Equation (11.2) implies that the two input voltages are also equal. This condi-
tion is termed the concept of the virtual short circuit. In addition, because of
the large input resistance of the op amp, the latter is assumed to take no cur-
rent for most calculations.
11.2 INVERTING CONFIGURATION
An op amp circuit connected in an inverted closed loop configuration is shown
in Figure 11.4.
I
1
I
2
Z
1
Z
2
V
o
V
in
Z
in
V
a
A
Figure 11.4 Inverting Configuration of an Op Amp
Using nodal analysis at node A, we have
VV
Z
VV
Z
I
ain aO
−
+
−
+=
12
1
0
(11.3)
From the concept of a virtual short circuit,
VV
ab
==
0
(11.4)
and because of the large input resistance,
I
1
= 0. Thus, Equation (11.3) sim-
plifies to
V
V
Z
Z
O
IN
=−
2
1
(11.5)
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© 1999 CRC Press LLC
The minus sign implies that
V
IN
and
V
0
are out of phase by 180
o
. The input
impedance,
Z
IN
,
is given as
Z
V
I
Z
IN
IN
==
1
1
(11.6)
If
ZR
11
=
and
ZR
22
=
,
we have an inverting amplifier shown in Figure
11.5.
V
o
V
in
R
2
R
1
Figure 11.5 Inverting Amplifier
The closed-loop gain of the amplifier is
V
V
R
R
O
IN
=−
2
1
(11.7)
and the input resistance is
R
1
. Normally,
R
2
>
R
1
such that
VV
IN
0
>
.
With the assumptions of very large open-loop gain and high input resistance,
the closed-loop gain of the inverting amplifier depends on the external com-
ponents
R
1
,
R
2
, and is independent of the open-loop gain.
For Figure 11.4, if
ZR
11
=
and
Z
jwC
2
1
=
,
we obtain an integrator
circuit shown in Figure 11.6. The closed-loop gain of the integrator is
V
V jwCR
O
IN
=−
1
1
(11.8)
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V
o
V
in
C
R
1
I
C
I
R
Figure 11.6 Op Amp Inverting Integrator
In the time domain
V
R
I
IN
R
1
=
and
IC
dV
dt
C
O
=−
(11.9)
Since
II
RC
=
() () ()
Vt
RC
Vtd V
OIN
t
O
=− +
∫
1
0
1
0
τ
(11.10)
The above circuit is termed the Miller integrator. The integrating time con-
stant is
CR
1
.
It behaves as a lowpass filter, passing low frequencies and at-
tenuating high frequencies. However, at dc the capacitor becomes open cir-
cuited and there is no longer a negative feedback from the output to the input.
The output voltage then saturates. To provide finite closed-loop gain at dc, a
resistance
R
2
is connected in parallel with the capacitor. The circuit is shown
in Figure 11.7. The resistance
R
2
is chosen such that
R
2
is greater than
R
.
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V
o
V
in
C
R
1
R
2
Figure 11.7 Miller Integrator with Finite Closed Loop Gain at DC
For Figure 11.4, if
Z
jwC
1
1
=
and
ZR
2
=
,
we obtain a differentiator cir-
cuit shown in Figure 11.8. From Equation (11.5), the closed-loop gain of the
differentiator is
V
V
jwCR
O
IN
=−
(11.11)
V
o
V
in
C
R
1
I
R
I
C
Figure 11.8 Op Amp Differentiator Circuit
In the time domain
IC
dV
dt
C
IN
=
, and
()
Vt IR
OR
=−
(11.12)
Since
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() ()
It It
CR
=
we have
()
()
Vt CR
dV t
dt
O
IN
=−
(11.13)
Differentiator circuits will differentiate input signals. This implies that if an
input signal is rapidly changing, the output of the differentiator circuit will ap-
pear “ spike-like.”
The inverting configuration can be modified to produce a weighted summer.
This circuit is shown in Figure 11.9.
R
1
R
2
R
F
R
n
I
n
I
F
V
1
V
2
V
n
I
1
I
2
V
o
Figure 11.9 Weighted Summer Circuit
From Figure 11.9
I
V
R
I
V
R
I
V
R
n
n
n
1
1
1
2
2
2
== =
, , ,
(11.14)
also
III I
FN
=++
12
(11.15)
VIR
OFF
=−
(11.16)
Substituting Equations (11.14) and (11.15) into Equation (11.16) we have
© 1999 CRC Press LLC
© 1999 CRC Press LLC
V
R
R
V
R
R
V
R
R
V
O
FF F
N
N
=− + +
1
1
2
2
(11.17)
The frequency response of Miller integrator, with finite closed-loop gain at dc,
is obtained in the following example.
Example 11.1
For Figure 11.7, (a ) Derive the expression for the transfer function
V
V
jw
o
in
()
.
(b) If
C
= 1 nF and
R
1
= 2KΩ, plot the magnitude response for
R
2
equal to
(i) 100 KΩ, (ii) 300KΩ, and (iii) 500KΩ.
Solution
ZR
sC
R
sC R
22
2
2
22
1
1
==
+
(11.18)
ZR
11
=
(11.19)
V
V
s
R
R
sC R
o
in
()
=
−
+
2
1
22
1
(11.20)
V
V
s
CR
s
CR
o
in
()
=
−
+
1
1
21
22
(11.21)
MATLAB Script
% Frequency response of lowpass circuit
c = 1e-9; r1 = 2e3;
r2 = [100e3, 300e3, 500e3];
n1 = -1/(c*r1); d1 = 1/(c*r2(1));
num1 = [n1]; den1 = [1 d1];
w = logspace(-2,6);
h1 = freqs(num1,den1,w);
f = w/(2*pi);
© 1999 CRC Press LLC
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d2 = 1/(c*r2(2)); den2 = [1 d2];
h2 = freqs(num1, den2, w);
d3 = 1/(c*r2(3)); den3 = [1 d3];
h3 = freqs(num1,den3,w);
semilogx(f,abs(h1),'w',f,abs(h2),'w',f,abs(h3),'w')
xlabel('Frequency, Hz')
ylabel('Gain')
axis([1.0e-2,1.0e6,0,260])
text(5.0e-2,35,'R2 = 100 Kilohms')
text(5.0e-2,135,'R2 = 300 Kilohms')
text(5.0e-2,235,'R2 = 500 Kilohms')
title('Integrator Response')
Figure 11.10 shows the frequency response of Figure 11.7.
Figure 11.10 Frequency Response of Miller Integrator with Finite
Closed-Loop Gain at DC
© 1999 CRC Press LLC
© 1999 CRC Press LLC
11.3 NON-INVERTING CONFIGURATION
An op amp connected in a non-inverting configuration is shown in Figure
11.11.
Z
2
Z
1
I
1
V
o
V
a
V
in
Z
in
A
Figure 11.11 Non-Inverting Configuration
Using nodal analysis at node A
V
Z
VV
Z
I
aaO
12
1
0
+
−
+=
(11.22)
From the concept of a virtual short circuit,
VV
IN a
=
(11.23)
and because of the large input resistance (
i
1
= 0 ), Equation (11.22) simplifies
to
V
V
Z
Z
O
IN
=+
1
2
1
(11.24)
The gain of the inverting amplifier is positive. The input impedance of the
amplifier
Z
IN
approaches infinity, since the current that flows into the posi-
tive input of the op-amp is almost zero.
© 1999 CRC Press LLC
© 1999 CRC Press LLC
If Z
1
=
R
1
and Z
2
=
R
2
, Figure 11.10 becomes a voltage follower with gain.
This is shown in Figure 11.11.
V
o
V
in
R
2
R
1
Figure 11.12 Voltage Follower with Gain
The voltage gain is
V
V
R
R
O
IN
=+
1
2
1
(11.25)
The zero, poles and the frequency response of a non-inverting configuration
are obtained in Example 11.2.
Example 11.2
For the Figure 11.13 (a) Derive the transfer function. (b) Use MATLAB to
find the poles and zeros. ( c ) Plot the magnitude and phase response, assume
that
C
1
= 0.1uF,
C
2
= 1000 0.1uF,
R
1
= 10KΩ, and
R
2
= 10 Ω.
V
o
V
in
R
2
R
1
V
1
C
1
C
2
Figure 11.13 Non-inverting Configuration
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Solution
Using voltage division
V
V
s
sC
RsC
IN
11
11
1
1
()
=
+
(11.26)
From Equation (11.24)
V
V
s
R
sC
O
1
2
2
1
1
()
=+
(11.27)
Using Equations (11.26 ) and (11.27), we have
V
V
s
sC R
sC R
O
IN
()
=
+
+
1
1
22
11
(11.28)
The above equation can be rewritten as
()
V
V
s
CR s
CR
CR s
CR
O
IN
=
+
+
22
22
11
11
1
1
(11.29)
The MATLAB program that can be used to find the poles, zero and plot the
frequency response is as follows:
diary ex11_2.dat
% Poles and zeros, frequency response of Figure 11.13
%
%
c1 = 1e-7; c2 = 1e-3; r1 = 10e3; r2 = 10;
% poles and zeros
b1 = c2*r2;
a1 = c1*r1;
num = [b1 1];
den = [a1 1];
disp('the zero is')
z = roots(num)
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© 1999 CRC Press LLC
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