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AC Electric Circuits

Passive Integrator and Differentiator Circuits


25 questions By Tony R. Kuphaldt

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  • Question 22 of 25

    Calculate the output voltage of this passive differentiator circuit 1 millisecond after the rising edge of each positive square wave pulse (where the square wave transitions from -5 volts to 5 volts):




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  • Question 23 of 25

    Calculate the output voltage of this passive differentiator circuit 150 microseconds after the rising edge of each “clock” pulse (where the square wave transitions from 0 volts to 5 volts):




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  • Question 24 of 25

    A passive differentiator is used to “shorten” the pulse width of a square wave by sending the differentiated signal to a “level detector” circuit, which outputs a “high” signal ( 5 volts) whenever the input exceeds 3.5 volts and a “low” signal (0 volts) whenever the input drops below 3.5 volts:





    Each time the differentiator’s output voltage signal spikes up to 5 volts and quickly decays to 0 volts, it causes the level detector circuit to output a narrow voltage pulse, which is what we want.

    Calculate how wide this final output pulse will be if the input (square wave) frequency is 2.5 kHz.

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