CE vs. CB Stages

The magnitude of the voltage gain of a CB stage with
source and base resistances is the same as that of a
CE stage with base resistance and emitter
degeneration.
Rin of CB Stage with Base Resistance (VA = ∞)

The input impedance of a CB stage with base
resistance is equal to 1/gm plus RB divided by (b+1).
This is in contrast to a degenerated CE stage, in
which the resistance in series with the emitter is
multiplied by (b+1) when seen from the base.
v
KCL   g m v  ix
r
r
v  
vx
r  RB
1


r
  g m  
v x   ix
 r
 r  RB 
v x r  RB 1
RB
Rin  


ix
b 1 gm b 1
Input Impedance Seen at Emitter vs. Base
Common Base Stage
Common Emitter Stage
Input Impedance Example

To find RX, we have to first find Req, treat it as the
base resistance of Q2 and divide it by (b+1).
1
RB
Req 

g m1 b  1
1
1  1
RB 


Rx 


g m 2 b  1  g m1 b  1 
Biasing of CB Stage

RE is necessary to provide a path for the bias
current IE to flow, but it lowers the input impedance.
1
 RE
g
1
RE
Rin 
|| RE  m

1
gm
 RE 1  g m RE
gm
vX 
Rin
RE
vin 
vin
Rin  RS
RE  1  g m RE RS
vout vout v X
Av 


vin v X vin
vout
RE
 g m RC 
vin
RE  1  g m RE RS
Reduction of Input Impedance Due to RE

The reduction of input impedance due to i1 is
undesirable because it shunts part of the input
current to ground instead of to Q1 (and RC).
 Choose RE >> 1/gm , i.e. ICRE >> VT
Creation of Vb


A resistive voltage divider lowers the gain.
To remedy this problem, a capacitor is inserted
between the base and ground to short out the
resistive voltage divider at the frequency of interest.
Example of CB Stage with Bias
Design a CB stage for Av = 10 and Rin = 50.
Rin = 50 ≈ 1/gm if RE >> 1/gm
 Choose RE = 500







VCC = 2.5V
IS = 5x10-16 A
b = 100
VA = ∞
Av = gmRC = 10  RC = 500
IC = gm·VT = 0.52mA
VBE=VTln(IC/IS)=0.899V
Vb = IERE + VBE = 1.16V
Choose R1 and R2 to provide Vb
and I1 >> IB, e.g. I1 = 52A
CB is chosen so that (1/(b+1))(1/wCB) is small compared to 1/gm
at the frequency of interest.
BJT Amplifiers (cont’d)
Emitter Follower
Emitter Follower
(Common Collector Amplifier)
Emitter Follower Core


When the input voltage (Vin) is increased by Vin, the
collector current (and hence the emitter current)
increases, so that the output voltage (Vout) is
increased.
Note that Vin and Vout differ by VBE.
Unity-Gain Emitter Follower

In integrated circuits, the follower is typically realized
as shown below.

The voltage gain is 1 because a constant collector current
(= I1) results in a constant VBE; hence Vout = Vin .
VA  
Av  1
Small-Signal Model of Emitter Follower

The voltage gain is less than 1 and positive.
VA  
v  vin  vout
KCL at emitter:
v
v
 g m v  out
r
RE
vin  vout
vout
 g m vin  vout  
r
RE
vout
1
RE


vin 1  r  1 R  1
E
gm
b  1 RE
Emitter Follower as a Voltage Divider
VA  
Emitter Follower with Source Resistance
VA  
vout
RE

vin R  1  RS
E
gm b  1
Input Impedance of Emitter Follower

The input impedance of an emitter follower is the
same as that of a CE stage with emitter
degeneration (whose input impedance does not
depend on the resistance between the collector and
VCC).
VA  
vx
Rin   r  (1  b ) RE
ix
Effect of BJT Current Gain


There is a current gain of (b+1) from base to emitter.
Effectively, the load resistance seen from the base is
multiplied by (b+1).
Emitter Follower as a Buffer

The emitter follower is suited for use as a buffer
between a CE stage and a small load resistance, to
alleviate the problem of gain degradation.

Av  g m RC Rspeaker

Rin1  r 2  (1  b2 ) Rspeaker
Av  gm RC Rin1 
Output Impedance of Emitter Follower


An emitter follower effectively lowers the source
impedance by a factor of b+1, for improved driving
capability.
The follower is a good “voltage buffer” because it
has high input impedance and low output
impedance.
 1
Rs 
 || RE
Rout   
 gm b 1 
Emitter Follower with Early Effect

Since rO is in parallel with RE, its effect can be easily
incorporated into the equations for the voltage gain
and the input and output impedances.
Av 
RE || rO
RS
1
RE || rO 

b 1 gm
Rin  r  b  1 RE || rO 
Rout
 Rs
1 
 
  || RE || rO
 b 1 gm 
Emitter Follower with Biasing


A biasing technique similar to that used for the CE
stage can be used for the emitter follower.
Note that VB can be biased to be close to VCC
because the collector is biased at VCC.
Supply-Independent Biasing

By putting an independent current source at the
emitter, the bias point (IC, VBE) is fixed, regardless
of the supply voltage value.
Summary of Amplifier Topologies


The three amplifier topologies studied thus far have
different properties and are used on different
occasions.
CE and CB stages have voltage gain with magnitude
greater than one; the emitter follower’s voltage gain is
at most one.
Amplifier Example #1

The keys to solving this problem are recognizing the
AC ground between R1 and R2, and using a
Thevenin transformation of the input network.
CE stage
Small-signal
equivalent circuit
Simplified small-signal
equivalent circuit
vout
R2 || RC
R1


R1 || RS
1
vin

 RE R1  RS
b 1 gm
Amplifier Example #2

AC grounding/shorting and Thevenin transformation
are needed to transform this complex circuit into a
simple CE stage with emitter degeneration.
vout
RC
R1


RS || R1
1
vin

 R2 R1  RS
b 1 gm
Amplifier Example #3


First, identify Req, which is the impedance seen at
the emitter of Q2 in parallel with the infinite output
impedance of an ideal current source.
Second, use the equations for a degenerated CE
stage with RE replaced by Req.
Rin  r 1  r 2  R1
1
R1
Req 

gm2 b  1
 RC
Av 
1
1
R1


g m1 g m 2 b  1
Amplifier Example #4
Note that CB shorts out R2 and provides a ground for
R1, at the frequency of interest.
 R1 appears in parallel with RC; the circuit simplifies to
a simple CB stage with source resistance.

RC || R1
Av 
1
 RS
gm
Amplifier Example #5

Note that the equivalent base resistance of Q1 is the
parallel connection of RE and the impedance seen at
the emitter of Q2.

RS 
1
1  1
 || RE 
Rin 



g m1 b  1  g m 2 b  1 
