this attenuation is undesirable and that we want the entire input signal
V
s
to appear across the
load. By inserting a voltage follower in between
R
s
and
R
L
, we can accomplish just that! Since
the current
i
p
= 0, the voltage
v
p
=
v
s
. Therefore, we have
v
s
=
v
p
=
v
n
=
v
o
(8.53)
218
CHAPTER 8.
OPERATIONAL AMPLIFIERS
8.4.2
Instrumentation Amplifier
Figure 8.22: Instrumentation amplifier
An instrumentation amplifier as shown in Figure
8.22
is used to amplify a small difference between
two signals. This type of circuit is often used in sensors to detect deviations from a nominal value.
If one can relate a physical quantity such as temperature, pressure, humidity, etc.
to a voltage,
then by using an instrumentation amplifier a sensor can be designed to detect when, for example,
room temperature/humidity/pressure/etc. deviates from an acceptable value. In Engr40M, we will
be using an instrumentation amplifier as part of a larger circuit that can monitor your heart rate!
To express
v
o
in terms of
v
1
and
v
2
, we first note that the the third amplifier is a difference
amplifier with inputs
v
o
1
and
v
o
2
.
Therefore, using the result we derived earlier, we have the
following equation:
v
o
=

R
4
R
5
v
o
1
+
R
4
+
R
5
R
5
R
4
R
4
+
R
5
v
o
2
(8.54)
This can be simplified to:
v
o
=
R
4
R
5
(
v
o
2

v
o
1
)
(8.55)
Next, we determine the intermediate output voltages
v
o
1
and
v
o
2
. We first write a KCL equation
at
v
n
1
, the negative input of the first amplifier and apply the opamp “golden rules:”
v
1

v
2
R
2
+
v
1

v
o
1
R
1
= 0
(8.56)
This can be simplified to:
v
o
1
=
v
1
+
R
1
R
2
(
v
1

v
2
)
(8.57)
Next, we write a KCL equation at
v
n
2
, the negative input of the second amplifier and apply the
opamp “golden rules:”
v
2

v
1
R
2
+
v
2

v
o
2
R
3
= 0
(8.58)
This can be simplified to:
v
o
2
=
v
2
+
R
3
R
2
(
v
2

v
1
)
(8.59)
8.5.
SUMMARY
219
Next, we subtract equation (
8.57
) from equation (
8.59
) to obtain:
v
o
2

v
o
1
=
v
2

v
1
+
R
3
R
2
(
v
2

v
1
) +
R
1
R
2
(
v
2

v
1
)
(8.60)
This can be simplified to:
v
o
2

v
o
1
=
R
1
+
R
2
+
R
3
R
2
(
v
2

v
1
)
(8.61)
Finally, we can obtain the overall expression for
v
o
in terms of
v
1
and
v
2
by substituting equation
(
8.61
) into equation (
8.55
), and we obtain the following:
v
o
=
R
4
R
5
R
1
+
R
2
+
R
3
R
2
(
v
2

v
1
)
(8.62)
In the case where all the resistors are equal except for
R
2
, the expression for the output voltage
becomes:
v
o
=
1 +
2
R
R
2
(
v
2

v
1
)
(8.63)
In this case, the gain of the instrumentation amplifier is set by
R
2
. This configuration can be very
useful if one desires to have a variable gain since
R
2
can be be implemented as a potentiometer.
8.5
Summary
The Golden Rules for Ideal OpAmps
•
A
=
∞
•
R
i
=
∞
•
R
o
= 0
•
i
p
=
i
n
= 0
•
v
n
=
v
p
(for negative feedback configurations)
The output voltage of an opamp is limited by the supply voltage for the opamp:

v
o
 ≤
V
dd
(8.64)
8.6
Analysis of nonideal Op Amp (bonus material)
The previous sections analyzed an ideal opamp behavior. Analyzing the behavior of a nonideal
opamp is only a little more difficult and is done in the next sections.
220
CHAPTER 8.
OPERATIONAL AMPLIFIERS
8.6.1
Noninverting Amplifier
Figure 8.23: A noninverting amplifier using the equivalent circuit model of an opamp
A noninverting amplifier can be used to amplify an input signal without inverting its sign. In order
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 Spring '19
 Hassan Kasfy