MULTISTAGE AMPLIFIERS
PART I - BJT AMPLIFIERS
ACKNOLEDGEMENT
The materials presented in these notes were partly taken from the
original notes “MULSTISTAGE AMPLIFIERS” by ENCIK SOHIFUL
ANUAR BIN ZAINOL MURAD, School of Microelectronic
1
 Many applications cannot be handled with singletransistor amplifiers in order to meet the
specification of a given amplification factor, input
resistance and output resistance
 As a solution – transistor amplifier circuits can be
connected in series or cascaded amplifiers
 This can be done either to increase the overall
small-signal voltage gain or provide an overall
voltage gain greater than 1 with a very low output
resistance
2
Multistage amplifier configuration
Cascade /RC coupling
.
.
R C1
RB
.
R C2
.
vo
Q2
vi
.
Q1
3
Multistage amplifier configuration
.
Cascode
R1
.
RL
.
vo
Q2
R2
vi
.
Q1
R3
4
Multistage amplifier configuration
Darlington/Direct coupling
.
.
R1
R2
.
vo
Q2
vi
.
Q1
5
Multistage amplifier configuration
Transformer coupling
.
T
vo
Q2
R1
vi
.
Q1
6
i) Cascade connection
- The most widely used method
- Coupling a signal from one stage to the
another stage and block dc voltage from one
stage to the another stage
- The signal developed across the collector
resistor of each stage is coupled into the
base of the next stage
- The overall gain = product of the individual
gain
7
Example 1
V CC
R1
15 kW
RC1
2.2 kW
b 1 = b 2 = 200
+20 V
ro1 = ro2 = 
R3
15 kW
RC2
2.2 kW
Ro
Ri
R2
4.7 kW
RE1
1 kW
R4
4.7 kW
RE2
1 kW
Draw the AC equivalent circuit and calculate Av,
Ri and Ro.
8
V CC +20 V
Solution
DC analysis
R1
15 kW
RC1
2.2 kW
The circuit under DC
condition (stage 1 and
stage 2 are identical)
Q1
R2
4.7 kW
RE1
1 kW
9
Applying Thevenin’s theorem, the
circuit becomes;
V CC
I BQ1 = I BQ 2 = 19.89 μA
RC1
2.2 kW
V BB
4.77 V
I CQ1 = I CQ 2 = 3.979 mA
bIB
R BB
IB
20 V
r 1 = r 2 = 1.307 kW
g m1 = g m 2 = 0.153 A/V
3.58 kW
(1+b )I B
RE1
1 kW
10
AC analysis
The small-signal equivalent circuit (mid-band);
Vi
R B1
B1
r1
B2
C1
+
v1
-
R C1
R B2
r2
g m 1v  1
E1
RB1 = R1 // R2
Vo
C2
+
v2
-
RC2
g m 2v  2
E2
RB 2 = R3 // R4
11
Vo =  g m 2v 2 RC 2
Vo
A2 =
=  g m 2 RC 2
v 2
v 2 =  g m1v 1 (RC1 // RB 2 // r 2 )
=  g m1Vi (RC1 // RB 2 // r 2 )
A1 =
v 1 = Vi 
v 2
=  g m 2 (RC1 // RB 2 // r 2 )
Vi
12
The small-signal voltage gain;
A = A1 A2 = g m1 g m 2 RC 2 (RC1 // RB 2 // r 2 )
Substituting values;
RB1 = RB 2 = R3 // R4 = 15 // 4.7 = 3.579 kW
RC1 // RB 2 // r 2 = 2.2 // 3.579 // 1.307 = 667 W
A = 0.153 0.153 2200  667 = 34350 V/V
13
The input resistance;
Rin = RB1 // r 1 = 3.579 // 1.307 = 0.957 kW
The output resistance;
Ro = RC 2 = 2.2 kW
14
Example 2
Assuming b1 = 170, b2 = 150 and VBE(ON) = 0.7 V,
calculate the voltage gain Av where;
vo
Av =
vs
15
V CC
DC analysis
+5 V
I1
RBB =
R1 R2
= 33.3 kW
R1 + R2
RE2
RC1
5 kW
IB2
VE2
VC1
IB1
R BB
V BB
VBB = VEE + (VCC
= 1.667
 R2 

 VEE )
 R1 + R2 
2 kW
IE2
Q2
IC1
VB1
Q1
VE1
RE1
IE1
VC2
RC2
IC2
1.5 kW
2 kW
V EE -5 V
16
The base-emitter loop of Q1
VBB + RBB I B1 + VBE1 + RE1I E1 + VEE = 0
RBB I B1 + (b1 + 1)RE1I B1 = VBB  VEE  VBE1
Substituting values;
33.3 103 I B1 + (170 + 1)2 103 I B1 = 1.667 + 5  0.7
I B1 = 7 A
I C1 = b1I B1 = 1.19 mA
I E1 = (b1 + 1)I B1 = 1.197 mA
17
VB1 = VBB  RBB I B1 = 1.9 V
VE1 = RE1I E1 + VEE = 2.606 V
For the RC1 – collector of Q1 – base of Q2 – RE2 loop
RE 2 I E 2 + VEB2 = RC1 I1
I 1 = I C1  I B 2
and
I E 2 = (b 2 + 1)I B 2
(b 2 + 1)RE 2 I B 2 + VEB2 = RC1 ( I C1  I B 2 )
18
Substituting values;
(150 + 1)2 103 I B 2 + 0.7 = 5 103 (1.19 103  I B 2 )
I B 2 = 17 μA
I C 2 = b 2 I B 2 = 2.565 mA
I E 2 = (b2 + 1)I B 2 = 2.582 mA
I1 = I C1  I B 2 = 1.19  0.017 = 1.173 mA
19
VC1 = VCC  RC1I1 = 5  5 1.173 = 0.865 V
VE 2 = VCC  RE 2 I E 2 = 5  2  2.582 = 0.164 V
VC 2 = RC 2 I C 2 + VEE = 1.5  2.565  5 = 1.175 V
20
AC analysis
 V 2
A1 
Vs
Vo
A2 
 V 2
21
V 2 = g m1V 1 (RC1 // r 2 )
 Rin 

V 1 = Vs 
 RS + Rin 
 V 2
=  g m1 (RC1 // r 2 )
V 1
V 1
Rin
=
Vs
RS + Rin
 Rin 
 V 2  V 2 V 1

A1 =
=

=  g m1 (RC1 // r 2 )
Vs
V 1 Vs
 RS + Rin 
Vo = g m 2V 2 (RC 2 // RL )
Vo
A2 =
=  g m 2 (RC 2 // RL )
 V 2
22
 Rin 
 g m 2 (RC 2 // RL )
A = A1 A2 =  g m1 (RC1 // r 2 )
 RS + Rin 
 Rin 
(RC1 // r 2 )(RC 2 // RL )
A = g m1 g m 2 
 RS + Rin 
Substituting values;
g m1 =
I C1 1.19
=
= 45.77 mA/V
VT
26
VT
26
r 1 = b1
= 170 
= 3714 W
I C1
1.19
23
gm2
I C 2 2.565
=
=
= 98.65 mA/V
VT
26
VT
26
r 2 = b 2
= 150 
= 1520 W
IC 2
2.565
Rin = R1 // R2 // r 1 = 100 // 50 // 3.714 = 3.342 kW
 Rin 
(RC1 // r 2 )(RC 2 // RL )
A = g m1 g m 2 
 RS + Rin 
 3.342 
= 45.77  98.65
(5 // 1.52 )(1.5 // 5)
 0.5 + 3.342 
A = 5286 V/V
24
ii) Cascode connection
- A cascode connection has one transistor on top of (in
series with) another
- The i/p applied to a C-E amp. (Q1) whose output is
used to drive a C-B amp. (Q2)
- The o/p signal current of Q1 is the i/p signal of Q2
- The advantage: provides a high i/p impedance with
low voltage gain to ensure the i/p Miller capacitance is
at a min. with the C-B stage providing good high freq.
operation
25
V CC
RC
CC2
R1
vo
CB
Q2
RL
R2
CC1
Q1
vs
R3
RE
CE
Cascode amplifier
26
V CC
DC analysis
May be performed
using the following
figure;
I1
R1
bI 
I1   1 B1 
1+ b2 
R2
b1 I B1
1+ b2
RC
Q2
b1 I B1
I B1
Q1
bI 
I1   1 B1   I B1
1+ b2 
R3
 b1 I B1 

b 2 
1+ b2 
(1 + b1 )I B1
RE
27
The equations are (assuming VBE = 0.7 V for both BJT’s);



b1I B1 
b1I B1
 + R3  I1 
R1I1 + R2  I1 
 I B1  = VCC
b2 +1 
b2 + 1





b1I B1
R3  I1 
 I B1  = 0.7 + RE (b1 + 1)I B1
b2 +1


The above equations may solved for the two unknown
currents namely I1 and IB1.
28
AC analysis
Q2
vo
Q1
vs
R BB
RBB = R2 // R3
R L'
RL '= RC // RL
The equivalent circuit under AC condition
29
C2
vo
B1
vs
R BB
r 1
RBB = R2 // R3
C1 E2
+
v 1
E1
r 2
g mv  1
g mv  2
R L'
+
v 2
B2
RL '= RC // RL
The ac equivalent circuit using hybrid- model for BJT
30
vo =  g m 2v 2 RL '
(1)
At node E2;
v 2
g m1v 1 =
+ g m 2 v 2
r 2
Or;
v 2
g m1r 2 v 1
=
1 + g m 2 r 2
Substituting in (1);
 g m1 g m 2 r 2 
 b2 
 RL ' v 1 =  g m1 
 RL ' vs
vo = 
 1 + b2 
 1 + g m 2 r 2 
31
The small-signal voltage gain;
 b2 
vo
 RL '
Av = =  g m1 
vs
 1 + b2 
When b2 >> 1
Av   g m1RL '
32
V CC
Example 3
R1
76 kW
CB
+15 V
RC
5 kW
CC2
Q2
R2
RL
Compute the
approximate smallsignal voltage gain
vo
5 kW
28 kW
CC1
b1 = b 2 = 150
Q1
VBE1 = VBE 2 = 0.7 V
vs
R3
37 kW
RE
3 kW
VT = 26 mV
CE
33
V CC
SOLUTION
+15 V
I1
DC analysis
R1
76 kW
bI 
I1   1 B1 
1+ b2 
R2
b1 I B1
1+ b2
RC
 b1 I B1 

1+ b2 
b 2 
Q2
28 kW
b1 I B1
I B1
Q1
bI 
I1   1 B1   I B1
1+ b2 
R3
5 kW
(1 + b1 )I B1
37 kW
RE
3 kW
The circuit under DC condition
34



b1I B1 
b1I B1
 + R3  I1 
R1I1 + R2  I1 
 I B1  = VCC
b2 +1 
b2 + 1



Substituting values;
150 I B1 
150 I B1



76kI1 + 28k I1 
+
37
k
I


I

 1
B1  = 15
150 + 1 
150 + 1



141I1  101.54I B1 = 15 103
(1)
35


b1I B1
R3  I1 
 I B1  = 0.7 + RE (b1 + 1)I B1
b2 +1


Or;


b1I B1

R3  I1 
 I B1   RE (b1 + 1)I B1 = 0.7
b2 + 1


Substituting values;
150 I B1


37 I1 
 I B1   3(150 + 1)I B1 = 0.7 10 3
150 + 1


37 I1  526.75I B1 = 0.7 103
36
I1 = 0.0189 103 + 14.24I B1
(2)
Substituting for I1 in (1)
(
)
141 0.0189 103 + 14.24I B1  101.54I B1 = 15 103
I B1
12.335  10 3
=
= 6.47 A
1906.3
I CQ1 = bI B1 = 150  6.47 A = 0.97 mA
I E 2 = I CQ1 = 0.97 mA
37
I B2 =
I E2
0.97
=
= 6.42 A
b + 1 151
I CQ 2 = bI B 2 = 150  6.42 A = 0.964 mA
38
AC analysis
C2
vo
B1
vs
R BB
r 1
C1 E2
+
v 1
E1
r 2
g mv  1
g mv  2
R L'
+
v 2
B2
Small-signal equivalent circuit using hybrid- model
39
(1)
vo =  g m 2v 2 RL '
At node E2;
Hence;
g m1v 1 =
v 2
v 2
+ g m 2 v 2
r 2
g m1r 2 v 1
=
1 + r 2 g m 2
Substituting for v2 in (1);
 g m1r 2 g m 2 
 RL ' v 1
vo = 
 1 + r 2 g m 2 
40
Or;
 b2 
 RL ' vi
vo =  g m1 
 1 + b2 
The voltage gain;
 b2 
vo
 RL '
Av  =  g m1 
vi
 1 + b2 
When b2   1;
Av   g m1RL '
41
Substituting values;
g m1 =
I CQ1
VT
0.97
=
= 37.3 mA/V
26
RL ' = RC // RL = 5k // 5k = 2.5 kW
Av  0.0373(2500)
Av = 93 V/V
42
iii) Darlington connection
C
Darlington pair
c1
Internal connection;
• Collectors of Q1 and
Q2;
• Emitter of Q1 and
base of Q2.
Provides high current
gain : IC  b2IB
IC
b1
B
c2
Q1
IB
e1
b2
Q2
e2
E
43
Currents in darlington pair
b 1I B 1
IB1
Q1
I E 1 = (1+b 1)I B 1 = I B 2
I C 1 + IC 2 = b 1 + b 2(1+b 1)I B 1
I C 2 = b 2I B 2 =b 2(1+b 1)I B 1
Q2
I E 2 = (1+b 2)I B 2 =(1+b 1)(1+b 2)I B 1
44
If b1 = b2 = b and assuming b is large;
bIB1
IB1
Q1
I E 1 = (1+b )I B 1 = I B 2
I C 1 + IC 2  b 2I B 1
I C 2 = b I B 2 =b (1+b )I B 1
Q2
I E 2 = (1+b )I B 2 =(1+b )2I B 1
45
Hybrid- model (assuming ro1 = ro2 = );
b1
C
c1
B
b1
c2
Q1
e1
b2
Q2
e2
E

r 1
c1
+
V 1
e1
g m 1V  1
b2
r 2
+
V 2
e2
c2
g m 2V  2
46
Darlington configuration provides;
• Increased current;
• High input resistance.
Darlington pair
configuration
47
Small-signal equivalent circuit
48
Input voltage source
is transformed into
current source
g m1V 1 = g m1r 1I i = b1I i
V 1 = I i r 1
Vi = V 1 + V 2
V 2 = (1 + b1 )I i r 2
I o = g m1V 1 + g m 2V 2
= b1 I i + b 2 (1 + b1 )I i
g m 2V 2
= g m 2 (1 + b1 )I i r 2
= b 2 (1 + b1 )I i
I b 2 = I i + b1I i = (1 + b1 )I i
49
V 1 = I i r 1
g m1V 1 = g m1r 1I i = b1I i
I b 2 = I i + b1I i = (1 + b1 )I i
V 2 = I b 2 r 2 = (1 + b1 )I i r 2
g m 2V 2 = g m 2 (1 + b1 )I i r 2 = b 2 (1 + b1 )I i
I o = g m1V 1 + g m 2V 2 = b1I i + b 2 (1 + b1 )I i
The current gain is;
Io
Ai = = b1 + b 2 (1 + b1 )  b1b 2
Ii
50
Vi = V 1 + V 2 = I i r 1 + (1 + b1 )I i r 2
The input resistance is;
Vi
Ri = = r 1 + (1 + b1 )r 2
Ii
EXERCISE 1
Show that the approximate expression for the input
resistance of the darlington configuration above is;
Ri  2b1r 2
Hints: use the relationships:
r =
bVT
I CQ
& I CQ1 
I CQ 2
b2
51
Example 4
b1 = b2 = 100
VA1 = VA2 = 
VBE1 = VBE 2 = 0.7 V
Determine the;
(a) Q-point for Q1 and Q2;
(b) voltage gain vo/vs;
(c) input resistance Ris;
(d) output resistance Ro
52
V CC
+10 V
(a) Determination of Q-points
Using Thevenin’s theorem;
VBB
RC
R1
 R2 
 = 2.72 V
= VCC 
 R1 + R2 
2.2 kW
335 kW
Q1
Q2
RBB =
R1 R2
= 91 kW
R1 + R2
R2
125 kW
RE2
1 kW
DC equivalent circuit
53
The circuit becomes;
V BB
+2.72 V
R BB
V CC
+10 V
RC
2.2 kW
b1 = b2 = 100
IC1
IB1
IC2
Q1
91 kW
IE2
RE2
I B 2 = I E1 = (b + 1)I B1
VBE1 = VBE 2 = 0.7 V
Q2
IB2
VA1 = VA2 = 
1 kW
I E 2 = (b + 1)I B 2 = (b + 1) I B1
2
54
RBB I B1 + 2VBE + RE 2 (b + 1) I B1 = VBB
2
Substituting values;
91kI B1 + 2  0.7 + 1k (100 + 1) I B1 = 2.72
2
1.32
I B1 =
10 3 = 0.128 μA
10292
I C1 = 12.8 μA
I E1 = I B 2 = 12.93 μA
I C 2 = 1.293 mA
I E 2 = 1.3 mA
VE 2 = 1.3 1 = 1.3 V
VE1 = 1.3 + 0.7 = 2 V
VC1 = VC 2 = 10  2.2(1.293 + 0.0128) = 7.127 V
55
VCE1 = VC1  VE1 = 7.127  2 = 5.127 V
VCE 2 = VC 2  VE 2 = 7.127  1.3 = 5.827 V
(a) The Q-points are;
Q1 :
I CQ1 = 12.8 μA; VCEQ1 = 5.127 V
Q2 :
I CQ 2 = 1.293 mA; VCEQ 2 = 5.827 V
56
(b) The small-signal voltage gain (mid-band);
R is
vo
vs
R BB
Q1
Q2
RC
Ro
The equivalent circuit under AC condition
57
Using the hybrid- model of transistor, the equivalent
circuit becomes;
R is
Ro
io
ii
vs
r 1
vo
+
v 1

g m 1v  1
R BB
r 2
+
v 2

RC
g m 2v  2
58
g m1 =
I CQ1
gm2 =
I CQ 2
r 1 =
r 2 =
=
VT
VT
b1
g m1
b2
gm2
12.8 A
= 0.492 mA/V
26 mV
1.293 mA
=
= 49.73 mA/V
26 mV
100
=
= 203.25 kW
3
0.492 10
100
=
= 2 kW
3
49.73 10
59
vo = (g m1V 1 + g m 2V 2 )RC
V 2
 V 1

 r 2

= 
+ g m1V 1 r 2 = 
+ g m1r 2 V 1
 r 1

 r 1

r 2
(
)
V 2 = 1 + b1
V 1
r 1
Substituting for V2 in the expression for vo and
simplifying;
 b1 + (1 + b1 )b 2 
vo = 
 RCV 1
r 1


60
vs = V 1 + V 2
Substituting for V2;
r 2
vs = V 1 + (1 + b1 ) V 1
r 1
V 2 = (1 + b1 )
r 2
V 1
r 1
 b1 + (1 + b1 )b 2 
vo = 
 RCV 1
r 1


 b1 + (1 + b1 )b 2 

RCV 1

r 1
vo


Av = =
r 2
vs
V 1 + (1 + b1 ) V 1
r 1
Simplifying;

vo
b1 + (1 + b1 )b 2 RC
Av = = 
vs
r 1 + (1 + b1 )r 2
61
Substituting values;

100 + 100 + 100 2.2
=
2
Av
(203.25 + 2 + 100  2)
Av = 55.4 V/V
62
R is
R ib
ii
+
r 1
R BB
Ro
io
ib
vo
+
v 1

g m 1v  1
vs
r 2
-
Ris = RBB // Rib
RC
+
v 2

g m 2v  2
vs
Rib =
ib
63
vs = V 1 + V 2
V 1 = ib r 1
V 2 = (ib + g m1V 1 )r 2 = (ib + g m1r 1ib )r 2
= (1 + b1 )r 2ib
vs = r 1ib + (1 + b1 )r 2ib
Rib =
vs
= r 1 + (1 + b1 )r 2
ib
Substituting values;
Rib = 203 + (1 + 100)2 = 405 kW
64
Ris = RBB // Rib =
91 405
91 + 405
Ris = 73.6 kW
Ro = RC = 2.2 kW
65
EXERCISE 2
Find; (a) ICQ1 and ICQ2
(b) Av = vo/vs
(c) Rib and Ro
I CQ 1
I CQ 2
b1 = b 2 = 100
VA1 = VA 2 = 
Answers:
(a) 2.08 mA & 69.9 mA
(b) 0.99 V/V
(c) 480 kW & 0.469 W
66