Module – 7
Unit – 7
Differential and Operational Amplifiers
Review Question:1. How is basic structure of a differential amplifier different from, for example, a
conventional RC coupled common – emitter amplifier?
2. In what different configurations can a differential amplifier be used?
3. How are the two inputs of an differential amplifier different? Explain inverting and
non-inverting nature of inputs.
4. Symmetry in construction of two halves of differential amplifier is emphasized. Give
reasons.
5. What is tail current? Efforts are made and several circuits suggested for the
constancy of tail current. Discuss.
6. The input impedance of differential amplifier is much higher (~ M Ω) than a
conventional common – emitter amplifiers. Explain.
7. Define an ideal operational amplifier.
8. Draw the approximate block diagram of an op amp giving various stages of the
amplifier.
9. An op amp is rarely used in open loop (i.e. without feedback) for linear amplifying
applications. Why?
10. Inverting input is a ‘virtual ground’ in op amp. What does it mean and what is its
significance?
11. What reasons would you assign for very wide use of op amps in analog and digital
circuits?
12. Define common mode rejection ratio(CMRR). Give its significance in device
performance.
13. Define ‘slew rate’. When does it start showing its effect on amplifier performance.
14. How does input off-set voltage in an op amp arise? And how can it be corrected?
Problems:
7.1 Estimate dc emitter current in each transistor of differential amplifier shown in fig. How
much is dc voltage from each collector to ground? How much is Vout?
+12V
RC
vi1
RC
16k
+
Vout
16k
+
VCE2
VCE1
-
-
IT
RE
24k
-12V
vi2
Solution:The tail current through 24k resistor is,
VEE
IT
VBE
RE
or , IT

12V
24 k
VEE
RE
0.5 mA
The emitter current, IE, in each transistor is,
IE
IE
1
0.5 mA
IT
2
2
0.25 mA
0.25 mA
Since IC = IE, voltage summation in the output circuit gives,
VCC = ICRC + VCE1
VCE1 = VCC – ICRC
= 12 – 0.25X10-3 X 16X103
Or, VCE1 = 8.0 V = VCE2 (due to symmetry)
Then,
Vout = VCE1 – VCE2 = 8 - 8 = 0V
7.2 Design an inverter amplifier with gain of 120 and input impedance of 5kΩ. Give the
circuit.
Solution:-
R1
vi
RF
+VCC
vo
+
-VEE
Figure shows the circuit for an inverting amplifier.
Since for an inverting amplifier, the input impedance Zi is,
Zi = R1 = 5kΩ (desired)
Therefore, R1 = 5kΩ
Further, the gain AV of inverting amplifier is,
AV
RF
R1
And, AV desired is 120, R1 = 5kΩ
Therefore,
RF = AV.R1 = 120 X 5k
or, RF = 600 kΩ
7.3 Find out the voltage gain of the non-inverting amplifier shown in fig.
+VC
C
vo
+
vi
a
1
k
R
1
-
~
VEE
99
k
R
2
Solution:The voltage gain of a non-inverting amplifier is,
AV
1
1
or AV
RF
R1
99 k
1k
100
7.4 In the amplifier circuit shown in fig., if open loop gain and open loop band width of the
op amp respectively are 105 and 10 HZ, Calculate the bandwidth of feedback amplifier (in
fig.).
+VCC
vo
+
vi
-
~
a
-VEE
99k
1k
R1
R2
Solution:If open loop band width is f2, the band width with feedback, f2(FB) is given by
f2(FB) = f2(1 + AB) ≈ f2.AB because AB>>1
Now, A = 105 (given)
And the gain of feedback network, B in the circuit shown in fig. is
B
R1
R1
RF
1k
1k 99k
1
100
Then,
f2(FB) = 10 X 105 X 10-2 HZ
or, f2(FB) = 10kHZ
10
2
7.5 For the summing amplifier shown in fig., estimate the values of resistors R1,R2 and R3
so that the output V0 is,
V0 = - (3V1 + V2 + 0.2V3)
What is the approximate value of the compensating resistor R?
v1
RF(=30k)
R1
a
v2
R2
+VCC
a
v3
R3
vo
-
a
+
-VEE
R
Solution:The output voltage, V0, for the summing amplifier is,
V0
RF
V1
R1
RF
V2
R2
RF
V3
R3
Thus for the desired output,
RF
R1
3 , or
or , R1
30k
R1
3
10 k
Similarly,
RF
R2
1 , or R2
or , R2
RF
30 k
30 k
And,
RF
R3
or R3
0.2 or , R3
RF
0.2
30 k
0.2
150 k
And,
R = R1
R2
or, R = 7.0kΩ
R3
= 10k 30k 150k
7.6 Determine the output voltage in the circuit shown in fig. If V a= 5V, Vb= -2V and Vc = 3b
RF
1k
2k
R1
+VCC
va
1k
vo
+
a
vb
a
vc
a
R1
1k
R2
1k
R3
-VEE
Solution:In the amplifier circuit shown in fig. Since the resistors R1, R2, and R3 are all equal to 1kΩ,
the voltage V1, at non-inverting input terminal will be average of the three voltages, V a, Vb,
and Vc.
Thus,
V1
Va
Vb
3
Vc
5V
2V
3
3V
2V
And the gain for non-inverting amplifier, AV, is
AV
V0
V1
1
or , V0
1
or , V0
6V
RF
R
RF
V1
R
1
2k
1k
2V
7.7 Differential gain Ad, of an op amp measures 100. In the measurement of commonmode gain experiment when 1.0V is applied common to both the inputs, output voltage
measured is 0.01V. How much is common-mode rejection ratio (CMRR)?
Solution:By definition, common mode rejection ratio (CMRR) is
CMRR (in dB)
20log10
Ad
Acm
Where Ad is gain in differential mode which is given as 100.
And, the gain in common mode, ACM is,
ACM
V0
Vi ( c m )
0.01V
1.0V
10
2
Therefore,
CMRR
100
10 2
20log10 (104 )
20log10
or, CMRR = 20 X 4 = 80 dB
CMRR = 80 dB
7.8 Figure shows a low-pass filter. Calculate the value of feedback resistor RF so that
band-pass gain is 100. Also calculate the value of resistor R to get cut-off frequency of
2kHZ.
RF
1k
+VCC
R
vo
+
vi
a
-VEE
~
0.2 µF
Solution:The gain in band-pass region is that of non-inverting amplifier and it is,
AV
1
RF
R1
AV
100, R1
100
1
or , RF
1k , then
RF
1k
99k
The cut-off frequency, f, for low-pass fitter is given by
1
2 RC
1
or , R
2 fC
or , R 398
(Pr actically R
f
1
2 3.14 2 103 0.2 10
400 )
6