Module – 3
Unit – 3
Small Signal BJT Amplifiers
Review Questions:
1. Define h – parameters for a transistor. Why are these called
hybrid parameters? What are their units?
2. Out of four h – parameters, two are most important. Which
are these? And why the other two have less significance?
3. hi and Zi both represent input impedance in h - and Z
systems of parameters but they are most equal. Why?
4. What are r – parameters and how are they superior to h –
parameters?
5. Common – base (CB) amplifier has limited applications.
Why?
6. Among BJT amplifiers, common-emitter amplifier is most
favoured. Give reasons.
7. What is an emitter follower? Discuss its main applications.
8. What are, in general, performance parameters of an
amplifier?
Problems:
3.1 Calculate dynamic emitter resistance r’e (also called ac emitter resistance) of the
transistor in the circuit of fig.(VBE = 0.7V)
+5V(+ VCC)
2k
IC
+
VCE
VBE
12k
k
IE
4.3k
-5V(- VEE)
Solution:The dynamic emitter resistance, re' , is expressed as,
r 'e
25mv
IE
Where IE is dc emitter current.
The dc emitter current in the circuit (i.e, voltage drop across resistor RE, divided by the
resistance) is
IE
VEE VBE
RE
5 0.7
4.3 10 3
Therefore,
r 'e
25 mV
1 mA
That is,
re' = 25 Ω
25
1 mA
3.2
How much is the voltage gain of the amplifier (fig) if the dynamic emitter
resistance, r’e, is 25 Ω. The current gain of the transistor is 80.
How much is input impedance of the amplifier.
The coupling and bypass capacitors may be assumed of negligible impedance at the
signal frequency. Take VBE = 0.7V
+9V
RC
12k
vi
C1
5k
C2
v0
A
B
RL
6k
k
2.3k
C3
5k
Solution:The voltage gain AV, for a common emitter amplifier with resistor RE by- passed is
AV
rc
r 'e
Where rc is effective (ac) impedance seen by the collector and it is
rc = Rc RL = 5k 5k = 2.5 kΩ
And re' = 25Ω (given)
Therefore,
AV
2.5 10 3
25
100
First we find out impedance between base (point A in Fig 3.2) and ground, Z i(base) and
we know, it is given by
Zi(base) = β re'
= 80 x 25
or Zi(base) = 2 kΩ
And, input impedance of the amplifier Zi(amp) is,
Zi(amp) = RB
Zi(base)
Where RB is effective biasing resistance, and
RB = R1 R2 = 12k
Therefore,
Zi(amp) = 4k 2k
or Zi(amp) = 1.33 kΩ
6k = 4 kΩ
3.3 Calculate the value of resistor RC so that the voltage gain of the amplifier in fig is
100. Capacitors C1, C2, and C3 may be assumed short at signal frequency.
+12V (+ VCC)
RC
vi
C2
vo
C1
10k
k
RE
5k
-12V
(- VEE)
RL
C3
4k
Solution:In case, RE is by-passed, the voltage gain Av of the amplifier is,
AV
rc
r 'e
Where rc is the effective (ac) resistance seen by the collector of the transistor, and re' is
the dynamic resistance of the emitter.
We know,
r 'e
25mv
IE
Now,
IE
VEE
VBE
12.0 0.7
5 10 3
RE
2.26 mA
Therefore,
r 'e
25 mV
2.26 mA
11
0.011 k
And,
Rc = RC RL
4 RC
RC 4
Where resistances have been taken in kΩ
Therefore,
AV
rc
r 'e
4 Rc
1
Rc 4 0.011
As Av = 100 (given), Rc (in kΩ) is,
100
or Rc
( Rc
4 Rc
4) (0.011)
1.51 k
3.4 Find out the smallest value of load RL in the amplifier circuit shown in fig so that
the voltage gain is at least 40. The dynamic emitter resistance of the transistor is 25 Ω.
The coupling and by-pass capacitors may be assumed short at signal frequency.
+12k
RC
3k
vo
8k
vi
RL
4k
k
4k
RE
Solution:The voltage gain of the common emitter amplifier with resistor RE by-passed (see fig.)
is expressed as
rc
r 'e
AV
40 (required gain)
Where rc is effective ac impedance seen by the collector, Which is
rc = RC
RL
RC RL
RC RL
Taking resistances in kΩ
rc
3R L
3 RL
And,
AV
or RL
3R L
3 RL
1.5 k
1
25 10
3
40
3.5 The silicon transistor in the common- base amplifier has the current gain α of 0.98.
Find the input impedance and voltage gain of the amplifier in fig. (VBE = 0.7V)
-9V (-VEE)
RE
6k
+9V (+Vcc)
RC
4k
VBE
~
vi
4k
RL
Solution:The dc voltage sources have to be grounded for ac analysis of the amplifier. Then the
input impedance of the CB amplifier (in fig.) is
Zi(amp) = RE
r’e ≈ r’e
Where r’e is dynamic emitter resiatance and r’e << RE
Also,
25mV
IE
r 'e
Where IE is dc emitter current in the circuit
VEE
IE
VBE
RE
9 0.7
6 10 3
or , I E
1.38 mA
Therefore,
r 'e
25 mV
1.38mA
18
Thus, the input impedance Zi(amp) is,
Zi(amp) ≈ r’e = 18Ω
The low value of input impedance is the main reason for limited applications of CB
amplifier.
The voltage gain of CB amplifier is expressed as,
rc
AV
r 'e
Since, rc = RC RL = 4k 4k
Then ,
AV
AV
0.98
2000
18
108.8
108.8
= 2kΩ = 2000 Ω
3.6 The transistor in the amplifier circuit shown in fig, has h – parameters, hie = 2kΩ and
hfl = 80. The values of hoe and hre are negligible. Calculate the voltage gain and input
impedance Zi(amp) of the amplifier. Capacitors C1, C2, and C3 may be assumed short at
signal frequency due to small impedances.
+15V
R1
vi
RC
6k
20k
C2
vo
C1
R2
RL
20k
k
2k
RE
C3
6k
Solution:The magnitude of voltage gain with h-parameters hoe and hre dropped, and emitter
resistance RE by-passed by capacitor C3 is
AV
h fe . Z i
hie
80 3k
2k
120
Because the ac load at collector, Zl, is
Zl = RC RL = 6k 6k = 3kΩ
Further, the impedance between base and ground , Zi(base) is,
Zi(base) = hie = 2kΩ
And, input impedance of amplifier, Zi(amp) is,
Zi(amp) = R1 R2 Zi(base)
= 20k 20k
= 10k Zi(base)
= 10k
2k
= 1.6kΩ
or Zi(amp) = 1.6kΩ
Zi(base)
3.7
For the amplifier circuit shown in fig. calculate the voltage gain and input
impedance of the amplifier when by-pass capacitor C3 is removed from the circuit.
+15V
R1
vi
RC
6k
20k
C2
vo
C1
R2
RL
20k
k
2k
RE
C3
6k
Solution:With by-pass capacitor C3 (in fig.) removed, the gain of a amplifier falls and input
impedance of the amplifier increases.
In case RE is not by-passed, the magnitude of voltage gain AV is
AV
Zl
RE
3k
2k
1.5
And,
Zi(base) = hie + (1 + hfe) RE
= 2k + (1 + 80) X 2k
Or, Zi(base) = 164 kΩ
And, input impedance of amplifier, Zi(amp) is
Zi(amp) = R1
R2
Zi(base)
= 20k
20k
10k
164k
or Zi(amp) = 9.4 kΩ
164 k
= 9.4 kΩ
3.8
The emitter follower (common collector amplifier) shown in fig. uses a transistor
with h-parameters hie = 4.5 kΩ, hfe = 120. Other parameters hoe and hre have negligible
effect on amplifier performance. Calculate voltage gain and input impedance of the
amplifier. The coupling and by-pass capacitors may be assumed short at signal
frequency.
+9V
60k
30k
~
3k
Zi (base)
Zi(amp)
RE
RL
6k
Solution:Neglecting the effect of hoe and hre on amplifier performance, the voltage gain of emitter
follower may be expressed as,
AV
h fc
Ze
hic
h fc . Z e
Where Ze is the effective load seen by the emitter, and it is
RL = 3k 6k = 2 kΩ
Ze = RE
And using hfc ≈ hfe and hic = hie,
We have,
h fe
AV
AV
hie
Ze
h fe Z e
120 2 k
4k
(120 2 k )
0.98
240
244
The input impedance as seen at the base w.r.t ground is,
Zi(base) = hfc. Ze = hfe. Ze
= 120 X 2 kΩ = 240 kΩ
The input impedance of the amplifier (that is, after taking the effect of biasing resistors),
Zi(amp) = RB
Zi(base)
And the effective base resistance RB is,
RB= R1
R2 = 60 k
30k = 20 kΩ
Therefore,
Zi(amp) = 20k 240k = 18.46 kΩ
or Zi(amp) = 18.46 kΩ