Download TUTORIAL 3: MULTISTAGE BJT AMPLIFIERS

TUTORIAL 3: MULTISTAGE BJT AMPLIFIERS - Additional
CASCADE CONFIGURATION
QUESTION
3.
Figure 3 shows an amplifier in cascode configuration where a common-base stage (Q2) is
being driven by a common-emitter stage (Q1).
(a)
Draw the circuit under dc condition and determine the quiescent currents ( I CQ1 and
I CQ 2 ) of both transistors.
(b)
Draw the small-signal equivalent circuit of the complete amplifier and perform an ac
analysis to show that the approximate overall small-signal voltage gain is given by the
expression;
Av 
vo
  g m1 RC // RL 
vi
(c)
If the peak-to-peak value of the input voltage vi is 30 mV, determine the approximate
peak-to-peak value of the output voltage vo .
Assume 1   2    150 , VBE1active  VBE 2active  VBE active  0.7 V , VA1  VA2  VA   ,
VT  26 mV and VCC  15 V .
Figure 3
SOLUTION
3.
(a)
The circuit under dc condition and
quiescent currents.
V CC
V CC
The currents are as labeled in the figure;
Taking the VCC–R1–R2–R3–GROUND loop gives us;
R1
76 k



I 
I
R1 I1  R2  I1  B1   R3  I1  B1  I B1   15
  1
 1



I1
I1 
1
Taking the R3–B1– E1–RE–GROUND loop;


I
R3  I1  B1  I B1   0.7  RE   1I B1
 1


Or;


I
R3  I1  B1  I B1   RE   1I B1  0.7
 1


Substituting values;
150 I B1


37 I1 
 I B1   3151I B1  0.7  10 3
151


Rearranging;
I1  0.0189  103  14.24I B1
Substituting (2) in (1);


Q2
 IB1
IB1
Rearranging;
141I1  101.54I B1  15  103
 2 I B1
 1
 I B1
 1
I
I 1  B1
 1
R2
28 k
Substituting values;
150 I B1 

76 I1  28 I1 

150  1 

150 I B1


37 I1 
 I B1   15  10 3
150  1


RC
5 k
141 0.0189  103  14.24I B1  101.54I B1  15  103
2
 I B1
 I B1
 1
R3
37 k
Q1
 + 1)I B 1
RE
3 k
12.335  10 3
 6.47 A
1906.3
I B1 
I CQ1  I B1  150  6.47 A  0.97 mA
Answer
I E 2  I CQ1  0.97 mA
I B2 
I E2
0.97

 6.42 A
  1 151
I CQ 2  I B 2  150  6.42 A  0.964 mA
(b)
Answer
The small-signal equivalent circuit and voltage gain
C2
.
vo
g m 2v  2
B1
.
C1
E2
vi
R 2 //R 3
r 1
+
v 

r 2
g m 1v  1
E1
vo   g m 2 v 2 RC // RL 
At node E2;
g m1v 1 
v 2
 g m 2 v 2
r 2
Thus;
v 2 
g m1r 2 v 1
1  r 2 g m 2
Substituting for v 2 in (3)
v 

B2
3
R C //R L
g r g 
vo   m1  2 m 2 RC // RL v 1
 1  r 2 g m 2 
  
  g m1  2 RC // RL vi
 1  2 
Where;
 2  g m 2 r 2
Av 
and
v 1  vi
  
vo
  g m1  2 RC // RL 
vi
 1  2 
When  2  1
Av   g m1 RC // RL 
(c) The approximate peak-to-peak value of the output voltage.
Av  0.03735000 // 5000  93 V/V
vo pp   Av vi pp   93  0.03  2.79 V
Answer
QUESTION
4.
Determine the input impedance Ri and output impedance Ro of the two-stage amplifier
shown in Figure 4.
Assume VCC  5 V , VEE  5 V , 1  127 ,  2  200 , VBE active for both transistors is 0.7
V and VT  26 mV .
Figure 4
SOLUTION
4.
 R2 
  VEE
VBB  VCC  VEE 
 R1  R2 
 47 
 5  5
  5  1.8 V
 100  47 
RB 

R1 R2
R1  R2
100  47
 32 k
100  47
Taking the base-emitter loop of Q1;
VBB  RB I B1  VBE1  RE1I E1  VEE  0
Or;
RB I B1  1  1RE1I B1  VBB  VEE  VBE1
Substituting values;
32  103 I B1  127  12  103 I B1  1.8  5  0.7
2.6
 9 μA
288  103
I B1 
I C1  1 I B1  127  9  10 6  1.15 mA
For the RC1 - collector of Q1 - base of Q2 - RE 2 loop;
RE 2 I E 2  VEB2  RC1I1
I1  I C1  I B 2 and I E 2   2  1I B 2
 2  1RE 2 I B 2  VEB2  RC1  I C1  I B 2 
Substituting values;
200  12  103 I B 2  0.7  5  103 1.15  103  I B 2 
5.05
 12.4 μA
407  103
I B2 
I C 2   2 I B 2  200  12.4  10 6  2.48 mA
I C1 1.15

 44.2 mA/V
VT
26
g m1 
r 1 
1
g m1
g m2 
r 2 

127
 2.87 k
0.0442
I C 2 2.48

 95.4 mA/V
VT
26
2
g m2

200
 2.1 k
0.0954
Rib  r 1  1  1 RE1  2.87  1  1272  258.87 k
The input resistance of the amplifier is;
Ri  RB Rib  32 258.87  28.5 k
The output resistance of the amplifier is;
Ro  RC 2  1.5 k