EE101: BJT circuits (Part 2)
M. B. Patil
mbpatil@ee.iitb.ac.in
www.ee.iitb.ac.in/~sequel
Department of Electrical Engineering
Indian Institute of Technology Bombay
M. B. Patil, IIT Bombay
Common-emitter amplifier
10 V
10 k
coupling
capacitor
RC
R1
3.6 k
coupling
capacitor
6V
CC
1.8 V
CB
RL
vs
R2
2.2 k
RE
1k
bypass
capacitor
RC
R1
CE
VCC
load
resistor
AND
1.1 V
R2
RE
DC circuit
VCC
RC
vs
R1
R2
RL
AC circuit
* We have already analysed the DC (bias) circuit of this amplifier and found that
VB = 1.8 V , VE = 1.1 V , VC = 6 V , and IC = 1.1 mA.
M. B. Patil, IIT Bombay
Common-emitter amplifier
10 V
10 k
coupling
capacitor
RC
R1
3.6 k
coupling
capacitor
6V
CC
1.8 V
CB
RL
vs
R2
2.2 k
RE
1k
bypass
capacitor
RC
R1
CE
VCC
load
resistor
AND
1.1 V
R2
RE
DC circuit
VCC
RC
vs
R2
R1
RL
AC circuit
* We have already analysed the DC (bias) circuit of this amplifier and found that
VB = 1.8 V , VE = 1.1 V , VC = 6 V , and IC = 1.1 mA.
* We now analyse the AC (small-signal) circuit to obtain vb , ve , vc , ic .
M. B. Patil, IIT Bombay
Common-emitter amplifier
10 V
10 k
coupling
capacitor
RC
R1
3.6 k
coupling
capacitor
6V
CC
1.8 V
CB
RL
vs
R2
2.2 k
RE
1k
bypass
capacitor
RC
R1
CE
VCC
load
resistor
AND
1.1 V
R2
RE
DC circuit
VCC
RC
vs
R2
R1
RL
AC circuit
* We have already analysed the DC (bias) circuit of this amplifier and found that
VB = 1.8 V , VE = 1.1 V , VC = 6 V , and IC = 1.1 mA.
* We now analyse the AC (small-signal) circuit to obtain vb , ve , vc , ic .
* We will then get the complete solution by simply adding the DC and AC results,
e.g., iC (t) = IC + ic (t).
M. B. Patil, IIT Bombay
Common-emitter amplifier
10 V
10 k
coupling
capacitor
RC
R1
3.6 k
coupling
capacitor
6V
CC
1.8 V
CB
RL
vs
R2
2.2 k
RE
1k
bypass
capacitor
RC
R1
CE
VCC
load
resistor
AND
1.1 V
R2
RE
DC circuit
VCC
RC
vs
R2
R1
RL
AC circuit
* We have already analysed the DC (bias) circuit of this amplifier and found that
VB = 1.8 V , VE = 1.1 V , VC = 6 V , and IC = 1.1 mA.
* We now analyse the AC (small-signal) circuit to obtain vb , ve , vc , ic .
* We will then get the complete solution by simply adding the DC and AC results,
e.g., iC (t) = IC + ic (t).
* We will assume that CB , CC , CE are large enough so that, at the signal
frequency (say, 1 kHz), they can be replaced by short circuits.
M. B. Patil, IIT Bombay
Common-emitter amplifier
vs
RC
R1
R2
RL
Common-emitter amplifier
rb
B
vs
RC
R1
R2
RL
vs
R1
R2
Cµ
Cπ
rπ
gm vbe
E
C
ro
RC
RL
Common-emitter amplifier
rb
B
vs
RC
R1
R2
RL
vs
R1
R2
Cµ
Cπ
rπ
gm vbe
C
ro
RC
RL
E
* The parasitic capacitances Cπ and Cµ are in the pF range. At a signal frequency
of 1 kHz, their impedance is 1/ωC ∼ 1/(2π × 103 × 10−12 ), i.e., ∼ 100 MΩ.
→ Cπ and Cµ can be replaced by open circuits.
Common-emitter amplifier
rb
B
vs
RC
R1
R2
RL
vs
R1
R2
Cµ
Cπ
rπ
gm vbe
C
ro
RC
RL
E
* The parasitic capacitances Cπ and Cµ are in the pF range. At a signal frequency
of 1 kHz, their impedance is 1/ωC ∼ 1/(2π × 103 × 10−12 ), i.e., ∼ 100 MΩ.
→ Cπ and Cµ can be replaced by open circuits.
* For simplicity, we will assume rb to be small and ro to be large (this assumption
will only slightly affect the gain computation).
Common-emitter amplifier
rb
B
vs
RC
R1
R2
RL
vs
R1
R2
Cµ
Cπ
rπ
gm vbe
C
ro
RC
RL
E
* The parasitic capacitances Cπ and Cµ are in the pF range. At a signal frequency
of 1 kHz, their impedance is 1/ωC ∼ 1/(2π × 103 × 10−12 ), i.e., ∼ 100 MΩ.
→ Cπ and Cµ can be replaced by open circuits.
* For simplicity, we will assume rb to be small and ro to be large (this assumption
will only slightly affect the gain computation).
* The above considerations significantly simplify the AC circuit.
Common-emitter amplifier
rb
B
vs
RC
R1
R2
RL
vs
R1
R2
Cµ
Cπ
rπ
gm vbe
C
ro
RC
C
B
RL
vs
R1
rπ
gm vbe
R2
RC
RL
E
E
* The parasitic capacitances Cπ and Cµ are in the pF range. At a signal frequency
of 1 kHz, their impedance is 1/ωC ∼ 1/(2π × 103 × 10−12 ), i.e., ∼ 100 MΩ.
→ Cπ and Cµ can be replaced by open circuits.
* For simplicity, we will assume rb to be small and ro to be large (this assumption
will only slightly affect the gain computation).
* The above considerations significantly simplify the AC circuit.
M. B. Patil, IIT Bombay
Common-emitter amplifier
C
B
vs
10 k
2.2 k
R1
R2
3.6 k 10 k
rπ
gm vbe
RC
RL vo
E
vo = −(gm vbe ) × (RC k RL ) = −(gm vs ) × (RC k RL )
M. B. Patil, IIT Bombay
Common-emitter amplifier
C
B
vs
10 k
2.2 k
R1
R2
3.6 k 10 k
rπ
gm vbe
RC
RL vo
E
vo = −(gm vbe ) × (RC k RL ) = −(gm vs ) × (RC k RL )
vo
= −gm (RC k RL )
vs
→ ALV = voltage gain =
(superscript L is used because the gain includes the effect of RL .)
M. B. Patil, IIT Bombay
Common-emitter amplifier
C
B
vs
10 k
2.2 k
R1
R2
3.6 k 10 k
rπ
gm vbe
RC
RL vo
E
vo = −(gm vbe ) × (RC k RL ) = −(gm vs ) × (RC k RL )
vo
= −gm (RC k RL )
vs
→ ALV = voltage gain =
(superscript L is used because the gain includes the effect of RL .)
Since IC (bias current) = 1.1 mA, gm = IC /VT = 1.1 mA/25.9 mV = 42.5 mf.
M. B. Patil, IIT Bombay
Common-emitter amplifier
C
B
vs
10 k
2.2 k
R1
R2
3.6 k 10 k
rπ
gm vbe
RC
RL vo
E
vo = −(gm vbe ) × (RC k RL ) = −(gm vs ) × (RC k RL )
vo
= −gm (RC k RL )
vs
→ ALV = voltage gain =
(superscript L is used because the gain includes the effect of RL .)
Since IC (bias current) = 1.1 mA, gm = IC /VT = 1.1 mA/25.9 mV = 42.5 mf.
→ ALV = −42.5 mf × (3.6 k k 10 k) = −112.5
M. B. Patil, IIT Bombay
Common-emitter amplifier
C
B
vs
10 k
2.2 k
R1
R2
3.6 k 10 k
rπ
gm vbe
RC
RL vo
E
vo = −(gm vbe ) × (RC k RL ) = −(gm vs ) × (RC k RL )
vo
= −gm (RC k RL )
vs
→ ALV = voltage gain =
(superscript L is used because the gain includes the effect of RL .)
Since IC (bias current) = 1.1 mA, gm = IC /VT = 1.1 mA/25.9 mV = 42.5 mf.
→ ALV = −42.5 mf × (3.6 k k 10 k) = −112.5
For vs (t) = (2 mV ) sin ωt, the AC output voltage is,
M. B. Patil, IIT Bombay
Common-emitter amplifier
C
B
vs
10 k
2.2 k
R1
R2
3.6 k 10 k
rπ
gm vbe
RC
RL vo
E
vo = −(gm vbe ) × (RC k RL ) = −(gm vs ) × (RC k RL )
vo
= −gm (RC k RL )
vs
→ ALV = voltage gain =
(superscript L is used because the gain includes the effect of RL .)
Since IC (bias current) = 1.1 mA, gm = IC /VT = 1.1 mA/25.9 mV = 42.5 mf.
→ ALV = −42.5 mf × (3.6 k k 10 k) = −112.5
For vs (t) = (2 mV ) sin ωt, the AC output voltage is,
vo = ALV vs = −(112.5) (2 mV ) sin ωt = −(125 mV ) sin ωt
M. B. Patil, IIT Bombay
Common-emitter amplifier
C
B
vs
10 k
2.2 k
R1
R2
3.6 k 10 k
rπ
gm vbe
RC
RL vo
E
vo = −(gm vbe ) × (RC k RL ) = −(gm vs ) × (RC k RL )
vo
= −gm (RC k RL )
vs
→ ALV = voltage gain =
(superscript L is used because the gain includes the effect of RL .)
Since IC (bias current) = 1.1 mA, gm = IC /VT = 1.1 mA/25.9 mV = 42.5 mf.
→ ALV = −42.5 mf × (3.6 k k 10 k) = −112.5
For vs (t) = (2 mV ) sin ωt, the AC output voltage is,
vo = ALV vs = −(112.5) (2 mV ) sin ωt = −(125 mV ) sin ωt
The AC collector current is,
ic = gm vbe = gm vs = −42.5 mf × (2 mV ) sin ωt = −85 sin ωt µA.
M. B. Patil, IIT Bombay
Common-emitter amplifier
10 V
3.6 k
RC
10 k
R1
CC
VCC
CB
vs
R2
RL
2.2 k
1 k RE
CE
For vs (t) = (2 mV ) sin ωt, we can now obtain expressions for the instantaneous currents and
voltages:
vC (t) = VC + vc (t) = VC + vo (t) = 6 V − (125 mV ) sin ωt .
iC (t) = IC + ic (t) = 1.1 mA − 0.085 sin ωt mA .
M. B. Patil, IIT Bombay
Common-emitter amplifier
10 V
3.6 k
RC
10 k
R1
CC
VCC
CB
vs
R2
RL
2.2 k
1 k RE
CE
For vs (t) = (2 mV ) sin ωt, we can now obtain expressions for the instantaneous currents and
voltages:
vC (t) = VC + vc (t) = VC + vo (t) = 6 V − (125 mV ) sin ωt .
iC (t) = IC + ic (t) = 1.1 mA − 0.085 sin ωt mA .
Note that the above procedure (DC + AC analysis) can be used only if the small-signal
approximation (i.e., |vbe | VT ) is valid. In the above example, the amplitude of vbe is 2 mV ,
which is much smaller than VT .
M. B. Patil, IIT Bombay
Common-emitter amplifier
10 V
3.6 k
RC
10 k
R1
CC
VCC
CB
vs
R2
RL
2.2 k
1 k RE
CE
For vs (t) = (2 mV ) sin ωt, we can now obtain expressions for the instantaneous currents and
voltages:
vC (t) = VC + vc (t) = VC + vo (t) = 6 V − (125 mV ) sin ωt .
iC (t) = IC + ic (t) = 1.1 mA − 0.085 sin ωt mA .
Note that the above procedure (DC + AC analysis) can be used only if the small-signal
approximation (i.e., |vbe | VT ) is valid. In the above example, the amplitude of vbe is 2 mV ,
which is much smaller than VT .
For vs (t) = (20 mV ) sin ωt, for example, the small-signal approximation will not hold, and a
numerical simulation will be required to obtain the currents and voltages of interest.
M. B. Patil, IIT Bombay
Common-emitter amplifier
10 V
3.6 k
RC
10 k
R1
CC
VCC
CB
vs
R2
RL
2.2 k
1 k RE
CE
For vs (t) = (2 mV ) sin ωt, we can now obtain expressions for the instantaneous currents and
voltages:
vC (t) = VC + vc (t) = VC + vo (t) = 6 V − (125 mV ) sin ωt .
iC (t) = IC + ic (t) = 1.1 mA − 0.085 sin ωt mA .
Note that the above procedure (DC + AC analysis) can be used only if the small-signal
approximation (i.e., |vbe | VT ) is valid. In the above example, the amplitude of vbe is 2 mV ,
which is much smaller than VT .
For vs (t) = (20 mV ) sin ωt, for example, the small-signal approximation will not hold, and a
numerical simulation will be required to obtain the currents and voltages of interest.
In practice, such a situation is anyway not prevalent (because it gives rise to distortion in the
output voltage) except in special types of amplifiers.
M. B. Patil, IIT Bombay
Common-emitter amplifier with partial bypass
coupling
capacitor
R1
coupling
capacitor
RC
R1
RC
CC
VCC
CB
RE1
R2
RL
VCC
load
resistor
vs
R2
RE
RE2
CE
bypass
capacitor
DC circuit
M. B. Patil, IIT Bombay
Common-emitter amplifier with partial bypass
coupling
capacitor
R1
coupling
capacitor
RC
R1
RC
CC
VCC
CB
RE1
R2
RL
VCC
load
resistor
vs
R2
RE
RE2
CE
bypass
capacitor
DC circuit
* For DC computation, CE is open, and the DC analysis is therefore identical to
our earlier amplifier, with RE ← RE 1 + RE 2 .
M. B. Patil, IIT Bombay
Common-emitter amplifier with partial bypass
coupling
capacitor
R1
coupling
capacitor
RC
R1
RC
CC
VCC
CB
RE1
R2
RL
VCC
load
resistor
vs
R2
RE
RE2
CE
bypass
capacitor
DC circuit
* For DC computation, CE is open, and the DC analysis is therefore identical to
our earlier amplifier, with RE ← RE 1 + RE 2 .
* Bypassing a part of RE (as opposed to all of it) does have an impact on the
voltage gain (see next slide).
M. B. Patil, IIT Bombay
Common-emitter amplifier with partial bypass
coupling
capacitor
R1
rπ
vs
CB
vs
R2
β ib
ib
CC
RE1
C
B
coupling
capacitor
RC
RL
VCC
load
resistor
R1
R2
gm vbe
E
RC
RL
RE1
RE2
CE
AC circuit
bypass
capacitor
Again, assume that, at the frequency of operation, CB , CC , CE can be replaced by
short circuits, and the BJT parasitic capacitances by open circuits.
M. B. Patil, IIT Bombay
Common-emitter amplifier with partial bypass
coupling
capacitor
R1
rπ
vs
CB
vs
R2
β ib
ib
CC
RE1
C
B
coupling
capacitor
RC
RL
VCC
load
resistor
R1
R2
gm vbe
E
RC
RL
RE1
RE2
CE
AC circuit
bypass
capacitor
Again, assume that, at the frequency of operation, CB , CC , CE can be replaced by
short circuits, and the BJT parasitic capacitances by open circuits.
vs
vs = vbe = ib rπ + (β + 1) ib RE 1 → ib =
.
rπ + (β + 1) RE 1
M. B. Patil, IIT Bombay
Common-emitter amplifier with partial bypass
coupling
capacitor
R1
rπ
vs
CB
vs
R2
β ib
ib
CC
RE1
C
B
coupling
capacitor
RC
RL
VCC
load
resistor
R1
R2
gm vbe
E
RC
RL
RE1
RE2
CE
AC circuit
bypass
capacitor
Again, assume that, at the frequency of operation, CB , CC , CE can be replaced by
short circuits, and the BJT parasitic capacitances by open circuits.
vs
vs = vbe = ib rπ + (β + 1) ib RE 1 → ib =
.
rπ + (β + 1) RE 1
vo = −β ib × (RC k RL ) →
vo
β (RC k RL )
=−
.
vs
rπ + (β + 1) RE 1
M. B. Patil, IIT Bombay
Common-emitter amplifier with partial bypass
coupling
capacitor
R1
rπ
vs
CB
vs
R2
β ib
ib
CC
RE1
C
B
coupling
capacitor
RC
RL
VCC
load
resistor
R1
R2
gm vbe
E
RC
RL
RE1
RE2
CE
AC circuit
bypass
capacitor
Again, assume that, at the frequency of operation, CB , CC , CE can be replaced by
short circuits, and the BJT parasitic capacitances by open circuits.
vs
vs = vbe = ib rπ + (β + 1) ib RE 1 → ib =
.
rπ + (β + 1) RE 1
vo = −β ib × (RC k RL ) →
vo
β (RC k RL )
=−
.
vs
rπ + (β + 1) RE 1
Note: RE 1 gets multiplied by (β + 1).
M. B. Patil, IIT Bombay
Frequency response of common-emitter amplifier
rb
B
Cµ
C
CB
103
CC
rπ
ro
gm vbe
vs
R1
R2
RC
E
RE
CE
RL
Gain
Cπ
102
101
101
102
103
104
106
105
107
108
Frequency (Hz)
M. B. Patil, IIT Bombay
Frequency response of common-emitter amplifier
rb
B
Cµ
C
CB
103
CC
rπ
ro
gm vbe
vs
R1
R2
RC
E
RE
CE
RL
Gain
Cπ
102
101
101
102
103
104
106
105
107
108
Frequency (Hz)
* CB , CE , CC are large capacitances → 1/ωC is perceptibly large only at low
frequencies.
M. B. Patil, IIT Bombay
Frequency response of common-emitter amplifier
rb
B
Cµ
C
CB
103
CC
rπ
ro
gm vbe
vs
R1
R2
RC
E
RE
CE
RL
Gain
Cπ
102
101
101
102
103
104
106
105
107
108
Frequency (Hz)
* CB , CE , CC are large capacitances → 1/ωC is perceptibly large only at low
frequencies.
* Cπ , Cµ are small capacitances → 1/ωC is perceptibly small only at high
frequencies.
M. B. Patil, IIT Bombay
Frequency response of common-emitter amplifier
rb
B
Cµ
C
CB
103
CC
rπ
ro
gm vbe
vs
R1
R2
RC
E
RE
CE
RL
Gain
Cπ
102
101
101
102
103
104
106
105
107
108
Frequency (Hz)
* CB , CE , CC are large capacitances → 1/ωC is perceptibly large only at low
frequencies.
* Cπ , Cµ are small capacitances → 1/ωC is perceptibly small only at high
frequencies.
* In the intermediate range (called “mid-band”), the large capacitances behave
like short circuits, and the small capacitances behave like open circuits. In this
range, the gain is independent of frequency.
M. B. Patil, IIT Bombay
General representation of an amplifier
source
resistance
Ro
Rs
source
(signal)
vs
vi
Rin
AV vi
vo
load
resistance
RL
amplifier
* An amplifier is represented by a voltage gain, an input resistance, and an output
resistance.
M. B. Patil, IIT Bombay
General representation of an amplifier
source
resistance
Ro
Rs
source
(signal)
vs
vi
Rin
AV vi
vo
load
resistance
RL
amplifier
* An amplifier is represented by a voltage gain, an input resistance, and an output
resistance.
* The above representation involves AC quantities only, i.e., it describes the AC
equivalent circuit of the amplifier.
M. B. Patil, IIT Bombay
General representation of an amplifier
source
resistance
Ro
Rs
source
(signal)
vs
vi
Rin
AV vi
vo
load
resistance
RL
amplifier
* An amplifier is represented by a voltage gain, an input resistance, and an output
resistance.
* The above representation involves AC quantities only, i.e., it describes the AC
equivalent circuit of the amplifier.
* The DC bias of the circuit can affect parameter values in the AC equivalent
circuit (AV , Rin , Ro ). For example, for the common-emitter amplifier,
AV ∝ gm = IC /VT , IC being the DC (bias) value of the collector current.
M. B. Patil, IIT Bombay
General representation of an amplifier
source
resistance
Ro
Rs
source
(signal)
vs
vi
Rin
AV vi
vo
load
resistance
RL
amplifier
* An amplifier is represented by a voltage gain, an input resistance, and an output
resistance.
* The above representation involves AC quantities only, i.e., it describes the AC
equivalent circuit of the amplifier.
* The DC bias of the circuit can affect parameter values in the AC equivalent
circuit (AV , Rin , Ro ). For example, for the common-emitter amplifier,
AV ∝ gm = IC /VT , IC being the DC (bias) value of the collector current.
* Suppose we are given an amplifier as a “black box” and asked to find AV , Rin ,
and Ro . What experiments would give us this information?
M. B. Patil, IIT Bombay
Voltage gain AV
source
resistance
Rs
source
(signal)
vs
il
Ro
vi
Rin
vo
AV vi
load
resistance
RL
amplifier
If RL → ∞, il → 0, and vo → AV vi .
M. B. Patil, IIT Bombay
Voltage gain AV
source
resistance
Rs
source
(signal)
vs
il
Ro
vi
Rin
vo
AV vi
load
resistance
RL
amplifier
If RL → ∞, il → 0, and vo → AV vi .
We can remove RL (i.e., replace it with an open circuit), measure vi and vo , then use
AV = vo /vi .
M. B. Patil, IIT Bombay
Input resistance Rin
source
resistance
Rs
source
(signal)
vs
ii
Ro
vi
Rin
vo
AV vi
load
resistance
RL
amplifier
Measurement of vi and ii yields Rin = vi /ii .
M. B. Patil, IIT Bombay
Output resistance Ro
Rs
Rs
Ro
vs
vi
Rin
AV vi
Ro
vo
RL
vi
Rin
io
vo
Method 1:
If vs → 0, AV vi → 0.
Now, connect a test source vo , and measure io .
Clearly, Ro = vo /io .
This method works fine on paper, but it is difficult to use experimentally.
M. B. Patil, IIT Bombay
Output resistance Ro
Rs
Ro
vs
vi
Rin
AV vi
vo
RL
Method 2:
vo =
RL
AV vi .
RL + Ro
M. B. Patil, IIT Bombay
Output resistance Ro
Rs
Ro
vs
vi
Rin
AV vi
vo
RL
Method 2:
vo =
RL
AV vi .
RL + Ro
If RL → ∞, vo1 = AV vi .
M. B. Patil, IIT Bombay
Output resistance Ro
Rs
Ro
vs
vi
Rin
AV vi
vo
RL
Method 2:
vo =
RL
AV vi .
RL + Ro
If RL → ∞, vo1 = AV vi .
1
1
If RL = Ro , vo2 = AV vi = vo1 .
2
2
M. B. Patil, IIT Bombay
Output resistance Ro
Rs
Ro
vs
vi
Rin
AV vi
vo
RL
Method 2:
vo =
RL
AV vi .
RL + Ro
If RL → ∞, vo1 = AV vi .
1
1
If RL = Ro , vo2 = AV vi = vo1 .
2
2
Procedure:
Measure vo1 with RL → ∞ (i.e., RL removed).
Vary RL to obtain vo = vo1 /2.
The corresponding RL is the same as Ro .
M. B. Patil, IIT Bombay
Common-emitter amplifier
3.6 k
coupling
capacitor
RC
coupling
capacitor
R1
CC
vs
CB
vs
C
B
10 k
RL
R2
2.2 k
RE
1k
CE
vi
R1
R2
rπ
VCC
load
resistor
RC
gm vbe
vo
RL
E
amplifier
bypass
capacitor
M. B. Patil, IIT Bombay
Common-emitter amplifier
3.6 k
coupling
capacitor
RC
coupling
capacitor
R1
CC
vs
CB
vs
C
B
10 k
RL
R2
2.2 k
RE
1k
CE
vi
R1
R2
rπ
VCC
load
resistor
RC
gm vbe
vo
RL
E
amplifier
bypass
capacitor
AV =
vo
, with RL → ∞.
vi
AV =
−gm vbe RC
= −gm RC = −42.5 mf × 3.6 k=153.
vi
M. B. Patil, IIT Bombay
Common-emitter amplifier
3.6 k
coupling
capacitor
RC
coupling
capacitor
R1
CC
vs
CB
vs
C
B
10 k
RL
R2
2.2 k
RE
1k
CE
vi
R1
R2
rπ
VCC
load
resistor
RC
gm vbe
vo
RL
E
amplifier
bypass
capacitor
AV =
vo
, with RL → ∞.
vi
AV =
−gm vbe RC
= −gm RC = −42.5 mf × 3.6 k=153.
vi
The input resistance of the amplifier is, by inspection, Rin = (R1 k R2 ) k rπ .
rπ = β/gm = 100/42.5 mf = 2.35 k → Rin = 1.24 k.
M. B. Patil, IIT Bombay
Common-emitter amplifier
3.6 k
coupling
capacitor
RC
coupling
capacitor
R1
CC
vs
CB
vs
C
B
10 k
RL
R2
2.2 k
RE
1k
CE
vi
R1
R2
rπ
VCC
load
resistor
RC
gm vbe
vo
RL
E
amplifier
bypass
capacitor
AV =
vo
, with RL → ∞.
vi
AV =
−gm vbe RC
= −gm RC = −42.5 mf × 3.6 k=153.
vi
The input resistance of the amplifier is, by inspection, Rin = (R1 k R2 ) k rπ .
rπ = β/gm = 100/42.5 mf = 2.35 k → Rin = 1.24 k.
The output resistance is RC (by “Method 1” seen previously).
M. B. Patil, IIT Bombay