Page 200 - Op Amps Design, Applications, and Troubleshooting

Voltage-Controlled Oscillator 183

A 3 independently. First, let us consider the +V W signal input. The voltage gain of
this signal is computed in the same manner as in Chapter 2 for a simple inverting
amplifier. That is,

In this case, the generic R/ is replaced with physical resistor R 6. Notice that
we have ignored the effects of D 5 and D 6. Recall that the output of a closed-loop op
amp will go to whatever level is required to bring the differential input voltage
back to near 0. By inserting a forward-biased diode in the feedback loop, the out-
put is forced to rise an additional 0.6 volts (the forward voltage drop of the diode).
By paralleling two diodes in opposite polarities (D 5 and D 6), we force the output to
be 0.6 volts larger than it would otherwise have been. The actual output voltage,
then, will be the normal expected output plus a fixed 0.6-volt potential that causes
the output to be more positive during positive output times and more negative
during negative output times. The reason for D 5 and D 6 will be evident in a
moment.
Since the voltage gain for the +V m signal is -1, the range of output voltages
for A 3 as a result of +V m is

v 0l = -1 x 1 V - 0.6 V = -1.6 V, and

v 02 = -1 x 5 V ~ 0.6 V = -5.6 V

Now let us consider the effects of the second A 3 signal input, which comes from
the output of A 2. The voltage gain for this input is computed in a similar manner.

The input voltage (output from A 2) is the ±V SAT levels for A 2. Since diode D 3 will
block the positive level, we need only calculate the effects of the -V$ AT input. The
resulting output voltage from A 3 as a result of this input is computed as shown:
v 0 = -2 x (-V SAT + 0.6) + 0.6

= -2 x (-13 V + 0.6 V) + 0.6 V
= +25.4 V

This computed value exceeds the limits of A 3 because it only has a ±15-volt power
supply. This means that the output of A 3 will be driven to its +V SAT level (+13
volts).