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TUTORIAL 3 – Questions & Answers
Q1 : The circuit shown in Figure 1 has transistor parameters  = 120 and
VA = .
(a) Determine the small signal parameters, gm and r for both
transistors.
(b) Determine the overall small-signal voltage gain, Av = vo/vs.
(c) Determine the input resistance Ris and the output resistance Ro.
TUTORIAL 3 – Questions & Answers
Q1 (cont’d)
Figure 1
TUTORIAL 3 – Questions & Answers
Q1 – Solution
(a) The small signal parameters, gm and r for both transistors.
DC analysis : Stage 1
RB1  R1 R2  67.3 12.7  10.68 k
 R2 

VBB1  VCC 
 R1  R2 
 12.7 
 12
  1.905 V
 67.3  12.7 
TUTORIAL 3 – Questions & Answers
Q1 – Solution
(a) (Cont’d)
VBB1  VBE1
I B1 
RB1  1  1 RE1
1.905  0.7

10.68k  1  1202k
 4.77 μA
I C1  1 I B1
 120  4.77μ
 0.572 mA
TUTORIAL 3 – Questions & Answers
Q1 – Solution
(a) (Cont’d)
I C1 0.572m
g m1 

 22 mA/V
VT
26m
r 1 

1
g m1
1VT
I C1
120  26m

0.572m
r 1  5.45 k
VA

ro1 


I C1 0.572m
TUTORIAL 3 – Questions & Answers
Q1 – Solution
(a) (Cont’d)
DC analysis : Stage 2
RB 2  R3 R4  15 45  11.25 k
VBB 2
 R4 

 VCC 
 R3  R4 
 45 
 12
9V
 15  45 
TUTORIAL 3 – Questions & Answers
Q1 – Solution
(a) (Cont’d)
I B2
VBB 2  VBE 2

RB 2  1   2 RE 2
9  0.7

11.25k  1  1201.6k
 40.5 μA
IC 2  2 I B2
 120  40.5μ
 4.86 mA
TUTORIAL 3 – Questions & Answers
Q1 – Solution
(a) (Cont’d)
gm2
I C 2 4.86m


 0.187 A/V
VT
26m
r 2 

2
gm2
 2VT
IC 2
120  26m

4.86m
r 2  642 
VA

ro 2 


I C 2 4.86m
TUTORIAL 3 – Questions & Answers
Q1 – Solution
(b) The overall small-signal voltage gain, Av = vo/vs.
First, consider the 2nd stage;
 V 2


Vo  
 g m 2V 2 RE 2 // RL 
 r 2

 1



 g m 2 RE 2 // RL V 2
 r 2

V 2
 1   2 RE 2 // RL 
r 2
TUTORIAL 3 – Questions & Answers
Q1 – Solution
(b) (Cont’d)
and;
Vo '  V 2  Vo
 1

 V 2  
 g m 2 RE RL V 2
 r 2

V 2
 r 2  1   2 RE RL 
r 2
TUTORIAL 3 – Questions & Answers
Q1 – Solution
(b) (Cont’d)
Vo
Av 2 
Vo '
V 2
1   2 RE 2 RL 
r 2

r 2  1   2 RE 2 RL Vr 2
2
TUTORIAL 3 – Questions & Answers
Q1 – Solution
(b) (Cont’d)
1   2 RE 2 RL 
Av 2 
r 2  1   2 RE 2 RL 
1  1201.6 0.25

0.642  1  1201.6 0.25
 0.976
TUTORIAL 3 – Questions & Answers
Q1 – Solution
(b) (Cont’d)
To find Av1, first
find Rib2 and Ri2;
TUTORIAL 3 – Questions & Answers
Q1 – Solution
(b) (Cont’d)
Rib 2  r 2  1   2 RE 2 RL 
 0.642  1  120 1.6 0.25
 26.8 k
TUTORIAL 3 – Questions & Answers
Q1 – Solution
(b) (Cont’d)
Ri 2  R3 R4 Rib2  15 45 26.8  7.9 k
Next , replace the 2nd stage with its equivalent input impedance, Ri2;
TUTORIAL 3 – Questions & Answers
Q1 – Solution
(b) (Cont’d)
Vo '   gm1V 1 RC1 Ri 2 
Vs  V 1
Vo '
Av1 
  g m1 RC1 Ri 2   22m10 7.9 k  97.1
Vs
Av  Av1 Av 2  97.1 0.976  94.8
TUTORIAL 3 – Questions & Answers
Q1 – Solution
(c) The input and the output resistances
TUTORIAL 3 – Questions & Answers
Q1 – Solution
(c) (Cont’d)
Ris  RB1 r 1   R1 R2 r 1   67.3 12.7 5.45  3.61 k
TUTORIAL 3 – Questions & Answers
Q1 – Solution
(c) (Cont’d)
The small-signal equivalent circuit of the complete amplifier;
TUTORIAL 3 – Questions & Answers
Q1 – Solution
(c) (Cont’d)
The equivalent circuit may be simplified as follows;
TUTORIAL 3 – Questions & Answers
Q1 – Solution
(c) (Cont’d)
To find Ro, set Vs to zero and apply Vx at the output port;
TUTORIAL 3 – Questions & Answers
Q1 – Solution
(c) (Cont’d)
Since Vs is set to zero, RB1 and r1 are shorted, V1 and gmV1 are both
equal to zero. The equivalent circuit becomes;
I e 2  I x  I b 2  g mV 2
TUTORIAL 3 – Questions & Answers
Q1 – Solution
(c) (Cont’d)
 1

Vx
V 2
 I x  
 g m V 2  I x  1   2 
RE 2
r 2
 r 2

 r 2 

V 2  Vx 
 r 2  Ro1 
Vx
Vx
 I x  1   2 
RE 2
r 2  Ro1
TUTORIAL 3 – Questions & Answers
Q1 – Solution
(c) (Cont’d)
 1
 r 2  Ro1  1   2 RE 2 

1  2  
Ix  

Vx  
Vx
r 2  Ro1 RE 2 
 RE 2 r 2  Ro1 

TUTORIAL 3 – Questions & Answers
Q1 – Solution
(c) (Cont’d)
I x  r 2  Ro1  1   2 RE 2 


r 2  Ro1 RE 2 
Vx 


Vx 
r 2  Ro1 RE 2


I x  r 2  Ro1  1   2 RE 2 
TUTORIAL 3 – Questions & Answers
Q1 – Solution
(c) (Cont’d)
 r 2  Ro1 

RE 2 


Vx
r 2  Ro1 

1  2 
  RE 2 ||
Rx 

1   2 
I x  r 2  Ro1  R 
E2
 1   2 

TUTORIAL 3 – Questions & Answers
Q1 – Solution
(c) (Cont’d)
Substituting values;
Ro1  RC1 R3 R4  10 15 45  5.3 k

0.642  5.3
Rx  1.6 ||
 1.6 || 0.049  48 
1  120
TUTORIAL 3 – Questions & Answers
Q2 : For the circuit in Figure 2, assume 1 = 120, 2 = 150, VA =  and
VBE (ON) = 0.7 V. Find:
(a) The small-signal parameters, gm and r for both transistors.
(b) The overall small-signal voltage gain Av = vo/vs.
TUTORIAL 3 – Questions & Answers
Q2 (cont’d)
Figure 2
TUTORIAL 3 – Questions & Answers
Q2 – Solution
(a) The small-signal parameters, gm and r for
both transistors.
VBB
 R2 
(VCC  VEE )  VEE
 
 R1  R2 
 45 
5  (5)  5


 80  45 
 1.4 V
RBB
R1 R2
80(45)


 28.8 K
R1  R2 80  45
TUTORIAL 3 – Questions & Answers
Q2 – Solution
(a) (Cont’d)
To find IB1, perform KVL around the baseemitter loop of Q1;
VBB  RBB I B1  VBE1  RE1 I E1  V 
Rearrange the equation,
RBB I B1  1  1I B1 RE1  VBB  V   VBE1
TUTORIAL 3 – Questions & Answers
Q2 – Solution
(a) (Cont’d)
Substituting values;
28.8 103 I B1  120  12 103 I B1
 1.4  5  0.7
Hence;
I B1  10.7A
and;
I C1  1I B1  1.285mA
TUTORIAL 3 – Questions & Answers
Q2 – Solution
(a) (Cont’d)
To find IB2;
The voltage drop across the resistor RC1
equals the voltage drop across RE2 plus
the voltage VEB2.
RE 2 I E 2  VEB2  RC1I1
where;
I1  I C1  I B 2
I E 2  2  1I B 2
TUTORIAL 3 – Questions & Answers
Q2 – Solution
(a) (Cont’d)
 2  1RE 2 I B 2  VEB2
 RC1  I C1  I B 2 
Susbtituting values;
150  13 103 I B 2  0.7

 4 103 1.285 103  I B 2

TUTORIAL 3 – Questions & Answers
Q2 – Solution
(a) (Cont’d)
I B 2  9.716 μA
I C 2   2 I B 2  1.457 mA
Thus, finding the transistor
parameters;
g m1
I C1 1.285


 49.42 mA/V
VT
26
r 1 
1VT
I C1
120(26)

 2428
1.285
TUTORIAL 3 – Questions & Answers
Q2 – Solution
(a) (Cont’d)
g m2
I C 2 1.457


 56.0 mA/V
VT
26
r 2 
 2VT
IC2

150(26)
 2676.7
1.457
TUTORIAL 3 – Questions & Answers
Q2 – Solution
(b) The overall small-signal voltage gain Av = vo/vs.
A  A1 A2   g m1 RC1 // r 2  g m 2 RC 2 // RL 
A  g m1 g m 2 RC1 // r 2 RC 2 // RL 
 49.4(56)( 4 2.676)(1.5 4.5)
 4990 V/V
TUTORIAL 3 – Questions & Answers
Q3 : Figure 3 shows an amplifier in cascode configuration where a commonbase stage (Q2) is being driven by a common-emitter stage (Q1).
(a) Draw the circuit under dc condition and determine the quiescent
currents (ICQ1 and ICQ2) 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 is 30 mV,
determine the approximate peak-to-peak value of the output
voltage .
Assume; 1  2    150;
VBE1active  VBE 2active  VBE active  0.7 V;
VA1  VA2  VA  ; VT  26 mV and VCC  15 V
TUTORIAL 3 – Questions & Answers
Q3 (cont’d)
Figure 3
TUTORIAL 3 – Questions & Answers
Q3 – Solution
TUTORIAL 3 – Questions & Answers
Q4 : Determine the input impedance Ri and output impedance Ro of the twostage 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.
TUTORIAL 3 – Questions & Answers
Q4 (Cont’d)
Figure 4
TUTORIAL 3 – Questions & Answers
Q4 – Solution
VBB  VCC
 R2 
  VEE
 VEE 
 R1  R2 
 47 
 5  5
  5  1.8 V
 100  47 
R1 R2
RB 
R1  R2
100  47

 32 k
100  47
TUTORIAL 3 – Questions & Answers
Q4 – Solution
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
I B1
2.6

 9 μA
3
288  10
TUTORIAL 3 – Questions & Answers
Q4 – Solution
I C1  1 I B1  127  9  10 6  1.15 mA
For the RC1 - collector of Q1 - base of Q2 - loop;
RE 2 I E 2  VEB2  RC1I1
I 1  I C1  I B 2
I E 2   2  1I B 2
 2  1RE 2 I B 2  VEB2  RC1  I C1  I B 2 
TUTORIAL 3 – Questions & Answers
Q4 – Solution
Substituting values;
200  12 103 I B 2  0.7

 5 103 1.15 103  I B 2
I B2

5.05

 12.4 μA
3
407  10
IC 2  2 I B2
 200 12.4 10 6  2.48 mA
TUTORIAL 3 – Questions & Answers
Q4 – Solution
g m1
I C1 1.15


 44.2 mA/V
VT
26
r 1 
g m2
1
g m1

127
 2.87 k
0.0442
I C 2 2.48


 95.4 mA/V
VT
26
r 2 
2
g m2
200

 2.1 k
0.0954
TUTORIAL 3 – Questions & Answers
Q4 – Solution
Rib  r 1  1  1 RE1  2.87  1  1272  258.87 k
TUTORIAL 3 – Questions & Answers
Q4 – Solution
Ri  RB Rib  32 258.87  28.5 k
TUTORIAL 3 – Questions & Answers
Q4 – Solution
Ro  RC 2  1.5 k