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?
Question 36.3: In Fig. 36-13a, the input voltage is a sine wave with a peak...
Since +15 V is applied to a 3 : 1 voltage divider, the reference voltage is
v_{ref} = +5 VThis 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
=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.
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