This chapter discusses small-signal modeling and linear amplification using transistors. The goals are to understand transistors as linear amplifiers, small-signal models, and amplifier characteristics. A simple common-emitter BJT amplifier circuit is presented and analyzed using DC and AC equivalent circuits. Key points include defining the Q-point, constructing small-signal models, and calculating voltage gain. Capacitor selection criteria are provided to maintain linearity in the amplifier.

1. Chapter 13
Small-Signal Modeling and Linear
Amplification

2. Chapter Goals
Understanding of concepts related to:
• Transistors as linear amplifiers
• dc and ac equivalent circuits
• Use of coupling and bypass capacitors to modify dc and ac equivalent
circuits
• Small-signal voltages and currents
• Small-signal models for diodes and transistors
• Identification of common-emitter amplifiers
• Amplifier characteristics such as voltage gain, input and output
resistances and linear signal range
• Rule-of-thumb estimates for voltage gain of common-emitter
amplifiers.

3. Introduction to Amplifiers
• The BJT is an an excellent amplifier when biased in the forward-active
region.
• The FET can be used as an amplifier if operated in the saturation
region.
• In these regions, the transistors can provide high voltage, current and
power gains.
• DC bias is provided to stabilize the operating point in the desired
operation region.
• The DC Q-point also determines
– The small-signal parameters of the transistor
– The voltage gain, input resistance, and output resistance
– The maximum input and output signal amplitudes
– The overall power consumption of the amplifier

4. A Simple BJT Amplifier
The BJT is biased in the forward active region by dc voltage sources VBE
and VCC = 10 V. The DC Q-point is set at, (VCE, IC) = (5 V, 1.5 mA) with IB
= 15 µA.
Total base-emitter voltage is: be
v
BE
V
BE
v +=
Collector-emitter voltage is: This produces a load line.C
R
C
i
CE
v −=10

5. BJT Amplifier (continued)
An 8 mV peak change in vBE gives a 5
µA change in iB and a 0.5 mA change
in iC.
The 0.5 mA change in iC gives a 1.65
V change in vCE .
If changes in operating currents and
voltages are small enough, then IC
and VCE waveforms are undistorted
replicas of the input signal.
A small voltage change at the base
causes a large voltage change at the
collector. The voltage gain is given
by:
The minus sign indicates a 1800
phase shift between input and
output signals.
˜Av =
˜vce
˜v
be
=
1.65∠180
0.008∠0
=206∠180=−206

6. A Simple MOSFET Amplifier
The MOSFET is biased in the saturation region by dc voltage sources VGS and
VDS = 10 V. The DC Q-point is set at (VDS, IDS) = (4.8 V, 1.56 mA) with VGS =
3.5 V.
Total gate-source voltage is: gsv
GS
V
GS
v +=
A 1 V p-p change in vGS gives a 1.25 mA p-p change in iDS and a 4 V p-p change
in vDS. Notice the characteristic non-linear I/O relationship compared to the BJT.

7. A Practical BJT Amplifier using
Coupling and Bypass Capacitors
• AC coupling through capacitors is
used to inject an ac input signal and
extract the ac output signal without
disturbing the DC Q-point
• Capacitors provide negligible
impedance at frequencies of interest
and provide open circuits at dc.
In a practical amplifier design, C1 and
C3 are large coupling capacitors or dc
blocking capacitors, their reactance (XC
= |ZC| = 1/ωC) at signal frequency is
negligible. They are effective open
circuits for the circuit when DC bias is
considered.
C2 is a bypass capacitor. It provides a
low impedance path for ac current from
emitter to ground. It effectively
removes RE (required for good Q-point
stability) from the circuit when ac
signals are considered.

8. DC and AC Analysis -- Application of
Superposition
• DC analysis:
– Find the DC equivalent circuit by replacing all capacitors by open
circuits and inductors (if any) by short circuits.
– Find the DC Q-point from the equivalent circuit by using the
appropriate large-signal transistor model.
• AC analysis:
– Find the AC equivalent circuit by replacing all capacitors by short
circuits, inductors (if any) by open circuits, dc voltage sources by
ground connections and dc current sources by open circuits.
– Replace the transistor by its small-signal model (to be developed).
– Use this equivalent circuit to analyze the AC characteristics of the
amplifier.
– Combine the results of dc and ac analysis (superposition) to yield the
total voltages and currents in the circuit.

9. DC Equivalent for the BJT Amplifier
• All capacitors in the original amplifier circuit are replaced by open
circuits, disconnecting vI, RI, and R3 from the circuit and leaving RE
intact. The the transistor Q will be replaced by its DC model.
DC Equivalent Circuit

10. AC Equivalent for the BJT Amplifier
• The coupling and bypass capacitors are replaced by short circuits. The DC
voltage supplies are replaced with short circuits, which in this case connect
to ground.

11. AC Equivalent for the BJT Amplifier
(continued)
R
B
=R
1
R
2
=10kΩ30kΩ
R=R
C
R
3
=4.3kΩ100kΩ
• By combining parallel resistors into equivalent RB and R, the equivalent AC
circuit above is constructed. Here, the transistor will be replaced by its
equivalent small-signal AC model (to be developed).

12. Hybrid-Pi Small-signal AC Model for
the BJT
• The hybrid-pi small-signal
model is the intrinsic low-
frequency representation of the
BJT.
• The small-signal parameters are
controlled by the Q-point and
are independent of the geometry
of the BJT.
Transconductance:
gm =
I
C
V
T
≅40I
C
Input resistance:
rπ=
βoV
T
I
C
=
βo
gm
Output resistance:
ro=
V
A
+V
CE
I
C

13. Small-signal Current Gain and
Amplification Factor of the BJT




























−
∂
∂
−
==
int
1
1
poQC
i
F
F
C
I
Frmgo
β
β
β
πβ
βo > βF for iC < IM, and βo < βF for
iC > IM, however, βo and βF are
usually assumed to be about
equal.
The amplification factor is given by:
For VCE << VA,
µF represents the maximum voltage
gain an individual BJT can provide,
independent of the operating point.
µ
F
≅
V
A
V
T
≅40V
A
µF ≡
vce
vbe
,vce =−rogmvbe
µ
F
=gmro=
I
C
V
T
V
A
+V
CE
I
C
=
V
A
+V
CE
V
T

14. Example βo Calculation for 2N2222A
Choose the Q-point at about (5 V, 5 mA) for this analysis. Notice the slope of
the DC current gain characteristic in this region. Ideally, the slope would be
zero.

15. βo=gmrπ=
β
F
1−I
C
1
β
F
∂β
F
∂i
C












Q − point














βo≅
β
F
1−I
C
1
β
F
∆β
F
∆I
C












Q − point














∆β
F
∆I
C
≅
200−100
10−2 −10−3
=5.6x103 at about IC = 5 mA and 25 °C
βo≅
180
1−5x10−3 5.6x103
180
























=
180
1−0.15



=212 for βF = 180
Given the tolerances usually encountered in forward current gain, the
assumption of βF = βo seems reasonable for preliminary analysis and
initial designs.
From Figure 3 for the 2N2222A BJT at the chosen Q-point…

16. Equivalent Forms of the Small-signal
Model for the BJT
• The voltage-controlled current source gmvbe can be transformed into a
current-controlled current source,
• The basic relationship ic=βib is useful in both dc and ac analysis when
the BJT is biased in the forward-active region.
v
be
=i
b
rπ =i
b
βo
gm
∴gmv
be
=gmi
b
rπ =βoi
b
ic =gmv
be
+
vce
ro
≅gmv
be
=βoi
b

17. Small Signal Operation of BJT






















































++++=
=+=∴
...
3
6
1
2
2
11
expexp
T
V
be
v
T
V
be
v
T
V
be
v
C
I
T
V
be
v
T
V
BE
V
S
Ici
C
I
C
i






















=
T
V
BE
v
S
I
C
i exp


































+++=−=∴ ...
3
6
1
2
2
1
T
V
be
v
T
V
be
v
T
V
be
v
C
I
C
I
C
ici
For linearity, ic should be directly proportional to vbe.
v
be
<<2V
T
=50 mV∴ic ≅I
C
v
be
V
T










=
I
C
V
T
v
be
=gmv
be
If we limit vbe to 5 mV, the relative change in ic compared to IC that
corresponds to small-signal operation is:
ic
I
C
=
gmv
be
I
C
=
v
be
V
T
≤
0.005
0.025
=0.200
for

18. Small-Signal Analysis of the Complete
C-E Amplifier: AC Equivalent
• The AC equivalent circuit is
constructed by assuming that all
capacitances have zero
impedance at signal frequency
and the AC voltage source is at
ground.
• Assume that the DC Q-point has
already been calculated.

19. Small-Signal Analysis of Complete C-E
Amplifier: Small-Signal Equivalent
R
L
=ro R
C
R
3
Overall voltage gain from source vi
to output voltage vo across R3 is:
Av =
vo
v
i
=
vo
v
be










v
be
v
i








∴Av =−gmR
L
R
B
rπ
R
I
+ R
B
rπ( )














vo=−gmv
be
R
L
and
v
be
=
v
i
R
B
rπ






R
I
+ R
B
rπ





















21. Capacitor Selection for the CE Amplifier
Zc =
1
jωC
Capacitive Reactance Xc ≡Zc =
1
ωC
where ω=2πf
X
c1
<<R
B
rπ∴Make X
c1
≤0.01 R
B
rπ





for < 1% gain error.
X
c2
⇒0 ∴Make X
c2
≤1Ωfor <1% gain error.
X
c3
<<R
3
∴Make X
c3
≤0.01 R
3





for <1% gain error.
The key objective in design is to make the capacitive reactance
much smaller at the operating frequency f than the associated
resistance that must be coupled or bypassed.

23. C-E Amplifier Input Resistance
• The input resistance, the total
resistance looking into the
amplifier at coupling capacitor C1,
represents the total resistance
presented to the AC source.
ππ
π
rRRr
B
RR
r
B
R
21
xi
xv
in
)(xixv
===
=

24. C-E Amplifier Output Resistance
• The output resistance is the total
equivalent resistance looking into the
output of the amplifier at coupling
capacitor C3. The input source is set to 0
and a test source is applied at the output.
C
Ror
C
RR
mg
or
C
R
≅==∴
++=
xi
xv
out
be
vxvxv
xi But vbe=0.
since ro is usually >> RC.

25. CE Amplifier Design Example
Using LabVIEW Virtual Instruments

30. Amplifier Power Dissipation
• Static power dissipation in amplifiers is determined from their DC
equivalent circuits.
P
D
=V
CE
I
C
+V
BE
I
B
Total power dissipated in C-B
and E-B junctions is:
where
Total power supplied is:
P
S
=V
CC
I
C
+I
2





where I
2
=I
1
+I
B
BE
V
CB
V
CE
V +=
I
1
=
V
CC
R
1
+R
2
and I
B
=
V
EQ
−V
BE
R
EQ
+β
F
+1





R
E
The difference is the power dissipated by the bias resistors.