Download cont`d - UniMAP Portal

EKT 104
ANALOGUE ELECTRONIC CIRCUITS
Multistage Amplifiers
Syllabus
Cascade connection, cascode connection, Darlington
connection, transformer coupling.
1
CASCADED 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
2
 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
3
(i) Cascade /RC coupling
.
.
R C1
RB
.
R C2
.
vo
Q2
C
vi
.
Q1
4
(ii) Cascode
.
R1
.
RL
.
vo
Q2
R2
vi
.
Q1
R3
5
(iii) Darlington/Direct coupling
.
.
R1
R2
.
vo
Q2
vi
.
Q1
6
(iv)
Transformer coupling
.
T
vo
Q2
R1
vi
.
Q1
7
Cascade connection
RC coupling
- Coupling capacitor couples the signal from one stage
to another stage and block dc voltage from one
stage to 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
8
i) Cascade connection
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.
9
Example 1 – Solution
V CC +20 V
DC analysis
The circuit under DC
condition (stage 1 and
stage 2 are identical)
R1
15 kW
RC1
2.2 kW
Q1
R2
4.7 kW
RE1
1 kW
10
Example 1 – Solution (cont’d)
V CC
Applying Thevenin’s theorem, the
circuit becomes;
4.7
VBB =
VCC = 4.7 V
(4.7 + 15)
15(4.7)
RBB = 15 4.7 =
= 3.57 KW
(15 + 4.7)
I BQ1 = I BQ 2 = 19.89 μA
RC1
2.2 kW
V BB
4.77 V
3.58 kW
g m1 = g m 2 = 0.153 A/V
(1+b )I B
RE1
1 kW
I CQ1 = I CQ 2 = 3.979 mA
r 1 = r 2 = 1.307 kW
bIB
R BB
IB
20 V
VT
r =
I BQ
gm =
I CQ
VT
11
Example 1 – Solution (cont’d)
AC analysis
The small-signal equivalent circuit (mid-band);
Vi
R B1
B1
r1
B2
C1
+
v1
-
R C1
g m 1v  1
R B2
r2
+
v2
-
E1
RB1 = R1 // R2
Vo
C2
RC2
g m 2v  2
E2
RB 2 = R3 // R4
12
Example 1 – Solution (cont’d)
Vo =  g m 2v 2 RC 2
A2 =
Vo
=  g m 2 RC 2
v 2
v 2 =  g m1v 1 (RC1 // RB 2 // r 2 )
=  g m1Vi (RC1 // RB 2 // r 2 )
v 1 = Vi
v 2
A1 =
=  g m1 (RC1 // RB 2 // r 2 )
Vi
13
Example 1 – Solution (cont’d)
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
14
Example 1 – Solution (cont’d)
The input resistance;
Rin = RB1 // r 1 = 3.579 // 1.307 = 0.957 kW
The output resistance;
Ro = RC 2 = 2.2 kW
15
Direct couple (DC)
The first stage is directly coupled to
the next stage without going through
a coupling capacitor
A common-emitter stage
driving another commonemitter stage
16
Direct couple (DC)
The biasing network must be
be suitably designed otherwise
the dc bias of one stage will
upset the bias voltage of the
other
17
Example 2
Perform a dc analysis
and hence calculate the
voltage gain Av where;
vo
Av =
vs
Assume b1 = 170, b2 =
150 and VBE(ON) = 0.7 V.
18
Direct couple (DC)
A common-emitter driving an
emitter-follower.
19
Example 2 - Solution
V CC
DC analysis
+5 V
I1
RBB
RE2
R1 R2
=
= 33.3 kW
R1 + R2
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
20
Example 2 – Solution (cont’d)
V CC
The base-emitter loop of Q1
I1
RE2
RC1
VBB + RBB I B1 + VBE1
+ RE1 I E1 + VEE = 0
VE2
VC1
IB1
V BB
= VBB  VEE  VBE1
5 kW
IB2
Rearranging;
RBB I B1 + (b1 + 1)RE1I B1
+5 V
R BB
2 kW
IE2
Q2
IC1
VB1
Q1
VE1
RE1
IE1
VC2
RC2
IC2
1.5 kW
2 kW
V EE -5 V
21
Example 2 – Solution (cont’d)
V CC
Substituting values;
+5 V
I1
RE2
33.3 103 I B1
RC1
+ (170 + 1)2 103 I B1
= 1.667 + 5  0.7
I B1 = 7 A
5 kW
IB2
VE2
VC1
IB1
V BB
R BB
2 kW
IE2
Q2
IC1
VB1
Q1
VE1
RE1
IE1
VC2
RC2
IC2
1.5 kW
2 kW
I C1 = b1I B1 = 1.19 mA
V EE -5 V
22
Example 2 – Solution (cont’d)
V CC
I E1 = (b1 + 1)I B1 = 1.197 mA
I1
RE2
RC1
VB1 = VBB  RBB I B1 = 1.9 V
VE1 = RE1I E1 + VEE
IB1
= 2.606 V
V BB
+5 V
5 kW
IB2
VE2
VC1
R BB
2 kW
IE2
Q2
IC1
VB1
Q1
VE1
RE1
IE1
VC2
RC2
IC2
1.5 kW
2 kW
V EE -5 V
23
Example 2 – Solution (cont’d)
V CC
For the RC1 – collector of Q1
– base of Q2 – RE2 loop;
RE 2 I E 2 + VEB2 = RC1I1
I1
RE2
RC1
5 kW
IB2
VE2
VC1
IB1
R BB
V BB
I E 2 = (b 2 + 1)I B 2
2 kW
IE2
Q2
IC1
VB1
I 1 = I C1  I B 2
and
+5 V
Q1
VE1
RE1
IE1
VC2
RC2
IC2
1.5 kW
2 kW
V EE -5 V
(b 2 + 1)RE 2 I B 2 + VEB2 = RC1 ( I C1  I B 2 )
24
Example 2 – Solution (cont’d)
V CC
+5 V
Substituting values;
I1
(150 + 1)2 103 I B 2 + 0.7
(
3
RE2
= 5 10 1.19 10  I B 2
3
)
RC1
5 kW
IB2
VE2
VC1
I B 2 = 17 μA
I C 2 = b 2 I B 2 = 2.565 mA
IB1
V BB
R BB
2 kW
IE2
Q2
IC1
VB1
Q1
VE1
RE1
IE1
VC2
RC2
IC2
1.5 kW
2 kW
V EE -5 V
25
Example 2 – Solution (cont’d)
V CC
I E 2 = (b2 + 1)I B 2 = 2.582 mA
I1
RE2
RC1
I1 = I C1  I B 2 = 1.19  0.017
= 1.173 mA
VC1 = VCC  RC1 I1
= 5  5 1.173
= 0.865 V
+5 V
5 kW
IB2
VE2
VC1
IB1
V BB
R BB
2 kW
IE2
Q2
IC1
VB1
Q1
VE1
RE1
IE1
VC2
RC2
IC2
1.5 kW
2 kW
V EE -5 V
26
Example 2 – Solution (cont’d)
V CC
VE 2 = VCC  RE 2 I E 2
+5 V
I1
= 5  2  2.582
= 0.164 V
RE2
RC1
5 kW
IB2
VE2
VC1
VC 2 = RC 2 I C 2 + VEE
= 1.5  2.565  5
= 1.175 V
IB1
V BB
R BB
2 kW
IE2
Q2
IC1
VB1
Q1
VE1
RE1
IE1
VC2
RC2
IC2
1.5 kW
2 kW
V EE -5 V
27
Example 2 – Solution (cont’d)
AC analysis
 V 2
A1 
Vs
Vo
A2 
 V 2
28
Example 2 – Solution (cont’d)
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
29
Example 2 – Solution (cont’d)
 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
30
Example 2 – Solution (cont’d)
 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 
31
Example 2 – Solution (cont’d)
Substituting values;
I C1 1.19
g m1 =
=
= 45.77 mA/V
VT
26
VT
26
r 1 = b1
= 170 
= 3714 W
I C1
1.19
32
Example 2 – Solution (cont’d)
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
33
Example 2 – Solution (cont’d)
 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
34
Example 3
Determine;
(a) Q-points
(b) Av;
(c) Ri;
(d) Ro.
The parameters for
Q1 and Q2 are;
b = 125; VBE(on) =
0.7 V; ro = .
35
Example 3 – Solution
DC analysis
The circuit under DC condition:
Applying Thevenin’s
theorem at the input;
RBB = R1 // R2
VBB
70  6
=
= 5.53 kW
70 + 6
 6 
= 10
  5 = 4.21 V
 70 + 6 
36
Example 3 – Solution (cont’d)
The circuit becomes as shown;
Taking the loop VBB – B1 –
E1 – V–;
RBB I B1 + (1 + b )RE1
= VBB  V   VBE1
Substituting values;
5.53I B1 + (1 + 125) 0.2I B1
= 4.21 + 5  0.7
I B1 = 2.93A
37
Example 3 – Solution (cont’d)
I C1 = bI B1 = 125  2.93A
= 0.366 mA
I E1 = (1 + b )I B1
= 125  2.93A
= 0.369 mA
38
Example 3 – Solution (cont’d)
Taking the loop V+ – RC1 – B2 –
E2 – RE2 – V–;
I 3 RC1 + VBE 2 + I E 2 RE 2
= V + V 
But;
I 3 = I C1 + I B 2
and;
I E 2 = (1 + b )I B 2
39
Example 3 – Solution (cont’d)
Hence;
(I C1 + I B 2 )RC1 + (1 + b )I B 2 RE 2
= V +  V   VBE 2
Substituting values;
(0.366 + I B 2 )5 + (1 + 125)I B 2 1.5
= 5 + 5  0.7
Or;
I B 2 = 38.5 A
40
Example 3 – Solution (cont’d)
Hence;
I C 2 = bI B 2 = 125  38.5 A
= 4.81 mA
I E 2 = (1 + b )I B 2
= (1 + 125) 38.5 A
= 4.85 mA
and;
I 3 = I C1 + I B 2 = 0.366 + 0.0385
= 0.405 mA
41
Example 3 – Solution (cont’d)
VC1 = V +  I 3 RC1
= 5  0.405  5
= 2.98 V
VE 1 = I E 1 RE 1 + V 
= 0.369  0.2  5
= 4.93 V
42
Example 3 – Solution (cont’d)
VC 2 = V + = 5 V
VE 2 = I E 2 RE 2 + V 
= 4.85 1.5  5
= 2.28 V
43
Example 3 – Solution (cont’d)
VCE1 = VC1  VE1
= 2.98 + 4.93
= 7.91 V
VCE 2 = VC 2  VE 2
= 5  2.28
= 2.72 V
44
Example 3 – Solution (cont’d)
(a) The Q-points are;
I C1 = 0.366 mA;
VCE1 = 7.91 V
and;
I C 2 = 4.81 mA;
VCE 2 = 2.72 V
45
Example 3 – Solution (cont’d)
The circuit under ac condition
46
Example 3 – Solution (cont’d)
Small–signal equivalent circuit
47
Example 3 – Solution (cont’d)
r 1 =
bVT
I C1
125  0.026
=
= 8.88 kW
0.000366
I C1 0.000366
g m1 =
=
= 14.08 mA/V
VT
0.026
r 2 =
gm2
bVT
IC 2
125  0.026
=
= 0.676 kW
0.00481
I C 2 0.00481
=
=
= 185 mA/V
VT
0.026
48
Example 3 – Solution (cont’d)
 V 2

Vo = 
+ g mV 2 (RE 2 // RL )
 r 2

V 2
= (1 + b )(RE 2 // RL )
r 2
Vo ' = V + Vo
V 2
= V 2 + (1 + b )(RE 2 // RL )
r 2
49
Example 3 – Solution (cont’d)
The voltage gain of the
second stage is;
Vo
Av 2 =
Vo '
V 2
(1 + b )(RE 2 // RL )
r 2
=
V 2
V 2 + (1 + b )(RE 2 // RL )
r 2
50
Example 3 – Solution (cont’d)
The voltage gain of the
second stage is;
(
1 + b )(RE 2 // RL )
=
r 2 + (1 + b )(RE 2 // RL )
Substituting values;
(
1 + 125)(1.5 // 10)
Av 2 =
0.676 + (1 + 125)(1.5 // 10)
= 0.996
51
Example 3 – Solution (cont’d)
Rib2 = r 2 + (1 + b )(RE 2 // RL )
= 0.676 + (1 + 125)(1.5 // 10)
= 165 kW
52
Example 3 – Solution (cont’d)
 V 1

Vs = V 1 + 
+ g m1V 1  RE1
 r 1

Vo ' =  g m1V 1 (RC1 // Rib2 )
Vo '
Av1 =
Vs
53
Example 3 – Solution (cont’d)
 g m1 (RC1 // Rib2 )
=
 1

1 +  + g m1  RE1
 r 1

b (RC1 // Rib2 )
Av1 = 
r 1 + (1 + b )RE1
54
Example 3 – Solution (cont’d)
Substituting values;
(b)
b (RC1 // Rib2 )
Av1 = 
r 1 + (1 + b )RE1
125(5 // 165)
=
= 17.8
8.88 + (1 + 125) 0.2
Av = Av1  Av 2 = 17.8  0.996 = 17.7
55
Example 3 – Solution (cont’d)
56
EXERCISE 1
The parameters for each transistor in FIG. E1 are b = 100, VA =  and
VBE = 0.7 V.
(a) Determine the small-signal parameters gm, r and ro for both
transistors.
(b) Determine the small-signal voltage gain Av1 for stage 1 and Av2 for
stage 2 separately, where
vo1
Av1 =
vs
and
vo
Av 2 =
vo1
(Assume the output of stage 1 is open to determine Av1)
(c) Determine the overall small-signal voltage gain Av, where;
vo
Av =
vs
57
EXERCISE 1 (cont’d)
FIG. E1
58