ELE 404 Electronic Circuits I
Module 3 : BJT Voltage Amplifiers
Dr. Fei Yuan
Dept. of Electrical, Computer, and Biomedical Eng.
Ryerson University
Toronto, ON, Canada
Email : fyuan@ryerson.ca
Copyright (c) Fei Yuan, 2022
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Module outline
Load line and maximum signal swing
Common-emitter (CE) amplifiers
1
2
CE amplifiers with a resistor load
CE amplifiers with a current-source load
Common-base (CB) amplifiers
1
2
CB amplifiers with a resistor load
CB amplifiers with a current-source load
Common-collector (CC) amplifiers (emitter
followers)
1
2
Multi-stage amplifiers
Current mirrors
Design considerations of BJT voltage
amplifiers
Examples
Practice problems
Emitter followers with a resistor load
Emitter followers with a current-source load
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Load line and maximum signal swing
Load line
iC =
VCC − vCE
RC
.
(1)
Eq.(1) can be represented by a load line.
The intersection of the load line and iC ∼ vCE
curve with vBE = VBias gives the DC operating
point.
Although DC operating point varies with either
VBias or RC , it is normally adjusted by varying
VBias as varying RC affects gain
Av = −gm (RC ||ro ).
Figure 1: Load line.
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Load line and maximum signal swing
Channel current
iC ≈ Is e
VBias +vin
Vt
= Is e
VBias
Vt
vin
vin
e V t = IC e V t ,
(2)
where IC is the DC component of iC .
If vin Vt (small-signal), making use of
vin
e Vt ≈ 1 +
vin
Vt
,
(3)
vin = IC + gm vin
(4)
we have
iC ≈ IC
1+
vin
Vt
= IC +
IC
Vt
where
gm =
IC
Vt
.
(5)
Figure 2: Load line.
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Load line and maximum signal swing
Since iC = IC + ic where ic is the AC component of iC ,
we have
ic ≈ gm vin
(6)
A linear relation between ic and vin exists under the
condition that vin Vt (small-signal).
Figure 3: Load line.
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Load line and maximum signal swing
Maximum signal swing
Transistors need to operate in active region to
have a large gm .
The boundary of active and saturation regions
and VCC are the lower and upper bounds of vo
signal swing =⇒ vo,total ,min = Vsat and
vo,total ,max = VCC .
The max. voltage swing
vo,AC ,max =
VCC − Vsat
2
.
(7)
Figure 4: Maximum signal swing obtained from a properly chosen
DC operating point.
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Load line and maximum signal swing
If DC operating point is too close to the
boundary of active and saturation regions, the
max. signal swing will be smaller.
Figure 5: Reduced maximum signal swing due to an improper DC
operating point.
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Load line and maximum signal swing
If DC operating point is too close to VCC , the
max. signal swing will be smaller.
Figure 6: Reduced maximum signal swing due to an improper DC
operating point.
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Common-emitter (CE) amplifiers
CE amplifier with resistor load
The emitter is the common node, i.e, the small-signal or AC
ground.
Rs is the internal resistance of the signal source vs .
R1 and R2 generate DC biasing voltage VB at the base.
Isolating capacitor C1 is sufficiently large such that it (i)
blocks any DC currents from flowing into the source and (ii)
provides a low-resistance path for signal vs to reach the gate
of the transistor.
Isolating capacitor C2 also needs to be sufficiently large
and serves similar purposes as those of C1 .
Figure 7: Common-emitter amplifier with a
resistor load.
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Common-emitter (CE) amplifiers
DC operating point
Assume β is given. To determine the DC operating point, write KCL
at the base
G2 VB + G1 (VB − VDD ) +
IC
β
= 0.
(8)
Since
VB
IC ≈ Is e Vt
(9)
Substitute (9) into (8)
G2 VB + G1 (VB − VDD ) +
Is
β
VB
e Vt = 0.
(10)
Figure 8: Common-emitter amplifier
with a resistor load.
VB can be determined from (10) provided that Is is available.
R1 and R2 need to be large in order to minimize the current flowing
through them → Minimize their power consumption.
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Common-emitter (CE) amplifiers
Optimal DC operating point : If Rc is known, find optimal R1 and R2
DC collector voltage
VC = Vsat +
1
2
(Vcc − Vsat ) =
1
2
(Vcc + Vsat ) .
(11)
DC collector current : Since VC = Vcc − Rc Ic , we have
Ic =
1
(Vcc − Vsat ) .
2Rc
(12)
DC base voltage
VB
Is e Vt =
1
2Rc
(Vcc − Vsat ) .
(13)
We have
VB = Vt ln
Vcc − Vsat
2Is Rc
.
(14) Figure 9: Common-emitter amplifier
with a resistor load.
This is the optimal VB that yields the max. output voltage swing of
1
(VCC − Vsat ).
2
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Common-emitter (CE) amplifiers
Case 1 : R1 and R2 are small
Since R1 and R2 are small, the current flowing through them is
large → IR1 , IR2 IB . We can neglect the impact of the input
resistance of the BJT or equivalently neglect IB . Since
VB =
R2
R1 + R2
Vcc ,
(15)
we have from (14)
R2
R1 + R2
Vcc = Vt ln
Vcc − Vsat
2Is Rc
.
Since R1 and R2 are both unknown, the values of R1 and R2
satisfying (16) are not unique.
(16)
Figure 10: Common-emitter amplifier
with a resistor load.
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Common-emitter (CE) amplifiers
Case 2 : R1 and R2 are large
Since R1 and R2 are large, → IR1 , IR2 IB are comparable to IB
−→ IB cannot be neglected.
KCL at node B
G2 VB + G1 (VB − Vcc ) +
Ic
β
= 0,
(17)
where Ic is known and given by (12) from which we have
VB =
G1 Vcc −
Ic
β
G1 + G2
.
(18)
We have from (14) and (18)
G1 Vcc −
Ic
β
G1 + G2
= Vt ln
Vcc − Vsat
2Is Rc
.
(19) Figure 11: Common-emitter amplifier
with a resistor load.
Since R1 and R2 are both unknown, the values of R1 and R2
satisfying (16) are not unique.
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Common-emitter (CE) amplifiers
Voltage gain of CE amplifier with resistor load (passive load)
Low-frequency behavior - The impact of capacitors is not
accounted for.
Small-signal analysis : KCL at output node
gC vo + go vo + gm vin = 0.
(20)
As a result,
⇒ Av =
vo
vin
= −gm (ro ||RC ).
(21)
Remarks
1
2
3
4
CE amplifier is an inverting amplifier.
gm ↑→ Av ↑.
Rc ↑→ Av ↑.
Input resistance : Rin = rbe , typical value 10K ∼ 100K.
Figure 12: Common-emitter amplifier with a
resistor load at low frequencies.
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Common-emitter (CE) amplifiers
Output impedance of CE amplifier with resistor load
(passive load)
Remove all independent sources in small-signal
equivalent circuit, and apply vx at the output terminal.
KCL at output node
g o v x + g C v x − ix = 0
⇒ Rout =
vx
ix
=
1
go + gC
(22)
Figure 13: Output resistance of common-emitter
= ro ||RC .
(23)
amplifier with resistor load at low frequencies.
Typical value of ro : Tens of KΩ ∼ hundreds of KΩ. In
order to have a large Rout , RC needs to be large. It is
costly to fabricate a large resistor on-chip.
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Common-emitter (CE) amplifiers
Voltage gain obtained earlier can be written as
Av = −gm (ro ||RC ) = −gm Rout .
(24)
Too boost Av , one can
1
2
3
Increase gm . Since gm =
Ic
Vt
, a large Ic is needed →
More power consumption as power consumption is given
Figure 14: Output resistance of common-emitter
by P = Ic VCC .
amplifier with resistor load at low frequencies.
Increase Rout . It is costly to fabricate a large resistor
on-chip.
How to increase Av without increasing gm or Rout ?
Replace the (resistor) passive load with an active load.
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Common-emitter (CE) amplifiers
Voltage gain of CE amplifier with current-source load
(active load)
Q1 operates in active mode for its large gm1 .
Q1 operates on the input voltage (information).
Q1 performs V-to-I conversion that maps vin to ic1 .
Q2 operates in active mode for its large ro2 .
vEB2 = VCC − Vb2 is constant → Current of Q2 is constant.
Q1 is represented by constant current source Icc2 in
parallel with its output resistance ro2 .
Figure 15: Common-emitter amplifier with
current-source load.
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Common-emitter (CE) amplifiers
KCL at output node
(go1 + go2 )vo + gm1 vin = 0.
⇒ Av =
vo
vin
= −gm1 (ro1 ||ro2 ).
(25)
(26)
Remarks :
Since ro1 , ro2 are large, a large Av is obtained without using
a large resistor.
The current of Q2 is re-used by Q1. No additional power
Figure 16: Common-emitter amplifier with
consumption.
current-source load.
Cost : Additional biasing voltage Vb2 .
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Common-emitter (CE) amplifiers
Output resistance Rout
Rout = ro1 ||ro2 .
(27)
Input resistance Rin
Rin = rbe .
(28)
Voltage gain
Av = −gm1 Rout .
17: Common-emitter amplifier with
(29) Figure
current-source load.
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Common-emitter (CE) amplifiers
BJT’s terminal resistance
Resistance looking into the base : rbe . Typical
value : 10 kΩ ∼ 100 kΩ → Base is a
high-impedance node.
Resistance looking into the collector : ro .
Typical value : 10 kΩ ∼ 100 kΩ → Collector is
a high-impedance node..
Resistance looking into the emitter :
rbe
re = 1+β
. Typical value : 100 Ω ∼ 1 kΩ →
Emitter is a low-impedance node.
Figure 18: BJT terminal resistances.
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Common-emitter (CE) amplifiers
CE amplifier with emitter degeneration
RE forms emitter degeneration.
Emitter degeneration forms a negative feedback mechanism that
stabilizes Ic1
Vb1 ↑→ VBE1 ↑→ Ic1 ↑→ VRE = RE Ic1 ↑→ VBE1 ↓→ Ic1 ↓ .
(30)
RE is small typically. If RE is large, the voltage gain will drop significantly Figure 19: Common-emitter
amplifier with current-source load
(to be seen shortly).
and emitter degeneration.
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Common-emitter (CE) amplifiers
Input resistance
vin = [rbe1 + RE (1 + β)] ib1 .
(31)
Since
β=
ic1
ib1
=
gm1 vbe1
gbe1 vbe1
= gm1 rbe1 ,
(32)
We have
Rin =
vin
ib1
= rbe1 + gm RE rbe1 + RE .
(33)
Since RE rbe1 typically, we have
Rin ≈ (1 + gm1 RE )rbe1 .
(34)
Emitter degeneration increases input resistance from rbe without
emitter degeneration to (1 + gm RE )rbe with emitter degeneration !
Figure 20: Common-emitter amplifier with
current-source load and emitter degeneration.
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Common-emitter (CE) amplifiers
Voltage gain
KCL at output node
go2 vo + go1 [vo − (1 + β)RE ib ] + gm1 rbe1 ib = 0,
(35)
⇒ (go1 + go2 )vo + [gm1 rbe1 − go1 RE (1 + β)] ib1 = 0.
(36)
Since
ib1 ≈
vin
rbe1 (1 + gm1 RE )
(37)
and gm1 rbe1 go1 RE (1 + β), we have
Av =
vo
vin
≈
−gm1 (ro1 ||ro2 )
.
1 + gm1 RE
(38)
The numerator is the voltage gain without emitter
degeneration −→ Emitter degeneration lowers voltage gain.
The larger RE , the more loss of voltage gain !
Figure 21: Common-emitter amplifier with
current-source load and emitter degeneration.
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Common-emitter (CE) amplifiers
Output resistance
KCL at 2
− Ix + go2 vx + go1 (vx − v1 ) + β ib1 = 0.
(39)
Since ib1 = −gbe1 v1 , we have
(gE + gbe1 + go1 + β go2 )v1 = go1 vx .
(40)
KCL at node 1
gE v1 + gbe1 v1 + go1 (v1 − vx ) − β(−gbe1 v1 ) = 0.
(41)
As a result,
(gE + go1 + gbe1 + β gbe1 )v1 = go1 vx .
Figure 22: Common-emitter amplifier with
current-source load and emitter degeneration.
(42)
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Common-emitter (CE) amplifiers
Substitute (42) into (40)
(go1 gE + go1 gbe1 + gE go2 + go2 gbe1 + go1 go2 + β go2 gbe1 )vx = (gE + gbe1 + go1 + β gbe1 )Ix (43)
Neglect go1 gbe1 , go2 gbe1 , go1 go2 as rbe1 is in kΩ ranges typically and note β =
have
(go1 gE + go2 gE + gm1 go2 )vx = (gE + gm1 )Ix .
ic1
ib1
= gm1 rbe1 , we
(44)
Further neglect gm1 go2 as gm1 gE typically. We have
Rout =
vx
Ix
= (ro1 ||ro2 )(1 + gm1 RE ).
(45)
Emitter degeneration increase output resistance !
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Common-emitter (CE) amplifiers
CE amplifiers with emitter degeneration possess four unique characteristics : (i) increased input
resistance, (2) reduced voltage gain, (iii) increased output resistance, and (iv) reduce power
consumption
A large output resistance is undesirable as it worsens the loading effect of the load resistor on the
gain of the amplifier.
A large input resistance is desirable as it reduces the loading effect of the amplifier on the input
source.
A reduced voltage gain is undesirable. Since the voltage gain of an amplifier is often measured
over a specific frequency range, the detrimental impact of emitter degeneration on voltage gain
can be removed if a shunt capacitor CE is added in parallel with RE . CE needs to be sufficiently
large such that it behaves as a short-circuit over the frequency range thereby eliminating RE .
26 / 82
Common-emitter (CE) amplifiers
A large input resistance in desirable
Increased input resistance is desirable as it reduces the
loading effect of the amplifier on the signal source.
vin
=
=
Rin
vs
Rs + Rin
1
vs
1 + RRs
in
≈
1−
Rs
Rin
vs
(46)
Figure 23: Common-emitter amplifier with
current-source load and emitter degeneration.
The larger the input resistance Rin , the more the input signal
vs will appear at the input of the amplifier.
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Common-emitter (CE) amplifiers
If the amplifier is a CE with emitter degeneration
vin
=
=
rbe (1 + gm RE )
Rs + rbe (1 + gm RE )
1
vs
1 + r (1+Rgs R )
be
≈
1−
vs
m E
Rs
rbe (1 + gm RE )
vs
(47) Figure 24: Common-emitter amplifier with
current-source load and emitter degeneration.
Since the input resistance of CE without emitter degeneration
is rbe , emitter degeneration lowers signal loss on Rs !
28 / 82
Common-emitter (CE) amplifiers
A large output resistance in undesirable
Increased output resistance is undesirable as it increases the
loading effect of load resistor RL of the amplifier.
vL
=
=
RL
vout
RL + Rout
1
vout
1 + RRout
L
≈
Rout
vout
1−
RL
(48) Figure 25: Common-emitter amplifier with
current-source load and emitter degeneration.
The larger the output resistance Rout , the less the output
voltage vout will be delivered to load resistor RL .
29 / 82
Common-emitter (CE) amplifiers
If the amplifier is a CE with emitter degeneration
vL
=
RL
vout
=
RL + ro1 ||ro2 (1 + gm RE )
1
v
ro1 ||ro2 (1+gm RE ) out
1+
R
≈
ro1 ||ro2 (1 + gm RE )
1−
vout
RL
L
Since the output resistance of the CE without emitter
degeneration is ro1 ||ro2 , emitter degeneration lowers the
voltage delivered to load resistor RL .
(49)
Figure 26: Common-emitter amplifier with
current-source load and emitter degeneration.
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Common-emitter (CE) amplifiers
Emitter degeneration lowers power consumption
If Vb1 is fixed and RE is present,
Vb1 = VBE1 + RE IE .
(50)
If Vb1 is fixed and RE is absent,
Vb1 = VBE1 .
(51)
It is evident that when RE is present, VBE1 will be smaller
VBE1
subsequently Ic1 = Is e Vt → Power consumption
P = Ic1 Vcc will also be smaller.
Figure 27: Common-emitter amplifier with emitter
degeneration.
31 / 82
Common-emitter (CE) amplifiers
The static power consumption
P = (IR1 + IE1 )Vcc .
(52)
IR1 can be reduced by increasing R1 and R2 while IE1 can be
reduced by increasing RE .
Large R1 and R2 and RE also yield a large input resistance (w/o
considering the effect of CE )
Rin ≈ R1 ||R2 ||rbe1 (1 + gm1 RE1 ).
28: Common-emitter amplifier with
(53) Figure
emitter degeneration.
32 / 82
Common-emitter (CE) amplifiers
Frequency-dependent emitter degeneration
Emitter degeneration lowers voltage gain. To preserve
voltage gain at the frequency of the input, shunt
capacitor CE that behaves as a short circuit at input
frequency is added.
CE results in a small gain at low frequencies where CE
behaves as an open-circuit and a large at high
frequencies where CE behaves as a short-circuit.
CE also lowers static (DC) power consumption without
sacrificing voltage gain.
Figure 29: Common-emitter amplifier with current-source load
and frequency-dependent emitter degeneration.
33 / 82
Common-base (CB) amplifiers
Voltage gain of CB amplifier with resistor load
Configuration : Base is connected to a constant DC biasing
voltage (a common node or AC ground) and functions as an
small-signal (AC) ground.
Vb biases the transistor in active mode → A large gm exists.
Vb needs to ensure that BJT operates in active mode for all
vin values.
For given Vb , the DC output voltage Vo is determined from
Vb
VCC = Vo + RC IS e Vt
1+
Vo
VA
.
(54)
Figure 30: Common-base amplifier.
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Common-base (CB) amplifiers
KCL at output node
GC vo + go (vo − vin ) + gm (0 − vin ) = 0
⇒ Av =
vo
vin
=
(gm ro + 1)RC
≈ gm (ro ||RC ).
ro + RC
(55)
(56)
CB amplifier is a non-inverting amplifier with the same
voltage gain as that of CE amplifier.
Figure 31: Common-base amplifier.
35 / 82
Common-base (CB) amplifiers
CE amplifier is an-inverting amplifier. Output
and input are out of phase.
CB amplifier is a non-inverting amplifier.
Output and input are in phase.
Figure 32: Top : Common-emitter amplifier. Bottom :
Common-base amplifier.
36 / 82
Common-base (CB) amplifiers
Input impedance of CB amplifier with resistor load
Remove all independent sources and apply vx at the input
node. KCL at emitter and collector
gbe vx + go (vx − vc ) − gm (0 − vx ) − ix = 0,
(57)
Gc vc + go (vc − vx ) + gm (0 − vx ) = 0.
(58)
⇒ Rin =
vx
ix
=
go + Gc
gbe (go + Gc ) + (go + gm )Gc
.
(59)
Since go ≈ 0,
Rin ≈
Since rbe =
β
gm
and re =
Rin ≈
1
gbe + gm
rbe
β+1
.
(60)
, we have
rbe
= re .
β+1
Figure 33: Input resistance of CB amplifier.
(61)
37 / 82
Common-base (CB) amplifiers
Output impedance of CB amplifier with resistor load
Remove all independent sources of the amplifier and apply a
test voltage source vx at the output node. KCL at output node
GC vo + go vo − ix = 0
⇒ Rout =
vx
ix
= ro ||RC .
(62)
(63)
CB amplifier has the same output resistance as that of CE
amplifier.
Figure 34: Output impedance of CB amplifier.
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Common-base (CB) amplifiers
Voltage gain of CB amplifier with current-source load
Q1 operates in active mode for its large gm1 . Q2 operates in
active mode for its large ro2 .
DC output voltage Vo is determined from IC1 ≈ IC2 (Note that
IC1 = IC2 + IB2 . Since IC2 IB2 , we have IC1 ≈ IC2 ).
Vo
Is e V t
VCC −Vb2
Vo
VCC − Vo
1+
1+
≈ Is e Vt
.
VA
VA
(64)
Figure 35: CB amplifier with a current-source
load.
39 / 82
Common-base (CB) amplifiers
KCL at the output node
go2 vo + go1 (vo − vin ) + gm1 (0 − vin ) = 0.
⇒ Av =
vo
vin
=
(gm1 ro1 + 1)ro2
.
ro1 + ro2
(65)
(66)
Make use of gm1 ro1 1
Av ≈ gm1 (ro1 ||ro2 ).
(67)
Figure 36: CB amplifier with a current-source
load.
40 / 82
Common-base (CB) amplifiers
CE and CB, which one should be used ?
For CE amplifier, Rin,CE ≈ R1 ||R2 ||rbe .
For CB amplifier, Rin,CB ≈ R1 ||R2 ||re .
Since re =
rbe
,
1+β
Rin,CB < Rin,CE . As a result, vin,CB < vin,CE .
CB amplifiers should not be used as a voltage amplifier in general
due to its low input resistance.
Figure 37: Comparison of CE and CB
amplifiers
41 / 82
Common-collector (CC) amplifiers (emitter followers)
Source follower with resistor load : DC biasing voltage
Configuration : The collector is connected to a constant
voltage and functions as a small-signal (AC) ground.
Base needs to have a proper DC voltage Vb to bias the
transistor in active mode.
Min. DC biasing voltage : Vb,min = VBE ,min .
DC output voltage Vo is determined from
Vb
Vo = RE Is e Vt
VCC − Vo
1+
.
VA
(68)
Figure 38: Emitter follower with a resistor load.
42 / 82
Common-collector (CC) amplifiers (emitter followers)
Voltage gain of source follower with resistor load
KCL at emitter
GE vo + go vo + gbe (vo − vin ) − gm (vin − vo ) = 0.
⇒ Av =
vo
vs
=
gm ro rbe RE
gm ro rbe RE + ro rbe + RE rbe + RE ro
(69)
.
(70)
Make use of gm ro rbe RE ro rbe , RE rbe , RE ro
Av ≈ 1.
(71)
The output (emitter) follows the input → Emitter follower.
Figure 39: Emitter follower with a resistor load.
43 / 82
Common-collector (CC) amplifiers (emitter followers)
Output impedance of emitter follower with resistor load
Remove all independent sources from the emitter follower and
apply a test voltage source vx at the output node.
KCL at the emitter
GE vx + go vx + gbe vx − gm (0 − vx ) − ix = 0.
(72)
We have
Rout =
vx
ix
=
ro rbe RE
gm ro rbe RE + ro rbe + RE rbe + RE ro
.
(73)
Make use of gm ro rbe RE ro rbe , RE rbe , ro rbe .
Rout ≈
1
gm1
.
(74) Figure 40: Output impedance of emitter
follower with a resistor load.
Emitter follower has a low output impedance.
44 / 82
Common-collector (CC) amplifiers (emitter followers)
Voltage gain of emitter follower with current-source load
Q1 and Q2 operate in active mode via proper biasing voltages
Vb1 and Vb2 .
KCL at emitter
go2 vo + go1 vo + gbe1 (vo − vin ) − gm1 (vin − vo ) = 0.
(75)
We have
Av =
vo
vs
=
gm1 ro1 ro2 rbe1 + ro1 ro2
gm1 ro1 ro2 rbe1 + ro1 rbe1 + ro1 ro2 + ro2 rbe1
.
(76)
Make use of gm1 ro1 ro2 rbe1 ro1 rbe1 , ro1 ro2 , ro2 rbe1 .
Av ≈ 1.
(77)
Figure 41: Emitter follower with a
current-source load.
45 / 82
Common-collector (CC) amplifiers (emitter followers)
Output impedance of emitter follower with current-source load
Remove all independent sources of the emitter follower and apply a
test voltage vx at the output node.
KCL at the source
go1 vx + go2 vx + gbe1 vx − gm1 (0 − vx ) − ix = 0.
(78)
We have
Rout
=
vx
ix
=
ro1 ro2 rbe1
gm1 ro1 ro2 rbe + ro1 rbe1 + ro2 rbe1 + ro1 ro2
.(79)
Make use of gm1 ro1 ro2 rbe1 ro1 rbe1 , ro2 rbe1 , ro1 ro2
Rout ≈
1
gm1
.
(80)
Figure 42: Emitter follower with a
current-source load.
46 / 82
Common-collector (CC) amplifiers (emitter followers)
A NPN-based source follower functions as a voltage
down-shifter as vo = vin − vBE1 . It shifts the voltage level
down by vBE1 .
A PNP-based source follower functions as a voltage up-shifter
as vo = vin + vEB1 . It shifts the voltage level down by vEB1 .
Figure 43: Emitter followers as voltage up /
down shifters.
47 / 82
Common-collector (CC) amplifiers (emitter followers)
Why emitter follower ?
Voltage-mode amplifiers : Both the input and output of the amplifiers are voltages.
A common-emitter amplifier with a resistor load can be represented by its Thevenin equivalent
circuit with VT = −gm Rc vin and RT = Rc ||ro (small-signal).
A common-base amplifier with a resistor load can be represented by its Thevenin equivalent circuit
with VT = gm Rc vin and RT = Rc ||ro (small-signal).
48 / 82
Common-collector (CC) amplifiers (emitter followers)
When the amplifier drives a resistor load RL , the voltage across the load :
VL =
RL
RL + Rout
VT =
1
1+
Rout
RL
VT ≈ VT
(81)
if Rout RL .
An emitter follower offers a low output resistance 1/gm !
49 / 82
Common-collector (CC) amplifiers (emitter followers)
If the load of a voltage amplifier is sufficiently large
(RL Rout ), there is no need to have an emitter follower
to lower the output resistance of the amplifier.
If the load of a voltage amplifier is sufficiently small
(Rl Rout ), an emitter follower that lowers the output
resistance of the amplifier is required. Otherwise, the
gain of the amplifier will diminish.
Figure 44: Emitter follower to minimize loading effect.
50 / 82
Common-collector (CC) amplifiers (emitter followers)
When load resistor RL is absent
Av ,w/o load = −gm1 (ro1 ||ro2 ).
(82)
When load resistor RL is present
Av ,w/ load = −gm1 (ro1 ||ro2 ||RL ).
(83)
If RL Rout , we will have Av ,w/ load = Av ,w/o load → RL
will have no loading impact on voltage gain.
If RL Rout , we will have Av ,w/ load Av ,w/o load →
RL will have a severe loading impact on voltage gain.
Solution : Add an emitter follower stage to lower the
output resistance from Rout1 = ro1 ||ro2 to
Rout2 = 1/gm4 ≈ re4 .
Figure 45: Loading impact in amplifiers.
51 / 82
Multi-stage amplifiers
Discrete multi-stage amplifiers
Cascading multiple CE amplifiers or CB amplifiers. Each
stage must be properly DC-biased such that (i) Q1 and
Q2 operate in active mode and (ii) Q2 has a large
voltage swing range as vo2 vo1 .
Isolation capacitors exist at the input and output and
between stages to (i) isolate the amplifier from both the
input source and the output load and (ii) DC-bias each
stage individually.
Since isolating capacitors function as short circuits, the
input resistance of the first-stage loads the signal source
and the input resistance of the second-stage loads the
first-stage.
Figure 46: Discrete two-stage BJT amplifiers.
52 / 82
Multi-stage amplifiers
DC operating point
Assume VBE1,2 =0.7 V and β is given.
KCL at node 1
g1 (V1 − Vcc ) + g2 V1 + IB1 = 0
(84)
KCL at node 3
g3 (V3 − Vcc ) + g4 V3 + IB2 = 0
(85)
KVL around the base-emitter of Q1
V1 = VBE1 + RE1 (1 + β)IB1
(86)
Figure 47: Discrete two-stage BJT amplifiers.
KVL around the base-emitter of Q2
V3 = VBE2 + RE2 (1 + β)IB2
(87)
4 unknowns (V1 , V3 , IB1 , IB2 ) in 4 equations.
53 / 82
Multi-stage amplifiers
Voltage gain
Assume Q1/Q2 operate in active mode.
Input resistance of stage 1
Rin1 = R1 ||R2 ||rbe1 (1 + gm1 RE1 ).
(88)
input voltage of Q1
v1 =
Rin1
Rin1 + Rs
vs .
(89)
Load resistance of stage 1 : Rc1 ||Rin2 where
Rin2 = R3 ||R4 ||rbe2 (1 + gm2 RE2 ).
Voltage gain of stage 1
Figure 48: Discrete two-stage BJT amplifiers.
Av 1 =
v2
v1
=
−gm1 (Rc1 ||Rin2 )
.
1 + gm1 RE1
(90)
Voltage gain of stage 2
Av 2 =
v4
v3
=
−gm2 Rc2
1 + gm2 RE2
.
(91)
54 / 82
Multi-stage amplifiers
Overall voltage gain
Av
=
=
v4
vs
=
v4
v3
×
v3
v1
×
v1
vs
−gm1 (Rc1 ||Rin2 )
×
Rin1 + Rs
1 + gm1 RE1
−gm2 Rc2
.
(92)
×
1 + gm2 RE2
Rin1
Figure 49: Discrete two-stage BJT amplifiers.
55 / 82
Multi-stage amplifiers
Integrated multi-stage amplifiers
No isolation capacitors exist at the input and output and between
stages due to the high cost of realizing them on chips.
The DC biasing voltage of Q1 is provided by the DC component of
the input : vin = vin,AC + vin,DC .
The DC biasing voltage of Q2 is provided by the DC component of
the output of stage 1 : vo1 = vo1,AC + vo1,DC .
If the load of the amplifier is a resistor, its load impact on stage 2
must be considered.
If vo1,DC < VCC /2, configuration (a) should be used, otherwise,
configuration (b) should be used to ensure Q2 operates in active
mode all the time.
Figure 50: Integrated two-stage BJT
amplifiers.
56 / 82
Current mirrors
Diode-connected BJT
Diode-connected BJT : Base and collector are tied
together −→ Ensure base-emitter pn-junction (diode) is
forward-biased. A small change of vbe gives rise to a
large change of emitter current −→ Resistance between
emitter and base (ON-resistance of the diode) is small !
KCL at collector
(go + gbe )vx + gm (vx − 0) − ix = 0.
(93)
we have
Rin =
vx
ix
=
1
go + gbe + gm
.
(94)
Make use of gm go , gbe
Rin ≈
1
gm
.
(95)
Figure 51: Diode-connected BJT.
Diode-connected BJT functions as a resistor with a low
resistance.
57 / 82
Current mirrors
A further look at Rin =
1
gm
1
gm
=
1
IC
Vt
=
Vt
IC
=
Vt
.
β Ib
(96)
Because
rbe =
Vt
IB
,
(97)
we have
1
gm
=
rbe
β
≈ re .
(98)
58 / 82
Current mirrors
Current mirrors without considering ro
A diode-connected BJT offers a low impedance → Current flows
into collector with less impedance.
Q1 and Q2 have the same vBE .
iC1 ≈ Is1 e
vBE
Vt
,
(99)
vBE
Vt
,
(100)
iC2 ≈ Is2 e
we have
iC2 =
Is2
Is1
iC1 .
(101)
Figure 52: BJT current mirror.
If Is2 = Is1 (when Q1 and Q2 are identical), iC2 = iC1 , it is a current
mirror. Otherwise, it is a current amplifier.
59 / 82
Current mirrors
In Module 2 (diodes), we showed that Is is proportional to emitter
area A. We there have
Is2
Is1
=
A2
A1
.
(102)
J = iC1 + iB1 + iB2 . Since iB1 , iB2 iC1 , we have
iC2 ≈
Is2
Is1
J.
(103)
Figure 53: BJT current mirror.
60 / 82
Current mirrors
A more accurate analysis of current mirrors without considering ro
Base currents iB1 and iB2 are accounted for while ro is not accounted for →
iB1 = iC1 /β1 and iB2 = iC2 /β2 .
h
J = iC1 + iB1 + iB2 = Is1 1 +
iC2 = Is2 e
vBE
Vt
1
β1
+
Is2
β2
i
e
vBE
Vt
,
(104)
.
we have
Is2
iC2 =
Is1 1 +
1
β1
+
Is2
J.
(105)
β2
Figure 54: BJT current
mirror.
If Q1 and Q2 are identical, i.e. β1 = β2 = β and Is1 = Is2 = Is , we have
iC2 =
1
1+
2
J≈
β
1−
2
β
J.
(106)
An intrinsic error exists due to finite β !.
61 / 82
Current mirrors
Because
iB3 =
iE3
=
β3 + 1
iB1 + iB2
β3 + 1
1
=
β3 + 1
iC1
β1
+
iC2
,
β2
(107)
we have
h
J = iC1 + iB3 = Is1 1 +
iC2 = Is2 e
vBE
Vt
1
β1 (β3 +1)
+
i
Is2
β2 (β3 +1)
e
vBE
Vt
,
(108)
.
As a result
Is2
iC2 =
Is1 1 +
1
β1 (β3 +1)
+
Is2
J.
(109)
β2 β3
Figure 55: BJT current
mirror.
If Q1, Q2, and Q3 are identical, i.e. β1 , β2 , β3 = β and Is1 , Is2 = Is , we have
iC2 =
1
1+
2
β(β+1)
J≈
1
1+
2
β2
J≈
1−
2
β2
J.
(110)
The intrinsic error caused by finite β is significantly reduced !
62 / 82
Current mirrors
Current mirrors with ro considered
When the impact of channel length modulation is
accounted for
iC1 ≈ Is1 e
vBE
Vt
iC2 ≈ Is2 e
vBE
Vt
vCE1
,
VA
vCE2
1+
,
VA
1+
(111)
(112)
Let vCE1 = vCE and vCE2 = vCE + ∆vCE , we have
iC2
iC1
=
Is2 1 +
Is1 1 +
vCE2
VA
vCE1
VA
=
Is2
Is1

∆vCE

1 +
VA
v
+ VCE
A
.
1
(113) Figure 56: BJT current mirror with output impedance
accounted for.
Since vCE1 6= vCE2 in general, the impact of vCE
mismatch cannot be neglected for precision current
mirrors.
63 / 82
Current mirrors
Synthesis of a small or a large resistor
A diode-connected BJT synthesizes a resistor of low
resistance 1/gm ≈ re .
A BJT in active mode synthesizes a resistor of large
resistance ro .
Figure 57: Synthesis of resistors.
64 / 82
Current mirrors
Current sources and current sinks
Use current mirrors to construct current sinks and
current sources from a master current source whose
value is stable, i.e. independent of temperature and
supply voltage fluctuation.
The currents of the current sinks and current sources are
set by the ratio of transistor size, specifically the area of
the emitter.
Figure 58: Current sources and current sinks.
65 / 82
Current mirrors
Two-stage amplifier consists of two CE amplifiers formed
by Q1 and Q2. Q1 has its DC biasing voltage VDC from
the input.
DC operating point is determined from Ic3 = Ic4
VA
Is4 e Vt = Is3 e
Vcc −VA
Vt
.
(114)
Note Is3 and Is4 are proportional to the area of the
emitter of Q3 and Q4, respectively.
Is3 = Is4 if Q3 and Q4 have the same emitter dimension
→ VA = Vcc /2 .
Figure 59: Current sources and current sinks.
Once VA is known, Ic5∼c8 can be determined.
66 / 82
Current mirrors
Voltage gain
vo1
vs
= −gm1 (ro1 ||ro7 ||Rin2 ),
(115)
where Rin2 = rbe2 .
vo2
vo1
vo2
vs
=
vo2
vo1
×
vo1
vs
= −gm2 (ro2 ||ro8 ).
(116)
Figure 60: Current sources and current sinks.
= gm1 gm2 (ro1 ||ro7 ||Rin2 )(ro2 ||ro8 ). (117)
67 / 82
Design considerations of BJT amplifiers
Norton equivalent circuit
An electric network at the output terminals can
be represented by its Norton equivalent circuit
consisting of Norton current source IN in
parallel with Norton conductance YN under the
following underlinetwo conditions :
1
Both networks have the same short-circuit
current at the output terminals : ISC = IN .
2
Both networks have the same open-circuit
voltage at the output terminals : VOC =
IN
YN
The Norton equivalent circuit of a network
should be used only if the output of the
network is a current.
.
Figure 61: Norton equivalent circuit of an electric network.
68 / 82
Design considerations of BJT amplifiers
An electric network and its Norton
equivalent circuit have the same behavior
at the output port
1
Both networks have the same short-circuit
current : IN = ISC .
2
Both networks have the same open-circuit
voltage : YN = 1/ZOC .
Figure 62: Norton equivalent circuit of an electric network.
69 / 82
Design considerations of BJT amplifiers
Thevenin equivalent circuit
An electric network can be represented by its
Thevenin equivalent circuit consisting of
Thevenin voltage source VT in series with
Thevenin impedance ZT under the following
two conditions :
1
Both networks have the same short-circuit
current : ISC = VT /ZT .
2
Both networks have the same open-circuit
voltage : VOC = VT .
Figure 63: Thevenin equivalent circuit of an electric network.
The Thevenin equivalent circuit of a network
should be used only if the output of the
network is a voltage.
70 / 82
Design considerations of BJT amplifiers
An electric network and its Thevenin
equivalent circuit have the same behavior
at the output nodes.
1
2
The same open-circuit voltage : VT = VOC
where VOC is the open-circuit voltage at the
output terminals of the electric network.
The same short-circuit current : If ISC is the
short-circuit impedance current at the output of
V
the electric network, then ISC = ZT from which
T
we obtain ZT =
VT
ISC
.
Figure 64: Thevenin equivalent circuit of an electric network.
71 / 82
Design considerations of BJT amplifiers
Loading impact of the first amplifier
vin1 =
Rin1
Rs + Rin1
vs .
(118)
To minimize the loading impact of the first amplifier,
Rin1 Rs is needed.
Loading impact of the 2nd amplifier
vin2 =
Rout1
Rout1 + Rin2
vout1 .
(119)
where vout1 is the output voltage of the first amplifier
without the 2nd amplifier. To minimize the loading impact
of the 2nd amplifier, Rin2 Rout1 is needed.
Figure 65: Loading effect in BJT amplifiers.
72 / 82
Design considerations of BJT amplifiers
Loading effect of the load
vo =
RL
RL + Rout2
vout2 .
(120)
where vout2 is the output voltage of the 2nd amplifier
without the load. To minimize the loading impact of the
load, RL Rout2 is needed.
Figure 66: Loading effect in BJT amplifiers.
73 / 82
Examples
Example 1 : Let β =200 and assume ro Rc . Also, C1,2 are large.
(i) Find DC voltage at the collector. (ii) Find the input resistance
Rin . (iii) Voltage gain Av = vvo .
s
Solution
C1 blocks DC currents from flowing to the source and (ii)
provides an AC path for the signal.
C2 (i) blocks DC current flowing to ground at emitter and (ii)
grounds the emitter in AC.
Rb must be sufficiently large as compared with Rs , why ?
(1) DC voltage at collector
Ic =
β
200
IE =
× 10−3 = 0.995 × 10−3 A,
β+1
201
(121)
we have
Figure 67: BJT amplifier : Example 1.
Vc = Rc Ic = 0.995V.
(122)
74 / 82
Examples
(2) Input resistance
gm =
Ic
Vt
=
0.995 × 10−3
26 × 10−3
= 0.0383 A/V,
(123)
= 5.22 kΩ.
(124)
we have
Rin1 = rbe =
β
gm
=
200
0.0383
and
Rin = Rb ||Rin1 = 3.43 kΩ.
(125)
Figure 68: BJT amplifier : Example 1.
75 / 82
Examples
(3) Voltage gain
vo = −gm (ro ||Rc )vin1 ≈ −gm Rc vin1 = −38.3vin1 .
(126)
Since
vin1 =
Rin1
Rin1 + Rs
vs = 0.839vs ,
(127)
we have
vo = 32.1vs → Av = 32.1.
(128)
Figure 69: BJT amplifier : Example 1.
76 / 82
Examples
Example 2 : Let β =100, Vc =1.5 V, Ic =1 mA, and VBE =0.7 V. Find
Rc and Rb .
Solution
IE = Ic + IB =
β+1
β
Ic = 1.01 × 10−3 A.
(129)
Since Vcc = Vc + Rc IE , we have
Rc =
Vcc − Vc
IE
≈ 1.49 × 103 Ω.
(130)
Figure 70: BJT amplifier : Example 2.
Since Vc = IB Rb + VBE , we have
Rb =
Vc − VBE
IB
= 8 × 104 .
(131)
77 / 82
Examples
Example 3 : Let β = ∞ and VBE1 = VBE2 = VBE3 = VBE . Show
Vcc
Io = 2R
.
E
Solution
Since β = ∞, we have IB =
Ic
β
=0. Because
VBE + Io RE = 2VBE + RI ,
(132)
Io RE = VBE + RI .
(133)
Vcc = RI + VBE + RE Io = 2RE Io ,
(134)
we have
Further because
Figure 71: BJT amplifier : Example 3.
we have
Io =
Vcc
2RE
.
(135)
78 / 82
Practice problems
Q3. Assume β =100, base current
IB is sufficiently small, ro is
Q1. Assume β =100, ro and C1,2 Q2. Assume β =100, ro and C1,2 sufficiently large, V =0.7 V, and
BE
are sufficiently large. Find Rin and are sufficiently large. Find Rin and all isolation capacitors are
vo
vo
v
at low frequencies.
at low frequencies.
sufficiently large. Find vo1 and
vs
vs
vo2
vin
Figure 72: Q1.
in
at low frequencies.
Figure 73: Q2.
Figure 74: Q3.
79 / 82
Practice problems
Q4. Assume β =100, Ic =0.5 mA,
Vc =1.5 V, VBE =0.7 V, and
capacitors are sufficiently large.
Find I and RB .
Q5. Assume that transistor
operates in active mode and
isolation capacitors are
sufficiently large. Find the input
resistance.
Q6. Assume β =100, VBE =0.7 V,
and ro is sufficiently large. Find
the DC collector current, DC
collector voltage, and voltage
gain.
Figure 75: Q4.
Figure 76: Q5.
Figure 77: Q6.
80 / 82
Practice problems
Q7.
1
Consider a common-emitter BJT amplifier with a resistor load. How do you define the maximum
output voltage swing for a given VBE ?
2
Consider a common-emitter BJT amplifier with a current-source load. How do you define the
maximum output voltage swing for a given VBE ?
3
Consider a common-emitter BJT amplifier with a resistor load Rc . How do you define the
maximum output voltage swing for a given Rc ?
4
What are the two design considerations when choosing the value of the isolation capacitors of
amplifiers ?
5
Consider a common-emitter amplifier with a current-source load. Under what condition that the
current-source load transistor can be modeled as a current source of a constant current in parallel
with a resistor ?
6
Consider a common-emitter amplifier with a current-source load. Can a NPN transistor be used as
the current-source load and why ?
81 / 82
Practice problems
Q8.
1
Consider a common-emitter amplifier with a current-source load. Determine the voltage gain of the
amplifier if the load of the amplifier is a resistor of (i) a small resistance and (ii) a large resistance.
Which one has a larger gain and why ? How do you ensure that the overall voltage gain of the
amplifier is independent of the load resistance ?
2
Consider a BJT operating in the active region. Find (i) the resistance looking into the base, (ii) the
resistance looking into the emitter, and (iii) the resistance looking into the collector of the BJT.
3
A common-collector amplifier (also known as emitter follower) does not provide any voltage gain.
Why such a configuration is of a practical importance ?
4
Show the schematic of a amplifier with the following constraints : (i) No resistor is allowed, (ii) a
large voltage gain, (iii) a large input resistance, (iv) a low output resistance, (v) two amplifying
stages, and (v) only one supply voltage and a current source are available. Find the input
resistance, the output resistance, and the voltage gain of the amplifier.
5
As a designer, you are required to design an amplifier to amplify the signal from a signal generator.
Let the signal generator be modeled as a voltage source VT in series with a resistor RT . You can
use either a common-emitter amplifier or a common-base amplifier, both with a current-source
load. Show the complete schematic of the entire circuit with the signal source included. Find the
output of the amplifiers and determine which one is larger.
82 / 82