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Operational Amplifiers

# Frequency Response

The open-loop gain of an operational amplifier varies with frequency. It can
be high at low frequencies but decreases rapidly above some frequency which
is characteristic of the particular amplifier. At high signal speeds, there is
also a limit on the maximum rate of change of the output voltage. This limit
is called the **slewing-rate limit**. These real amplifier characteristics
are considered in the next sections.

The **amplitude-response** and **phase-response** curves for a typical
operational amplifier are given in the figure below. The low-frequency gain
of the amplifier is 100 dB, or 100,000, and the low-frequency phase shift is
essentially zero. At some frequency (often below 10 Hz as in this case) the
gain begins to fall off, usually at 6 dB per octave. A slope of 6 dB per octave
is used because it provides the basis for stability when feedback is applied
properly; this comes about because the maximum phase shift is about -90° for all
gains from 100 to 0 dB (10^{5} to 1). Negative feedback means that there
is an additional -180° of phase shift outside the amplifier, and so the
maximum phase shift for all usable gains is -(180 + 90) = -270°. Since it
takes 360° of phase shift (input and output in phase) for an instability
in the form of oscillations to occur, a 6-dB-per-octave-slope operational
amplifier will be stable if negative feedback is applied properly.

A very important indicator of operational amplifier performance can be
extracted from the figure above by inspection, namely, the gain margin for a
particular application. First note that the gain without feedback is called
the *open-loop* gain, while the gain achieved by the application
of feedback is called the *closed-loop* gain. The gain margin is the
difference in gain between the open-loop and closed-loop gains. The open-loop
and closed-loop gains and gain margin for an amplifier with a closed-loop
gain of 20 dB (× 10) are shown at some frequency *f* in the figure
above. The significance of gain margin is as follows: The gain of an operational
amplifier circuit depends solely on the external feedback elements and not
on its internal transistors and resistors as long as there is adequate gain
margin. For example, for the 20 dB gain closed-loop plot shown in the
figure above, the gain margin varies with frequency as follows:

For most applications of operational amplifiers, a gain margin of 40 dB is sufficient, but even 20 dB is often adequate, depending on accuracy requirements. If 40 dB of gain margin is needed with a closed-loop gain of 20 dB, an amplifier having the response shown in the figure above can be used only up to 1 kHz, while if 20 dB is satisfactory, then the frequency can be extended to 10 kHz.

The following example illustrates the effect of finite bandwidth on the voltage follower.

**Example**: Find the magnitude of the gain at 100 kHz of a voltage
follower that uses the operational amplifier for which the frequency response
is plotted in the figure above.

**Solution**: Referring to the figure above, the follower has an input
voltage *v*_{1}, an output voltage *v*_{2} and
an open-loop gain of *A*_{O} (it is nonideal in this
respect). The output voltage is equal to the voltage difference between the
input terminals of the amplifier times *A*_{O}, or

Rearranging this equation gives the closed-loop gain *A*_{C},

If *A*_{O} is infinity, as with an ideal operational amplifier,
then the closed-loop gain *A*_{C} = 1. In this example, however,
|*A*_{O}| = 20 dB or 10 at 100 kHz. It is not correct to
substitute |*A*_{O}| = 10 into the previous equation because
an *incorrect* answer of 0.9091 will be obtained, which is more than
nine percent lower than the ideal gain of one.

In the region where the gain is falling off at 6 dB per octave,
*A*_{O} is a phasor quantity and must be substituted as such
into the previous equation. The curve in the amplitude response graph is a plot
of the magnitude of *A*_{O} or |*A*_{O}|; accordingly,

For this example, assume *θ* is nearly -90°, also,
|*A*_{O}| = 10, so that

Substituting *A*_{O} from the previous equation into the equation
*A*_{C} = 1/(1+1/*A*_{O}) gives

The magnitude of *A*_{C} is then

which is only about one-half percent below the ideal value of one.