PHYS225
Lecture 6
Electronic Circuits
Last lecture
– Transistors
• History
• Basic physics of operation
• Ebers-Moll model
– Small signal equivalent
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
(The Art of Electronics, Horowitz
– A terminal is attached to each layer
and Hill, 2 Ed.)
nd
(Introductory Electronics, Simpson, 2nd Ed.)
Transistors
Heat sink
Introduction to Transistors
• When a transistor is off it behaves like a two–diode
circuit
• A transistor operates (or 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 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 (where b ≈ 20 – 200)
I C  bI B
I E  b  1I B
(and thus IC ≈ IE)
• 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
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:
(BC junction forward biased)
(The Art of Electronics, Horowitz and Hill, 2nd Ed.)
VB
0.6 V
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 and Vout = 0
– Data with RE = 3.3k:
B
C
E
E
(The Art of Electronics,
Horowitz and Hill, 2nd
Ed.)
Vout
Vin
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
(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
• 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
– Output impedance (i.e. the impedance looking into the emitter
of the transistor):
Z out
Z source Z source


1 b
b
• 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:
– 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
(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)
+9.4
8.8
27.6
Vin
Vout
5
0
–3
4.4
–0.6
–3.6
18.8
8.8
2.8
–4.4
–5
–10
–5.0
–5.0
–5.0
0.0
0.0
0.0
Emitter Follower With Load
• 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
Transistors as Current Sources
• A transistor can be used as a current source
VE VB  0.6
I E  IC 

RE
RE
– 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
• Consider a transistor current source with a resistor RC as
load, and block unwanted DC at the base input (Vin is an
AC signal):
1
f 3dB 
2 Req C
C
1
2 f 3dB Req
Req  R1 R2 b RE
(Note DC quiescent
output voltage of 10 V)
– Now imagine we apply a base signal vB
– The emitter follows the wiggle so vE = vB
– Then the change in the emitter current is:
vE vB
iE 

 iC
RE RE
lower-case letters represent small changes
Common–Emitter Amplifier
– 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 change at the input gets
turned into a negative change at the output
• Input and output impedance:
– Zin = R1  R2  bRE ≈ 8k
– Zout = RC  (impedance looking into collector) = RC  (high Z
current source) ≈ RC = 10k
• Be careful to choose R1 and R2 correctly so that design is
not b dependent (R1  R2 << bRE)
Transistor topologies
•
•
•
•
Switch
Current source
Emitter follower
Common-emitter amplifier
• Common-base amplifier
– Useful at higher frequencies
• No “Miller effect”
• 5pf at 100MHz is 320Ω!
• Other interesting combinations also exist
– Usually can be replaced by OpAmps or other devices
Darlington Transistors
• Allow for much greater gain in a circuit
• β = β1 * β2