This document contains review questions and practice problems related to transistor (BJT) circuit analysis. The review questions cover topics like current gains, transistor biasing, I-V characteristics, and transistor operation. The practice problems involve calculating operating points, currents, and voltages in common emitter amplifier circuits using transistor parameters like current gain β and bias voltages. Solving the problems requires applying concepts like Kirchhoff's laws and transistor modeling.

1. Module 2
Unit 2
TRANSISTOR (BJT)
Review Questions
1 Define current gains α and β. How are they related?
2 In transistor design, the base has most critical features as
compared to emitter and collector. Discuss.
3 Why do we bias a transistor? What are the considerations in
choosing an appropriate biasing scheme?
4 Draw I - V characteristics (Collector characteristics) for a
transistor and discuss the salient features.
5 Explain ‘base width modulation’ and its influence on
transistor characteristics.
6 What is collector feedback bias? How does this biasing
provide stability to the circuit?
7 Give reasons for the wide use of ‘voltage divider bias’ in BJT
amplifiers.
8 Discuss the flow of three currents IE, IB and IC in a forward
biased emitter junction and reverse biased collectior
junction.

2. Problems
2.1 A transistor has current gain of 0.99 when used in common base (CB) configuration.
How much will be the current gain of this transistor in common emitter (CE)
configuration ?
Solution :- The current gain in common base circuit is written as α, and it has been
given equal to 0.99.
α and current gain in common emitter configuration β are related as





1
Therefore,
99
.
0
1
99
.
0


 = 99
or β = 99

3. 2.2 Determine the minimum value of current gain β required to put the transistor in
saturation when
Vin = +5V.
Assume, VBE(sat) = 0.8 V,
VCE(sat) = 0.12 V
+12V
V
Vin
80k
k
IB
VBE
VCE
IC
5k
+
-

4. Solution:- Let IB and IC be base current and collector current respectively (see fig.
above).
Taking +5V as input voltage Vin, the summation of voltages (Kirchhoff’s voltage law,
KVL) in the base-emitter loop gives
8 X 103
X IB + VBE – 5V = 0
We assume that the transistor is in saturation, so that
VBE = VBE(Sat) = 0.8 V
Therefore ,
80 X 103
X IB + 0.8 V – 5V = 0
or , mA
V
IB 0525
.
0
10
80
2
.
4
3



Summation of voltages in the collector loop gives,
5 X 103
X IC + VCE = 12 V
or mA
IC 36
.
2
10
5
2
.
0
12
3




And the condition for the transistor in saturation is that the minimum value of gain β ,
,
45
525
.
0
36
.
2
min 


mA
mA
I
I
B
C

or βmin = 45.

5. 2.3 The fixed bias circuit shown in figure uses a silicon transistor with VBE = 0.7V.
(a) Find the collector current, IC, and voltage VCE, if β of transistor is 60.
(b) Find IC and VCE if β changes to 80.
What conclusions may be drawn?
+VCC (9V)
60k
k
VBE
VCE
0.5k
+
-

6. Solution:-
(a) Summation of voltages in base-emitter loop results in
VCC = RBIB + VBE
or mA
R
V
V
I
B
BE
CC
B 138
.
0
10
60
7
.
0
9
3






and,
IC = βIB
= 60 X 0.138 mA
IC = 8.28 mA
Summation of voltages in the collector circuit gives,
RCIC +VCE = VCC
Or VCE = VCC – ICRC
= 9 - 8.28 X 10-3
X 0.5 X 103
= 9 – 4.14
Or VCE = 4.86 V
Thus the operating point values are
IC = 8.28 mA , and
VCE = 4.86 V
(b) Proceeding in the same way, and if the current gain β equals 80, then
Since IB = 0.138 mA, therefore
IC = βIB = 80 X 0.138 mA
or IC = 12.42 mA

7. And VCE = VCC- ICRC
= 9 – 12.42 X 10-3
X 0.5 X 103
= 9 – 6.21
VCE = 2.79 V
With β = 80, the operating point values are :
IC = 12,42 mA, and
VCE = 2.79 V
CONCLUSIONS:-
Changes in collector current IC when β changes from 60 to 80 are from 8.28 mA to
12.42 mA.
And VCE changes from 4.86V to 2.79V.
These changes are drastic and operation may shift to regions to give distorted output.

8. 2.4 Find out the operating point current ICQ, and voltage VCEQ in the circuit shown.
(VBE = 0.7 V, β of transistor is 200).
+9V(VCC)
50k
k
IC
RE
VCE
RC
2.5k
+
4.3k
-9V(VEE)
-

9. Solution:-
Considering the potential difference across the resistor RE, the emitter current IE (≈IC)
may be written as,
3
10
3
.
4
7
.
0
9






E
BE
EE
C
E
R
V
V
I
I
or IC = ICQ = 1.93 mA
Voltage summation in the collector circuit leads to,
VCC = RCIC + VCE
or VCE = VCC – RCIC
= 9 – 2.5 X 103
X 1.93 X 10-3
= 9 – 4.83
or VCE = VCEQ = 4.17V

10. 2.5 Calculate the approximate value for the base resistor RB which will forward bias the
emitter junction of silicon transistor (β = 100) in the circuit. Collectior – emitter voltage
VCE of 2.5 V reverse biases the collector.
(VBE = 0.7V)
+5V
RB
VBE
VCE
1.5k
+
IC
IE
1k
-

11. Solution:-
We assume that the emitter junction is in forward bias and VBE = 0.7V.
Summation of voltages in the base-emitter circuit results in,
RBIB +VBE + REIE = VCC
Since IC = IE,
and


 E
C
B
B
I
I
I
or
I 

Then above equation yields,
)
(
3
.
4
7
.
0
0
.
5
.
A
R
R
I
or
V
V
R
R
I
or
V
I
R
V
I
R
E
B
E
BE
CC
E
B
E
CC
E
E
BE
E
B









































Now, summation of voltages in the collector – emitter circuit gives,
RCIC + VCE + REIE = VCC
Because IC = IE
And substituting for other parameters
IE(RC + RE) = VCC – VCE = 5.0 – 2.5 = 2.5
   
mA
R
R
I
or
E
C
E 1
10
1
5
.
1
5
.
2
5
.
2
3






Substituting IE = 1mA in eqn
(A)
3
.
4
10
1
100
10
1 3
3









  B
R
RB = 330 X 103Ω
or RB = 330 kΩ

12. 2.6 The operating point values of current IC(=ICQ) and voltage VCE(=VCEQ) in the circuit
have magnitudes of 0.9 mA and 3.72 V respectively when the current gain β for the
transistor is 100. The transistor in the circuit is replaced by another one with β = 200.
Calculate the new values of ICQ and VCEQ. What do you infer?
2k
R2
+6V (VCC)
R1
0.7V
VCE
1k
+
IC
RC
1.3k
4k
-

13. Solution:-
Thevenized voltage Vth across R2 is
V
V
V CC
TH
2
6
6
2
10
4
10
2
10
2
3
3
3









That is VTH = 2V
And, as we have derived the relation earlier,

TH
E
BE
TH
CQ
E
R
R
V
V
I
I




Where Thevenized resistance RTH is
RTH =R1 R2 = 4k 2k = 1.33 kΩ,
Then with β= 200,
mA
I
or
I
I
I
CQ
cQ
C
E
995
.
0
10
006
.
0
10
3
.
1
7
.
0
2
3
3








Further, summation of voltage in the collector circuit under the condition
IC = IE leads to,
VCC = IC (RC + RE) + VCE
Or, VCE = VCEQ = VCC-IC (RC + RE)
= 6V – 0.995 mA (1k + 1.3k)
Or VCEQ = 3.71V

14. Inference:
The circuit is highly stable as the change in collector current is 0.995 – 0.99 = 0.005 mA
only (change is ≈0.5%) which is negligible. Similarly VCE changes only 0.3% which is
also negligible when β changes by 100% i.e from 100 to 200.

15. (2.7) Calculate the value of resistor RB in the circuit shown to put VCE at 3.0V
(VBE = 0.7V and β = 80)
IB RC
RE
+
+9V
RB
VBE
2 k
IC
IE
3k

16. Solution:-
Summation of voltages in the base-emitter circuit gives,
RBIB + VBE + REIE = VCC
Now, VBE = 0.7 V, and IC = IE.
Also,
80
C
C
B
I
I
I 


Substituting these in above equation, and taking resistances in kΩ,
)
(
3
.
8
.
3
80
.
3
.
8
7
.
0
9
3
80
.
A
I
k
I
R
or
V
V
I
k
I
R
C
C
B
BE
CC
C
C
B





















Summation of voltages in the collector circuit results in,
ICRC + VCE + IE X 3k = 9V
or IC X 2k + 3.0V + IC X 3k = 9V
or IC (2k +3k) = 9 - 3 = 6V
mA
k
V
I
or C 2
.
1
5
6


Substituting the value of IC in eqn (A)
k
R
k
mA
R
or
V
mA
k
R
B
B
B
313
3
.
313
015
.
0
7
.
4
3
.
8
2
.
1
3
80
2
.
1
.





