Neamen D. Microelectronics: Circuit Analysis and Design
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648 Part 2 Analog Electronics
EXERCISE PROBLEM
Ex 9.6: Consider the difference amplifier in Figure 9.24(a). (a) Design the circuit
with
R
2
= R
4
,
R
1
= R
3
, and such that the differential voltage gain is
A
d
= 50
.
For input voltages in the range of
−50
mV to
+50
mV, the maximum current in
R
2
is to be limited to
50 μ
A. (b) Using the results of part (a), what is the maximum
current in
R
3
? (Ans. (a) Set
R
2
= R
4
= 100
k
, then
R
1
= R
2
= 2
k
;
(b)
0.49 μ
A)
In the ideal difference amplifier, the output
v
O
is zero when
v
I1
= v
I2
. However,
an inspection of Equation (9.50(b)) shows that this condition is not satisfied if
R
4
/R
3
= R
2
/R
1
. When
v
I1
= v
I2
, the input is called a common-mode input signal.
The common-mode input voltage is defined as
v
cm
= (v
I1
+v
I2
)/2
(9.56)
The common-mode gain is then defined as
A
cm
=
v
O
v
cm
(9.57)
Ideally, when a common-mode signal is applied,
v
O
= 0
and
A
cm
= 0
.
A nonzero common-mode gain may be generated in actual op-amp circuits. This
is discussed in Chapter 14.
A figure of merit for a difference amplifier is the common-mode rejection ratio
(CMRR), which is defined as the magnitude of the ratio of differential gain to
common-mode gain, or
CMRR =
A
d
A
cm
(9.58)
Usually, the CMRR is expressed in decibels, as follows:
CMRR(dB) = 20 log
10
A
d
A
cm
(9.59)
Ideally, the common-mode rejection ratio is infinite. In an actual differential ampli-
fier, we would like the common-mode rejection ratio to be as large as possible.
EXAMPLE 9.7
Objective: Calculate the common-mode rejection ratio of a difference amplifier.
Consider the difference amplifier shown in Figure 9.24(a). Let
R
2
/R
1
= 10
and
R
4
/R
3
= 11
. Determine CMRR(dB).
Solution: From Equation (9.50(b)), we have
v
O
= (1 +10)
11
1 + 11
v
I2
−(10)v
I1
or
v
O
= 10.0833v
I2
−10v
I1
(9.60)
The differential-mode input voltage is defined as
v
d
= v
I2
−v
I1
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Chapter 9 Ideal Operational Amplifiers and Op-Amp Circuits 649
and the common-mode input voltage is defined as
v
cm
= (v
I1
+v
I2
)/2
Combining these two equations produces
v
I1
= v
cm
−
v
d
2
(9.61(a))
and
v
I2
= v
cm
+
v
d
2
(9.61(b))
If we substitute Equations (9.61(a)) and (9.61(b)) in Equation (9.60), we obtain
v
O
= (10.0833)
v
cm
+
v
d
2
−(10)
v
cm
−
v
d
2
or
v
O
= 10.042v
d
+0.0833v
cm
(9.62)
The output voltage is also
v
O
= A
d
v
d
+ A
cm
v
cm
(9.63)
If we compare Equations (9.62) and (9.63), we see that
A
d
= 10.042 and A
cm
= 0.0833
Therefore, from Equation (9.59), the common-mode rejection ratio, is
CMRR(dB) = 20 log
10
10.042
0.0833
= 41.6dB
Comment: For good differential amplifiers, typical CMRR values are in the range of
80–100 dB. This example shows how close the ratios
R
2
/R
1
and
R
4
/R
3
must be in
order to achieve a CMRR value in that range.
Computer Verification: A PSpice analysis was performed on the differential ampli-
fier in this example with a
μ
A-741 op-amp. For input voltages of
v
I1
=−50
mV and
v
I2
=+50
mV, the output voltage is
v
O
= 1.0043
V, which gives a differential
voltage gain of 10.043. For input voltages of
v
I1
= v
I2
= 5
V, the output voltage
is
v
O
= 0.4153
V, which gives a common-mode voltage gain of
A
cm
= 0.4153/5 =
0.0831
. The common-mode rejection ratio is then
CMRR = 10.043/0.0831 =
120.9 ⇒ 41.6dB
, which agrees with the hand analysis. This result demonstrates that
at this point, the nonideal characteristics of the
μ
A-741 op-amp do not affect these
results.
EXERCISE PROBLEM
*
Ex 9.7: In the difference amplifier shown in Figure 9.24(a),
R
1
= R
3
= 10
k
,
R
2
= 20
k
, and
R
4
= 21
k
. Determine
v
O
when: (a)
v
I1
=+1
V,
v
I2
=−1
V;
and (b)
v
I1
= v
I2
=+1
V. (c) Determine the common-mode gain. (d) Determine
the CMRR(dB). (Ans. (a)
v
O
=−4.032
V, (b)
v
O
= 0.0323
V, (c)
A
cm
= 0.0323
,
(d)
CMRR(db) = 35.9
dB)
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650 Part 2 Analog Electronics
Instrumentation Amplifier
We saw in the last section that it is difficult to obtain a high input impedance and a
high gain in a difference amplifier with reasonable resistor values. One solution is to
insert a voltage follower between each source and the corresponding input. However,
a disadvantage of this design is that the gain of the amplifier cannot easily be
changed. We would need to change two resistance values and still maintain equal
ratios between
R
2
/R
1
and
R
4
/R
3
. Optimally, we would like to be able to change the
gain by changing only a single resistance value. The circuit in Figure 9.26, called
an instrumentation amplifier, allows this flexibility. Note that two noninverting
amplifiers, A
1
and A
2
, are used as the input stage, and a difference amplifier, A
3
is the
second, or amplifying, stage.
9.5.4
–
+
–
+
+
–
R
3
R
4
R
3
R
4
A
3
R
2
R
2
R
1
A
2
A
1
v
O
v
I2
v
O1
v
O2
v
I1
Figure 9.26 Instrumentation amplifier
+
–
–
+
–
+
–
R
3
R
4
R
3
R
4
R
2
R
2
R
1
A
1
+
+
–
A
3
v
I1
– v
I2
R
1
i
1
=
A
2
v
O
v
I2
v
O1
= v
I1
+ i
1
R
2
v
O2
= v
I2
– i
1
R
2
v
I1
i
1
i
1
0
0
0
0
Figure 9.27 Voltages and currents in instrumentation amplifier
We begin the analysis using the virtual short concept. The voltages at the invert-
ing terminals of the voltage followers are equal to the input voltages. The currents and
voltages in the amplifier are shown in Figure 9.27. The current in resistor
R
1
is then
i
1
=
v
I1
−v
I2
R
1
(9.64)
The current in resistors R
2
is also i
1
, as shown in the figure, and the output voltages
of op-amps A
1
and A
2
are, respectively,
v
O1
= v
I1
+i
1
R
2
=
1 +
R
2
R
1
v
I1
−
R
2
R
1
v
I2
(9.65(a))
and
v
O2
= v
I2
−i
1
R
2
=
1 +
R
2
R
1
v
I2
−
R
2
R
1
v
I1
(9.65(b))
From previous results, the output of the difference amplifier is given as
v
O
=
R
4
R
3
(v
O2
−v
O1
)
(9.66)
Substituting Equations (9.65(a)) and (9.65(b)) into Equation (9.66), we find the out-
put voltage, as follows:
v
O
=
R
4
R
3
1 +
2R
2
R
1
(v
I2
−v
I1
)
(9.67)
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Chapter 9 Ideal Operational Amplifiers and Op-Amp Circuits 651
R
1
100 kΩ
pot
R
1v
R
1f
Figure 9.28 Equivalent
resistance R
1
in
instrumentation amplifier
Since the input signal voltages are applied directly to the noninverting terminals
of A
1
and A
2
, the input impedance is very large, ideally infinite, which is one desir-
able characteristic of the instrumentation amplifier. Also, the differential gain is a
function of resistor R
1
, which can easily be varied by using a potentiometer, thus pro-
viding a variable amplifier gain with the adjustment of only one resistance.
EXAMPLE 9.8
Objective: Determine the range required for resistor R
1
, to realize a differential gain
adjustable from 5 to 500.
The instrumentation amplifier circuit is shown in Figure 9.26. Assume that
R
4
= 2R
3
, so that the difference amplifier gain is 2.
Solution: Assume that resistance R
1
is a combination of a fixed resistance R
1f
and a
variable resistance
R
1v
, as shown in Figure 9.28. The fixed resistance ensures that the
gain is limited to a maximum value, even if the variable resistance is set equal to
zero. Assume the variable resistance is a 100 k
potentiometer.
From Equation (9.67), the maximum differential gain is
500 = 2
1 +
2R
2
R
1 f
and the minimum differential gain is
5 = 2
1 +
2R
2
R
1 f
+100
From the maximum gain expression, we find that
2R
2
= 249R
1 f
Substituting this R
2
value into the minimum gain expression, we have
1.5 =
2R
2
R
1 f
+100
=
249R
1 f
R
1 f
+100
The resulting value of
R
1 f
is
R
1 f
= 0.606
k
, which yields
R
2
= 75.5
k
.
Comment: We can select standard resistance values that are close to the values cal-
culated, and the range of the gain will be approximately in the desired range.
Design Pointer: An amplifier with a wide range of gain and designed with a
potentiometer would normally not be used with standard integrated circuits in elec-
tronic systems. However, such a circuit might be very useful in specialized test
equipment.
EXERCISE PROBLEM
Ex 9.8: For the instrumentation amplifier in Figure 9.26, the parameters are
R
4
= 90 k
,
R
3
= 30
k
, and
R
2
= 50
k
. Resistance
R
1
is a series combina-
tion of a fixed 2 k
resistor and a 100 k
potentiometer. (a) Determine the range
of the differential voltage gain. (b) Determine the maximum current in
R
1
for
input voltages in the range
−25
mV to
+25
mV. (Ans. (a)
5.94 ≤ A
d
≤ 153
,
(b)
25 μ
A)
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652 Part 2 Analog Electronics
Integrator and Differentiator
In the op-amp circuits previously considered, the elements exterior to the op-amp
have been resistors. Other elements can be used, with differing results. Figure 9.29
shows a generalized inverting amplifier for which the voltage transfer function has
the same general form as before, that is,
v
O
v
I
=−
Z
2
Z
1
(9.68)
where Z
1
and Z
2
are generalized impedances. Two special circuits can be developed
from this generalized inverting amplifier.
9.5.5
C
2
R
1
+–
v
O
= V
C
–
v
I
v
C
1
R
1
C
2
v
I
(t') dt'
∫
O
t
–
+
Figure 9.30 Op-amp integrator
Z
1
Z
2
v
O
v
I
v
1
= 0
v
2
= 0
0
–
+
Figure 9.29 Generalized inverting amplifier
Equation (9.70) is the output response of the integrator circuit, shown in Fig-
ure 9.30, for any input voltage
v
I
. Note that if
v
I
(t)
is a finite step function, output
v
O
will be a linear function of time. The output
v
O
will be a ramp function and will
eventually saturate at a voltage near either the positive or negative supply voltage.
We will use the integrator in filter circuits, which are covered in Chapter 15.
We will show in Chapter 14 that nonzero bias currents into the op-amp greatly
influence the characteristics of this circuit. A dc current through the capacitor will
cause the output voltage to linearly change with time until the positive or negative
In the first, Z
1
corresponds to a resistor and Z
2
to a capacitor. The impedances are
then
Z
1
= R
1
and
Z
2
= 1/sC
2
, where s again is the complex frequency. The output
voltage is
v
O
=−
Z
2
Z
1
v
I
=
−1
sR
1
C
2
v
I
(9.69)
Equation (9.69) represents integration in the time domain. If V
C
is the voltage
across the capacitor at
t = 0
, the output voltage is
v
O
= V
C
−
1
R
1
C
2
t
0
v
I
(t
) dt
(9.70)
where
t
is the variable of integration. Figure 9.30 summarizes these results.
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Chapter 9 Ideal Operational Amplifiers and Op-Amp Circuits 653
supply voltage is reached. In many applications, a transistor switch needs to be added
in parallel with the capacitor to periodically set the capacitor voltage to zero.
The second generalized inverting op-amp uses a capacitor for Z
1
and a resistor
for Z
2
, as shown in Figure 9.31. The impedances are
Z
1
= 1/sC
1
and
Z
2
= R
2
, and
the voltage transfer function is
v
O
v
I
=−
Z
2
Z
1
=−sR
2
C
1
(9.71(a))
The output voltage is
v
O
=−sR
2
C
1
v
I
(9.71(b))
Equation (9.71(b)) represents differentiation in the time domain, as follows:
v
O
(t) =−R
2
C
1
dv
I
(t)
dt
(9.72)
The circuit in Figure 9.31 is therefore a differentiator.
Differentiator circuits are more susceptible to noise than are the integrator cir-
cuits. Input noise fluctuations of small amplitudes may have large derivatives. When
differentiated, these noise fluctuations may generate large noise signals at the output,
creating a poor output signal to noise ratio. This problem may be alleviated by plac-
ing a resistor in series with the input capacitor. This modified circuit then differenti-
ates low-frequency signals but has a constant high-frequency gain.
EXAMPLE 9.9
Objective: Determine the time constant required in an integrator.
Consider the integrator shown in Figure 9.30. Assume that voltage V
C
across the
capacitor is zero at
t = 0
. A step input voltage of
v
I
=−1
V is applied at
t = 0
. De-
termine the time constant required such that the output reaches
+10
V at
t = 1
ms.
Solution: From Equation (9.70), we have
v
o
=
−1
R
1
C
2
t
0
(−1) dt
=
1
R
1
C
2
t
t
0
=
t
R
1
C
2
At
t = 1
ms, we want
v
O
= 10
V. Therefore,
10 =
10
−3
R
1
C
2
which means the time constant is
R
1
C
2
= 0.1
ms.
Comment: As an example, for a time constant of 0.1 ms, we could have
R
1
= 10
k
and
C
2
= 0.01 μ
F, which are reasonable values of resistance and capacitance.
R
2
C
1
v
O
v
I
–
+
Figure 9.31 Op-amp differentiator
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654 Part 2 Analog Electronics
EXERCISE PROBLEM
Ex 9.9: An integrator with input and output voltages that are zero at
t = 0
is driven
by the input signal shown in Figure 9.32. (a) For circuit parameters
R
1
= 10
k
and
C
2
= 0.1 μ
F, determine the output voltage at
t =
(i) 1 ms, (ii) 2 ms, (iii) 3 ms,
and (iv) 4 ms. (b) Repeat part (a) for circuit parameters
R
1
= 10
k
and
C
2
= 1 μ
F. (Ans. (a) (i)
−1
V, (ii) 0, (iii)
−1
V, (iv) 0; (b) (i)
−0.1
V, (ii) 0,
(iii)
−0.1
V, (iv) 0)
R
L
v
I
v
O
v
O1
i
D
i
L
–
+
Figure 9.33 Precision half-wave
rectifier circuit
Slope = 1
v
O
v
I
Figure 9.34 Voltage transfer characteristics of
precision half-wave rectifier
v
I
(V)
t (ms)
–1
1
1234
Figure 9.32 Figure for Exercise Ex 9.9
Nonlinear Circuit Applications
Up to this point in the chapter, we have used linear passive elements in conjunction
with the op-amp. Many useful circuits can be fabricated if nonlinear elements, such
as diodes or transistors, are used in the op-amp circuits. We will consider three sim-
ple examples to illustrate the types of nonlinear characteristics that can be generated
and to illustrate the general analysis technique.
Precision Half-Wave Rectifier
An op-amp and diode are combined as shown in Figure 9.33 to form a precision half-
wave rectifier. For
v
I
> 0
, the circuit behaves as a voltage follower. The output volt-
age is
v
O
= v
I
, the load current i
L
is positive, and a positive diode current is induced
such that
i
D
= i
L
. The feedback loop is closed through the forward-biased diode.
The output voltage of the op-amp,
v
O1
, adjusts itself to exactly absorb the forward
voltage drop of the diode.
For
v
I
< 0
, the output voltage tends to go negative, which tends to produce neg-
ative load and diode currents. However, a negative diode current cannot exist, so the
diode cuts off, the feedback loop is broken, and
v
O
= 0
.
The voltage transfer characteristics are shown in Figure 9.34. The rectification is
precise in that, even at small positive input voltages,
v
O
= v
I
and we do not observe
a diode cut-in voltage.
9.5.6
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Chapter 9 Ideal Operational Amplifiers and Op-Amp Circuits 655
A potential problem in this circuit exists for
v
I
< 0
. The feedback loop is broken
so that the op-amp output voltage
v
O1
will saturate near the negative supply voltage.
When
v
I
switches positive, it will take time for the internal circuit to recover, so the
response time of the output voltage may be relatively slow. In addition, for
v
I
< 0
and
v
O
= 0
, there is now a voltage difference applied across the input terminals of
the op-amp. Most op-amps provide input voltage protection so the op-amp will not
be damaged in this case. However, if the op-amp does not have input protection, the
op-amp may be damaged if the input voltage is larger than 5 or 6 V.
Log Amplifier
Consider the circuit in Figure 9.35. The diode is to be forward biased, so the input
signal voltage is limited to positive values. The diode current is
i
D
= I
S
(e
v
D
/V
T
−1)
(9.73(a))
If the diode is sufficiently forward biased, the
(−1)
term is negligible, and
i
D
∼
=
I
S
e
v
D
/V
T
(9.73(b))
The input current can be written
i
1
=
v
I
R
1
(9.74)
and the output voltage, since
v
1
is at virtual ground, is given by
v
O
=−v
D
(9.75)
Noting that
i
1
= i
D
, we can write
i
1
=
v
I
R
1
= i
D
= I
S
e
−v
O
/V
T
(9.76)
If we take the natural log of both sides of this equation, we obtain
ln
v
I
I
S
R
1
=−
v
O
V
T
(9.77(a))
or
v
O
=−V
T
ln
v
I
I
S
R
1
(9.77(b))
R
1
D
1
0
+–
v
I
v
O
i
D
v
1
= 0
i
1
v
2
= 0
0
v
D
–
+
Figure 9.35 Simple log amplifier
Equation (9.77(b)) indicates that, for this circuit, the output voltage is proportional
to the log of the input voltage. One disadvantage of this circuit is that the reverse-
saturation current I
S
is a strong function of temperature, and it varies substantially from
one diode to another. A more sophisticated circuit uses bipolar transistors to eliminate
the I
S
parameter in the log term. This circuit will not be considered here.
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656 Part 2 Analog Electronics
Antilog or Exponential Amplifier
The complement, or inverse function, of the log amplifier is the antilog, or exponen-
tial, amplifier. A simple example using a diode is shown in Figure 9.36. Since
v
1
is at
virtual ground, we can write for
v
I
> 0
i
D
∼
=
I
S
e
v
I
/V
T
(9.78)
and
v
O
=−i
2
R =−i
D
R
(9.79(a))
or
v
O
=−I
S
R · e
v
I
/V
T
(9.79(b))
i
D
v
1
= 0
v
O
v
I
R
i
2
–
+
Figure 9.36 A simple antilog, or exponential, amplifier
The output voltage is an exponential function of the input voltage. Again, there are more
sophisticated circuits that perform this function, but they will not be considered here.
Test Your Understanding
TYU 9.9 A current source has an output impedance of
R
S
= 100 k
. Design a
current-to-voltage converter with an output voltage of
v
O
=−10 V
when the signal
current is
i
S
= 100 μA
. (Ans. Figure 9.20 with
R
F
= 100 k)
TYU 9.10 Design the voltage-to-current converter shown in Figure 9.22 such that
the load current in a
500
load can be varied between 0 and l mA with an input volt-
age between 0 and
−5V
. Assume the op-amp is biased at
±10 V
. (Ans.
R
2
= 5k
;
for example, let
R
3
= 7k
,
R
1
= 10 k
,
R
F
= 14 k)
TYU 9.11 All parameters associated with the instrumentation amplifier in Fig-
ure 9.26 are as given in Exercise Ex 9.8, except that resistor R
2
associated with the A
1
op-amp is
R
2
= 50 k ± 5%
. (a) Determine the maximum and minimum possible
values of the common-mode gain. (b) Determine the maximum and minimum possi-
ble values of the differential-mode gain. (c) Determine the minimum CMRR(dB).
(Ans. (a)
A
cm
= 0
; (b)
A
d
(
min
)
= 5.87
,
A
d
(
max
)
= 156.75
; (c)
CMRR =∞)
TYU 9.12 Design the instrumentation amplifier in Figure 9.26 such that the variable
differential voltage gain is in the range of 5 to 500. The range of the input voltages
is between
−2
mV and
+2
mV, and the maximum current in
R
1
is to be limited
to
2 μ
A. Set the gain of the difference amplifier to 2.5. (Ans.
R
1
(
fixed
)
= 2
k
,
R
2
= 199
k
,
R
1
(
var
)
= 396
k
)
TYU 9.13 An integrator is driven by the series of pulses shown in Figure 9.37. At the
end of the tenth pulse, the output voltage is to be
v
O
=−5V
. Assume
V
C
= 0
at
t = 0
.
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Chapter 9 Ideal Operational Amplifiers and Op-Amp Circuits 657
Determine the time constant and values of
R
1
and
C
2
that will meet these specifica-
tions. (Ans.
τ = 20 μs
; for example, let
C
2
= 0.01 μF
,
R
1
= 2k)
9.6 OPERATIONAL TRANSCONDUCTANCE
AMPLIFIERS
Objective: • Discuss the operational transconductance amplifier.
The operational amplifiers considered up to this point have been voltage amplifiers.
The input signal is a voltage and the output signal is a voltage.
Another type of op-amp is an operational transconductance amplifier (OTA). This
op-amp is a voltage-input, current-output amplifier. Its circuit symbol is shown in Fig-
ure 9.38(a) and the equivalent circuit model is given in Figure 9.38(b). For the ideal
OTA, both the input and output impedance is infinite. (The output impedance of an
ideal current source is infinite.) The output current for the ideal circuit can be written as
i
O
= g
m
v
d
(9.80)
where
g
m
is called the unloaded transconductance, with units of amperes per volt.
The transconductance can be varied by changing the control current in the op-amp
circuit. The OTA can then be used to electronically program functions.
v
I
(V)
t
1
12
. . . 10
10 ms
. . .
Figure 9.37 Figure for Exercise TYU 9.13
+
–
+
–
+
–
v
d
v
C
+
–
v
d
i
cont
R
g
m
v
d
i
o
i
o
(
a
)(
b
)
Figure 9.38 (a) Circuit symbol of the OTA. (b) Equivalent circuit model of the OTA.
We will see examples of actual OTA circuits in Chapter 13.
One example of an OTA application is shown in Figure 9.39. This circuit is a
simple voltage-controlled amplifier. The output op-amp is configured as a current-to-
voltage converter. We see that
v
O
=−i
O
R
F
=−i
O
(25 k)
(9.81)
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