Page 198 - Op Amps Design, Applications, and Troubleshooting

Voltage-Controlled Oscillator 181

FIGURE 4.5 A voltage-controlled oscillator providing both triangle- and square-wave outputs.

and, therefore, the feedback current will be constant. The feedback current must
flow through Q.
You may recall from basic electronics theory that if a capacitor is charged
with a constant current source, the resulting voltage will increase linearly. Because
the charging current for Q is constant, we can expect a linear ramp of voltage
across Q. And because the left end of Q is connected to a virtual ground point, the
other end (output of the op amp) will reflect the linear ramp voltage. The input
voltage V IN is positive, so we know that the output ramp will be increasing in the
negative direction.
A 2 is configured as a voltage comparator circuit with the upper and lower
thresholds being established by diodes Dj and D 2. As long as the ramp voltage is
above the lower threshold point (established by Dt), the output of amplifier A 2
will remain at its negative limit (-V SAT).
Amplifier A 3 is connected as an inverting summing amplifier. One input
comes from A 2 and receives a gain of -2. The other input is provided by +V !N and
receives a gain of -1. As long as the output of A 2 is at its -~V SAT level, diode D 3 will
be forward-biased and this voltage will be coupled to the input of A 3, Clearly, this
high-negative voltage will drive amplifier A 3 into saturation. That is, the output of
A 3 will be at the + V SAT level regardless of the value of input voltage (+V^). It is this
+VSAT level on the output of A 3 that causes D 4 to remain in a reverse-biased state.
This circuit condition remains constant as long as the ramp voltage on the output
of AI is above the lower threshold voltage of A 2.
Once the decreasing ramp voltage from AI falls below the lower threshold
voltage of comparator AI, the output of A 2 changes to its +V SAT level. This reverse-
biases diode D 3 and causes A 3 to act as a simple inverting amplifier with regard to
the input voltage + V iN. A voltage level that is equal (but opposite polarity) to +V IN
is felt at the right end of R 4. Since R* is half as large as R a and has the same voltage
applied, we can expect the current flow through JR 4 to be twice as large as that
through RI and in the opposite direction.