Copyright 20 13 Dorling Kindersley (India) Pvt. Ltd

example: 13.10-6
A single Opamp, a resistor and a capacitor can form a good differentiator circuit. This circuit is shown 
in Fig. 13.10-16. (i) Derive the transfer function of the circuit and show that it is a differentiator. 
(ii) An ideal Opamp is too good to be true. A practical Opamp suffers from many non-idealities. 
The particular non-ideality that compromises the circuit performance will vary depending on circuit 
function. For example, we found that integrator circuit is severely compromised by offsets in a 
practical Opamp. We will see in this example that it is the limited gain and bandwidth of Opamp that 
compromises the performance of differentiator circuit.
An amplifier contains many capacitors – intentional as well as parasitic – in it and hence represents a
high-order dynamic circuit. However, some Opamps like IC 
741 can be modelled approximately as a single-time constant 
amplifier. That is, its gain function is of the form 
A
s
1
+
t
with 
A
≈
250,000 and 
t
≈
4 ms. Obtain the transfer function of 
differentiator circuit with 
R
= 
10k
W
and 
C
= 
1
m
F using IC 741 
and find its impulse response.
(iii) Suggest a method to modify the oscillatory impulse 
response to critically damped impulse response.
Fig. 13.10-16 
An Opamp 
differentiator 
circuit
C
+
+
–
–
v
S
(
t
)
+
R
v
o
(
t
)
–

---

13.42
Analysis of Dynamic Circuits by Laplace Transforms
Solution
(i) This is essentially an inverting amplifier structure. The transfer function 
= 
−
= −
Impedance in feedback path
Impedance in input line
sRC
. It is a differentiator since multiplication by 
s
in 
s
-domain 
is equivalent to differentiation in time-domain according to time-differentiation theorem on Laplace 
transforms. It is an inverting differentiator.
(ii) The Opamp is to be modelled as a dependent 
source that senses the voltage transform between 
non-inverting pin and inverting pin and produces a 
voltage transform 
AV s
s
d
( )
1
+
t
at its output with respect 
to ground, where 
V
d
(
s
) is the transform of voltage 
of non-inverting pin with respect to inverting pin. 
The 
s
-domain equivalent circuit incorporating this 
model for Opamp is shown in Fig. 13.10-17.
Let the node voltage transform at the inverting pin be 
V
1
(
s
). Writing the node equation at inverting 
pin, we get
sC V s
V s
R
V s
AV s
s
[ ( )
( )]
( )
( )
1
1
1
1
1
0
−
+
− −
+









 =
t
Simplifying this equation results in 
V s
V s
sC
sC
R
A
s
1
1
1
1
( )
( )
=
+
+
+




t
V
o
(
s
) is 
−
+
A
s
V s
1
1
t
( )
.
Therefore, 
V s
V s
AsC
s
sC
R
A
s
sRC
A
RC
s
s
A
R
o
( )
( )
(
)
=
−
+
+
+
+




= −
+
+ +
1
1
1
1
1
1
2
t
t
t
t
C
C
t
The DC gain 
A
of any practical Opamp is in thousands and hence 
A
+ 
1 
≈
A
. Therefore,
V s
V s
sRC
A
RC
s
s
A
RC
o
( )
( )
(
)
= −
+
+
t
t
t
2
1
Note that the order of the circuit is two. The Opamp contributes one extra order to the circuit. 
Substituting 
R
= 
10k
W
, 
C
= 
1
m
F, 
A
= 
250000 and 
t
= 
4ms, we get
V s
V s
s
s
s
s
s
o
( )
( )
(
.
)
(
.
)
(
= −
+
+
= −
+
0 01
79060
250
79060
0 01
79060
12
2
2
2
2
55
79057
62 5 10
125
79057
2
2
6
2
2
)
(
.
)
(
)
+
= −
×
+
+
s
s
Fig. 13.10-17 
Transformed equivalent 
circuit for differentiator 
circuit in Fig. 13.10-16 
C
–
–
–
+
+
+
+
R
V
1
(
s
)
V
0
(
s
)
V
d
(
s
)
–
V
(
s
) 1
sC
AV
d
(
s
)
1 + s
τ

---

Network Functions and Pole-Zero Plots 
13.43
Step response is the inverse of transfer function multiplied by 1/
s
. 
(
.
)
(
)
(
.
)
(
)
−
×
+
+
=
−
×
+
+
62 5 10
1
125
79057
62 5 10
79057
79057
125
6
2
2
6
2
s
s
779057
2
∴
≈ −
×
−
v t
e
t u t
o
t
( ) (
.
)
cos
( )
62 5 10
79057
6
125
V
Of course, if a unit step is really applied to this circuit, the output of Opamp will saturate. But what 
is to be noted is that the response is 
highly under-damped
. The oscillation is at 12.6 kHz and oscillation 
period is about 0.08 ms. But the time constant of damping exponential is 40 ms. It takes about 5 
time constants for an exponential transient to settle down. That implies that the 12.6 kHz transient 
oscillations will last for about 200 ms before they die down. That is a bad transient performance.
(iii) The solution is to add a little damping by means of a resistor in series with the input capacitor. 
This will make the differentiator imperfect however.