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## Q. 36.3

In Fig. 36-13a, the input voltage is a sine wave with a peak value of 10 V. What is the trip point of the circuit? What is the cutoff frequency of the bypass circuit? What does the output waveform look like?

## Verified Solution

Since +15 V is applied to a 3 : 1 voltage divider, the reference voltage is

$v_{ref} = +5 V$

This is the trip point of the comparator. When the sine wave crosses through this level, the output voltage switches states.

Formula (36-3), $f_c=\frac{1}{2\pi (R_1\parallel R_2)C_{BY}}$

the cutoff frequency of the bypass
circuit is

$f_c=\frac{1}{2\pi (200 k\Omega \parallel 100 k\Omega )(10 \mu F)}$

=0.239 Hz

This low cutoff frequency means that any 60-Hz ripple on the reference supply voltage will be heavily attenuated.

Figure 36-13b shows the input sine wave. It has a peak value of 10 V. The rectangular output has a peak value of approximately 15 V. Notice how the output voltage switches states when the input sine wave crosses the trip point of+5 V.