Lab 9: Op Amps Reading Assignment

Lab 9: Op Amps
Reading Assignment
1. HH Lab Manual Class 9 and Worked Examples (pg 187-199 of lab manual)
2. HH Text pg 222-224 (yes, only 2 pages!)
3. Hands-On Ch 8-8.1
Additional background:
Source: Curtis, Lab 9
Figure 1: Idealized circuits for a) integrator, b) differentiator, and c) generalized inverting amplifier.
A. The Integrator
We will first consider the simple integrating circuit shown in Figure 1a. For this
circuit, the inverting input, v−, is a virtual ground because the op-amp golden rules
tell us v−=v+. The second golden rule (no current into the inputs) tells us that the
current through Rin also goes through Cf. Thus i = dQC/dt = vin/Rin. Since
Q = −Cf vo, we get:
−Cf (dvo/dt) = vin/Rin (1)
This time domain treatment shows that the output is 1/(RinCf) times the integral of
the input voltage.
It is also instructive to consider the behavior in the frequency domain. First note
that a time-dependent signal can be written as the sum of sinusoidal signals
The integral of the signal has the form
Thus, to form the integral of a general waveform, we need a magnitude response
that scales as 1/ω and that has a 90◦ phase shift over the relevant frequency
range. Since the gain of the generalized inverting amplifier (shown in Figure 1c)
G = −Zf / Zin (5)
the gain of the circuit shown in Figure 1a is just −(jωRinCf)−1, so we see that each
term is weighted by the (jωN)−1 factor required in (4) to give the Fourier
components of the integral. This again shows that the output is proportional to
the integral of the input with the same −1/(RinCf) proportionality factor as above.
B. The Differentiator
The gain of the circuit shown in Figure 1b is
G(ω) = −Zf / Zin = −jωRfCin (6)
Using (3),
so to take the derivative, we need to multiply each Fourier coefficient by its
frequency, ω, and introduce a 90◦ phase shift. The factor of jω in (6) shows that
the circuit does exactly that. Thus the circuit in Fig. 1b is a differentiator. You
may wish to prove to yourself that a time-domain treatment of the circuit gives the
same results.
C. Complications
In practice, both these idealized circuits suffer from a similar problem. For the
integrator, we must realize that the input signal is likely to have a small DC offset.
Even a very small DC current will charge up the capacitor and cause the op-amp
to reach its maximum output voltage within a short time period. For a frequency
domain treatment of this problem, remember the gain of the integrator is
(jωRC)−1. Thus, any non-zero DC input (which corresponds to ω = 0) will have
infinite gain for an idealized op-amp. In reality, this means the op-amp output will
reach its maximum voltage very quickly. A practical op-amp integrator circuit
must be modified to keep the gain finite at low frequencies. Similarly, the
idealized differentiator has a gain of jωRC that becomes large at high
frequencies. This is both very difficult to achieve and makes the circuit subject to
high frequency noise. A practical circuit will cut off the divergence of the gain at
large so that the output is not dominated by high frequency noise. The op-amp
open-loop gain, A(ω), will eventually reduce the gain at high frequencies. This
implies that there is a peak in the gain somewhat like that in a resonant circuit.
The circuits you build in the following sections will demonstrate, at least to some
extent, how to cope with these problems.