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+ⅈ2πf
f
c
H
f
K
1
2
f
f
c
(1)
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+ⅈ2πf
f
c
Z
1
2
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+ⅈ2πf
f
c
Z
F
1
1
2
f
f
c
,
Z=1K
Z
1
K
. Consider the reciprocal of the feedback
impedance (its admittance):
Z
F
-1=1+ⅈ2πf
f
c
Z
F
1
2
f
f
c
. Thus, the admittance is a sum of admittances,
suggesting the parallel combination of a resistor (value =
1
Ω
1Ω
) 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
.
Thus, 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) 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
10
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
.
