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Introduction to Transistors
• A transistor is a device with three separate layers of
semiconductor material stacked together
– The layers are made of n–type or p–type material in the
order pnp or npn
– The layers change abruptly to form the pn or np junctions
– A terminal is attached to each layer
(The Art of Electronics, Horowitz
and Hill, 2nd Ed.)
(Introductory Electronics, Simpson, 2nd Ed.)
Introduction to Transistors
• Thus when a transistor is
off it behaves like a
two–diode circuit
• A transistor operates (or
(Lab 4–1)
turns on) when the base–emitter junction is forward
biased and the base–collector junction is reversed
biased (“biasing”)
(The Art of Electronics,
Horowitz and Hill, 2nd Ed.)
(Electronic Devices and
Circuits, Bogart, 1986)
Transistor Biasing (npn Transistor)
• Electrons are constantly
supplied to the emitter by
the battery with voltage VEE
• These electrons can:
1. Recombine with holes in
the base, giving rise to IB
2. Diffuse across base and be swept (by electric field at
base–emitter junction) into collector, then diffuse around
and eventually recombine with holes injected into
collector, giving rise to IC
• Since the base region is designed so thin, process 2
dominates (no time for #1 to occur as often)
– In an actual npn transistor, 98 or 99% of the electrons that
diffuse into the base will be swept into the collector
Current Flow Inside a Transistor
• Current flow for an npn transistor (reverse for pnp):
– From conservation of current
(IE = IB + IC) we can obtain the following
expressions relating the currents:
I C  bI B
(Electronic Devices and Circuits,
Bogart, 1986)
I E  b  1I B
(and thus IC ≈ IE)
where b ≈ 20 – 200
(depends on emitter current)
• b increases as IE increases (for very small IE) since there is less
chance that recombination will occur in the base
• b decreases slightly (10–20%) as IE increases beyond several mA
due to increased base conductivity resulting from larger number of
charge carriers in the base
• Thus b is not a constant for a given transistor!
• An average value of 100 is typically used
(Lab 4–5)
Transistor Current Amplification
• If the “input” current is IB and the “output” current is
IC, then we have a current amplification or gain
– Happens because base–emitter junction is forward-biased
– Forward bias ensures that the base–emitter junction
conducts (transistor is turned on)
– Reverse bias ensures that most of the large increase in
the base–emitter current shows up as collector current
Thus small gains in IB
result in large gains in IE
and hence IC
(Student Manual for The Art
of Electronics, Hayes and
Horowitz, 2nd Ed.)
Basic Transistor Switch Circuit
• Transistor “switch” circuit:
(Lab 4–9)
(BC junction forward biased)
VB
0.6 V
(The Art of Electronics, Horowitz and Hill, 2nd Ed.)
0.2 V VC
0 V VE
– With switch open, transistor is off and lamp is off
– With switch closed, IB = (10 – 0.6) V / 1k = 9.4 mA
– However, IC = bIB  940 mA (assuming b = 100)
•
•
•
•
When collector current IC = 100 mA, lamp has 10V across it
To get a higher current, collector would need to be below ground
Transistor can’t do this, so it goes into saturation
Collector voltage gets as close to emitter voltage as it can (about
0.2 V higher) and IC remains constant (IC is “maxed out”)
Emitter Follower
• Output “follows” the input: only
difference is a 0.6 V diode drop
– True for Vin > 0.6 V
– If Vin < 0.6 V, transistor turns off (no
current – “valve” is closed) and Vout = 0
– Data with RE = 3.3k:
Vin
Vout
(Lab 4–2)
B
C
E
E
(The Art of
Electronics, Horowitz
and Hill, 2nd Ed.)
Emitter Follower
• By returning the emitter
resistor to a negative
supply voltage, you can
obtain negative voltage
swings as well
– Data with RE = 3.3k:
(The Art of Electronics,
Horowitz and Hill, 2nd Ed.)
Emitter Follower Biasing
• You must always provide a DC path for base bias
current, even if it is just through a resistor to ground
– HW Problem 2.5
(The Art of Electronics, Horowitz and Hill, 2nd Ed.)
Emitter Follower Biasing
• With RB included in the previous circuit:
f = 1 kHz
Emitter Follower Biasing
• Without RB included in the previous circuit:
(Here there is no DC base bias current, so transistor is off.)
Emitter Follower Biasing
(Lab 4–4)
• To obtain symmetric output waveforms without
“clipping,” provide constant DC bias using a voltage
divider
– Capacitors block “outside” DC current, which may affect
quiescent (no input) values (“AC-coupled follower”)
(The Art of Electronics, Horowitz and Hill, 2nd Ed.)
Emitter Follower Impedance
• The usefulness of the emitter follower can be seen
by determining its input and output impedance:
– Input impedance (i.e. the impedance looking into the base
of the transistor):
Zin  Z load 1  b   b Z load
• Details of proof given in class
– Output impedance (i.e. the impedance looking into the
emitter of the transistor):
Z out
Z source Z source


1 b
b
• Details provided by you in the homework!
• Thus the input impedance is much larger than the
output impedance
Emitter Follower Impedance
• Thus the input and output “sees” what it wants to
see on the other side of the transistor:
(Student Manual for The Art
of Electronics, Hayes and
Horowitz, 2nd Ed.)
• Using an emitter follower, a given signal source
requires less power to drive a load than if the source
were to drive the load directly
– Very good, since in general we want
Zout (stage n) << Zin (stage n + 1) (by at least a factor of 10)
– An emitter follower has current gain, even though it has no
voltage gain
– The emitter follower has power gain
Emitter Follower Impedance
• When measuring the input and output impedance of
the emitter follower, it is useful to think about the
Thévenin equivalent circuit as “seen” at the input
and the output: (Lab 4–3)
– Input impedance seen by the source:
Vin
VB
Zsource
Zin
Z in
VB 
Vin
Z source  Z in
– Output impedance seen by the load:
Vout, no load
~
(Student Manual for The Art
of Electronics, Hayes and
Horowitz, 2nd Ed.)
Vout, load
Zout
Zload
Z load
Vout,load 
Vout, no load
Z out  Z load
Emitter Follower With Load (HW 2.2)
(The Art of Electronics,
Horowitz and Hill, 2nd Ed.)
• Consider the following circuit:
Vin
IE
Vout
– Vout and Vin waveforms:
Vin (V) Vout (V) IE (mA)
Vin
Vout
+9.4
5
0
8.8
4.4
–0.6
27.6
18.8
8.8
–3
–3.6
2.8
–4.4
–5
–5.0
–5.0
0.0
0.0
–10
–5.0
0.0
Emitter Follower With Load
• Thus the npn emitter follower can only “source”
current (supply current to something like a load)
• It cannot “sink” current (draw current from something
like a load)
• In this example, the transistor turns off when
Vin = –4.4 V (Vout = –5.0 V)
– Then IE = 0 and the base–emitter junction becomes reverse
biased
– As Vin increases further, a rather large reverse bias
develops across this junction which could result in
breakdown
• The output could swing more negative than –5 V by
reducing the RE = 1k resistor, but this increases
power consumption in both the resistor and transistor
Zener Diodes as Voltage Regulators
• Zener diodes “like” to break down at
a particular reverse bias:
– When reverse biased, they provide a
constant voltage drop over a wide range
of currents
• Zeners thus provide a means of
voltage regulation
(Student Manual for The Art
of Electronics, Hayes and
Horowitz, 2nd Ed.)
(The Art of
Electronics, Horowitz
and Hill, 2nd Ed.)
– We choose the specifications for the zener based on:
Vin,min  Vout
R
 I out (max)
Pzener
 Vin,max  Vout



 I out,min Vzener
Rmin


Example Problem 2.3
Design a +10 V regulated supply for load
currents from 0 to 100 mA; the input
voltage is +20 to +25 V. Allow at least
10 mA zener current under all (worst-case)
conditions. What power rating must the
zener have?
Solution details given in class.
Emitter Followers as Voltage Regulators
• However, the zener current can change significantly
depending on the load, affecting regulation
performance
• A better voltage regulator would incorporate an
emitter follower:
(HW 2.4)
(The Art of
Electronics, Horowitz
and Hill, 2nd Ed.)
– Here the zener current is more constant, relatively
independent of load current since changes in IE (or Iload)
produce only small changes in IB
• Load current determined from (VB – 0.6 V) / Rload
Transistors as Current Sources
• A transistor can be used as a
current source with the setup at
right: (Lab 4–6)
VC
VE VB  0.6
I E  IC 

RE
RE
VE
(The Art of Electronics, Horowitz and Hill, 2nd Ed.)
– Note that IC is independent of VC as long as VC > VE + 0.2 V
(i.e., the transistor is not saturated)
• The output voltage (Vload or VC) range over which Iload
(= IC) is (nearly) constant is called the output
compliance
Deficiencies of Current Sources
• The load current will still vary somewhat, even when
the transistor is “on” and not in saturation
• There are two kinds of effects that cause this:
– VBE varies somewhat with collector-to-emitter voltage for a
given collector current (Early effect), as does b
• DVBE ≈ –0.0001 DVCE
• We assume VBE = constant = 0.6 V in the basic transistor model
– VBE and b depend on temperature
• DVBE ≈ –2.1 mV/0C
• We neglect changes in b by assuming IC = IE
• To minimize DVBE from both effects, choose VE large
enough ( 1V) so that DVBE  10 mV will not result in
large fractional changes in the voltage across RE
– VE too large will result in decreased output compliance,
however (VC range for transistor “on” state decreases)
Common–Emitter Amplifier (Lab 4–7)
• Consider a transistor current
source with a resistor RC as
load, and block unwanted
DC at the base input (Vin is
an AC signal):
(Note DC quiescent
output voltage of 10 V)
(The Art of Electronics, Horowitz and Hill, 2nd Ed.)
1
1
f 3dB 
C
so
2 Req C
2 f 3dB Req where Req  R1 R2 b RE
– Now imagine we apply a base wiggle vB via the input signal
– The emitter follows the wiggle so vE = vB
– Then the wiggle in the emitter current is:
(lower-case letters
vE vB
iE 

 iC
represent small
RE RE
changes, or “wiggles”)
Common–Emitter Amplifier
– Now VC = VCC – ICRC so vC = –iCRC = –vB(RC / RE)
– Since vin = vB and vout = vC, we have a voltage amplifier,
with a voltage gain of:
vout
RC
G

vin
RE
– Minus sign means that a positive wiggle at the input gets
turned into a negative wiggle at the output
• Input and output impedance:
– Zin = R1  R2  bRE ≈ 8k (see figure on previous slide)
– Zout = RC  (impedance looking into collector) = RC  (high Z
current source) ≈ RC = 10k (see figure on previous slide)
• Be careful to choose R1 and R2 correctly so that
design is not b dependent (R1  R2 << bRE)