Op-amps not only have the circuit model shown in
Figure 1,
but their element values are very special.
-
The input resistance,
R
in
R
in
,
is typically large, on the order of
1 MΩ.
-
The output resistance,
R
out
R
out
,
is small, usually less than 100 Ω.
-
The voltage gain,
G
G,
is large, exceeding
105
10
5
.
The large gain catches the eye; it suggests that an op-amp could
turn a 1 mV input signal into a 100 V one. If you were to build
such a circuit--attaching a voltage source to node
a, attaching node
b
to the reference, and looking at the output--you would be
disappointed. In dealing with electronic components, you cannot
forget the unrepresented but needed power supply.
It is impossible for electronic components to yield voltages
that exceed those provided by the power supply or for them to
yield currents that exceed the power supply's rating.
Typical power supply voltages required for op-amp circuits are
±15V
±
15
V
. Attaching the 1 mv signal not only would fail
to produce a 100 V signal, the resulting waveform would be
severely distorted. While a desirable outcome if you are a rock
& roll aficionado, high-quality stereos should not distort
signals. Another consideration in designing circuits with
op-amps is that these element values are typical: Careful
control of the gain can only be obtained by choosing a circuit
so that its element values dictate the resulting gain, which
must be smaller than that provided by the op-amp.
The feedback configuration shown in
Figure 2
is the most common op-amp circuit for obtaining what is known
as an inverting amplifier.
R
F
R
out
R
out
-G
R
F
1
R
out
+1
R
in
+1
R
L
1R+1
R
in
+1
R
F
-1
R
F
v
out
=1R
v
in
R
F
R
out
R
out
G
R
F
1
R
out
1
R
in
1
R
L
1
R
1
R
in
1
R
F
1
R
F
v
out
1
R
v
in
(1)
provides the exact input-output relationship. In choosing element
values with respect to op-amp characteristics, we can simplify the
expression dramatically.
- Make the load resistance,
R
L
R
L
,
much larger than
R
out
R
out
.
This situation drops the term
1
R
L
1
R
L
from the second factor of
Equation 1.
-
Make the resistor,
R
R,
smaller than
R
in
R
in
,
which means that the
1
R
in
1
R
in
term in the third factor is negligible.
With these two design criteria, the expression
(
Equation 1)
becomes
R
F
R
out
-G
R
F
1R+1
R
F
-1
R
F
v
out
=1R
v
out
R
F
R
out
G
R
F
1
R
1
R
F
1
R
F
v
out
1
R
v
out
(2)
Because the gain is large and the resistance
R
out
R
out
is small, the first term becomes
-1G
1
G
,
leaving us with
-1G1R+1
R
F
-1
R
F
v
out
=1R
v
in
1
G
1
R
1
R
F
1
R
F
v
out
1
R
v
in
(3)
-
If we select the values of
R
F
R
F
and
R
R
so that
GR≫
R
F
≫
G
R
R
F
,
this factor will no longer depend on the op-amp's inherent
gain, and it will equal
-1
R
F
1
R
F
.
Under these conditions, we obtain the classic input-output
relationship for the op-amp-based inverting amplifier.
v
out
=-
R
F
R
v
in
v
out
R
F
R
v
in
(4)
Consequently, the gain provided by our circuit is entirely
determined by our choice of the feedback resistor
R
F
R
F
and the input resistor
R
R.
It is always negative, and can be less than one or greater
than one in magnitude. It cannot exceed the op-amp's inherent
gain and should not produce such large outputs that distortion
results (remember the power supply!). Interestingly, note that
this relationship does not depend on the load resistance. This
effect occurs because we use load resistances large compared
to the op-amp's output resistance. Thus observation means
that, if careful, we can place op-amp circuits in cascade,
without incurring the effect of
succeeding circuits changing the behavior (transfer function)
of previous ones; see
this problem.
As long as design requirements are met, the input-output
relation for the inverting amplifier also applies when the
feedback and input circuit elements are impedances (resistors,
capacitors, and inductors).
Let's design an op-amp circuit that functions as a lowpass
filter. We want the transfer function between the output and
input voltage to be
Hf=K1+ⅈf
f
c
H
f
K
1
f
f
c
where
K
K
equals the passband gain and
f
c
f
c
is the cutoff frequency. Let's assume that the inversion
(negative gain) does not matter. With the transfer function
of the above op-amp circuit in mind, let's consider some
choices.
-
Z
F
=K
Z
F
K
,
Z=1+ⅈf
f
c
Z
1
f
f
c
.
This choice means the feedback impedance is a resistor
and that the input impedance is a series combination of
an inductor and a resistor. In circuit design, we try to
avoid inductors because they are physically bulkier than
capacitors.
-
Z
F
=11+ⅈf
f
c
Z
F
1
1
f
f
c
,
Z=1K
Z
1
K
.
Consider the reciprocal of the feedback impedance (its
admittance):
Z
F
-1=1+ⅈf
f
c
Z
F
1
f
f
c
.
Since this admittance is a sum of admittances, this
expression suggests the parallel combination of a
resistor (value = 1 Ω) and a capacitor (value =
1
f
c
F
1
f
c
F
).
We have the right idea, but the values (like 1 Ω)
are not right. Consider the general
RC
R
C
parallel combination; its admittance is
1
R
F
+ⅈ2πfC
1
R
F
2
f
C
.
Letting the input resistance equal
R
R,
the transfer function of the op-amp inverting amplifier
now is
Hf=-
R
F
R1+ⅈ2πf
R
F
C
H
f
R
F
R
1
2
f
R
F
C
Thus, we have the gain equal to
R
F
R
R
F
R
and the cutoff frequency
1
R
F
C
1
R
F
C
.
Creating a specific transfer function with op-amps does not have
a unique answer. As opposed to design with passive circuits,
electronics is more flexible (a cascade of circuits can be built
so that each has little effect on the others; see
(Reference)) and gain (increase
in power and amplitude) can result. To complete our example,
let's assume we want a lowpass filter that emulates what the
telephone companies do. Signals transmitted over the telephone
have an upper frequency limit of about 3 kHz. For the
second design choice, we require
R
F
C=3.3×10-4
R
F
C
3.3-4
.
Thus, many choices for resistance and capacitance values are
possible. A 1 μF capacitor and a 330 Ω resistor,
10 nF and 33 kΩ, and 10 pF and 33 MΩ would
all theoretically work. Let's also desire a voltage gain of
ten:
R
F
R=10
R
F
R
10
,
which means
R=
R
F
10
R
R
F
10
.
Recall that we must have
R<
R
in
R
R
in
.
As the op-amp's input impedance is about 1 MΩ, we don't
want
R
R
too large, and this requirement means that the last choice for
resistor/capacitor values won't work. We also need to ask for
less gain than the op-amp can provide itself. Because the
feedback "element" is an impedance (a parallel resistor
capacitor combination), we need to examine the gain
requirement more carefully. We must have
|
Z
F
|R<105
Z
F
R
10
5
for all frequencies of interest. Thus,
R
F
|1+ⅈ2πf
R
F
C|R<105
R
F
1
2
f
R
F
C
R
10
5
.
As this impedance decreases with frequency, the design
specification of
R
F
R=10
R
F
R
10
means that this criterion is easily met. Thus, the first two
choices for the resistor and capacitor values (as well as many
others in this range) will work well. Additional
considerations like parts cost might enter into the
picture. Unless you have a high-power application (this isn't
one) or ask for high-precision components, costs don't depend
heavily on component values as long as you stay close to
standard values. For resistors, having values
r10d
r
10
d
,
easily obtained values of
r
r
are 1, 1.4, 3.3, 4.7, and 6.8, and the decades span 0-8.
What is special about the resistor values; why these rather
odd-appearing values for
r
r?
The ratio between adjacent values is about
2
2
.
When we meet op-amp design specifications, we can simplify our
circuit calculations greatly, so much so that we don't need
the op-amp's circuit model to determine the transfer
function. Here is our inverting amplifier.
When we take advantage of the op-amp's
characteristics—large input impedance, large gain, and
small output impedance—we note the two following
important facts.
-
The current
i
in
i
in
must be very small. The voltage produced by the dependent
source is
105
10
5
times the voltage
v
v.
Thus, the voltage
v
v
must be small, which means that
i
in
=v
R
in
i
in
v
R
in
must be tiny. For example, if the output is about 1 v,
the voltage
v=10-5V
v
10
5
V
,
making the current
i
in
=10-11A
i
in
10
11
A
. Consequently, we can ignore
i
in
i
in
in our calculations and assume it to be zero.
-
Because of this assumption—essentially no current
flow through
R
in
R
in
—the
voltage
v
v
must also be essentially zero. This means that in op-amp
circuits, the voltage across the op-amp's input is
basically zero.
Armed with these approximations, let's return to our original
circuit as shown in
Figure 5.
The node voltage
e
e
is essentially zero, meaning that it is essentially tied to
the reference node. Thus, the current through the resistor
R
R
equals
v
in
R
v
in
R
.
Furthermore, the feedback resistor appears in parallel with
the load resistor. Because the current going into the op-amp
is zero, all of the current flowing through
R
R
flows through the feedback resistor
(
i
F
=i
i
F
i
)!
The voltage across the feedback resistor
v
v
equals
v
in
R
F
R
v
in
R
F
R
.
Because the left end of the feedback resistor is essentially
attached to the reference node, the voltage across it equals
the negative of that across the output resistor:
v
out
=-v=-
v
in
R
F
R
v
out
v
v
in
R
F
R
.
Using this approach makes analyzing new op-amp circuits much
easier. When using this technique, check to make sure the
results you obtain are consistent with the assumptions of
essentially zero current entering the op-amp and nearly zero
voltage across the op-amp's inputs.
Let's try this analysis technique on a simple extension of
the inverting amplifier configuration shown in
Figure 6.
If either of the source-resistor combinations were not
present, the inverting amplifier remains, and we know that
transfer function. By superposition, we know that the
input-output relation is
v
out
=-
R
F
R
1
v
in
(
1
)
-
R
F
R
2
v
in
(
2
)
v
out
R
F
R
1
v
in
(
1
)
R
F
R
2
v
in
(
2
)
(5)
When we start from scratch, the node joining the three
resistors is at the same potential as the reference,
e≈0
e
0
,
and the sum of currents flowing into that node is
zero. Thus, the current
i
i
flowing in the resistor
R
F
R
F
equals
v
in
(
1
)
R
1
+
v
in
(
2
)
R
2
v
in
(
1
)
R
1
v
in
(
2
)
R
2
.
Because the feedback resistor is essentially in parallel
with the load resistor, the voltages must satisfy
v=-
v
out
v
v
out
.
In this way, we obtain the input-output relation given above.
What utility does this circuit have? Can the basic notion of the
circuit be extended without bound?
"Electrical Engineering Digital Processing Systems in Braille."