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Consider a symmetrical square wave of 20-V peak-to-peak, 0 average, and 2-ms period applied to a Miller integrator. Find the value of the time constant CR such that the triangular waveform at the output has a 20-V peak-to-peak amplitude.
  • a)
    0.25ms
  • b)
    0.50ms
  • c)
    2.5ms
  • d)
    5.0ms
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
Consider a symmetrical square wave of 20-V peak-to-peak, 0 average, an...
 According to the question 1/CR = 2.
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Consider a symmetrical square wave of 20-V peak-to-peak, 0 average, an...
Given data:
Symmetrical square wave of 20-V peak-to-peak, 0 average, and 2-ms period applied to a Miller integrator.
We need to find the value of the time constant CR such that the triangular waveform at the output has a 20-V peak-to-peak amplitude.

Miller Integrator:
Miller integrator is a type of operational amplifier integrator circuit with a capacitor connected between the output and the input terminals of the amplifier. The capacitor C is used to integrate the input signal and produces an output signal that is proportional to the time integral of the input signal.

Solution:
The triangular waveform at the output of a Miller integrator is given by the equation:

Vout = - (Vin/CR) * t + V0

Where, Vin is the input voltage, V0 is the initial output voltage, CR is the time constant of the circuit and t is the time.

For a symmetrical square wave with a 2-ms period, the frequency of the waveform is 1/2ms = 500 Hz. The peak-to-peak voltage of the square wave is 20V. The average value of the square wave is zero.

The triangular waveform at the output of the Miller integrator will have a maximum amplitude when the input signal changes polarity. At this point, the output voltage should be equal to the peak-to-peak voltage of the triangular waveform, which is 20V.

The time taken for the input voltage to change from 0V to 10V is half the period of the waveform, which is 1 ms.

Substituting the values in the equation for the output voltage, we get:

Vout = - (10/CR) * 0.001 + V0

When the input voltage changes from 0V to 10V, the output voltage changes from V0 to V0 - (10/CR) * 0.001.

Similarly, when the input voltage changes from 10V to 0V, the output voltage changes from V0 - (10/CR) * 0.001 to V0 + (10/CR) * 0.001.

The peak-to-peak voltage of the output triangular waveform is given by:

Peak-to-peak voltage = (V0 - (10/CR) * 0.001) - (V0 + (10/CR) * 0.001)

Peak-to-peak voltage = - (20/CR) * 0.001

For the peak-to-peak voltage of the output waveform to be 20V, we have:

(20/CR) * 0.001 = 20

CR = 0.5 ms

Therefore, the value of the time constant CR such that the triangular waveform at the output has a 20-V peak-to-peak amplitude is 0.5 ms.
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Consider a symmetrical square wave of 20-V peak-to-peak, 0 average, and 2-ms period applied to a Miller integrator. Find the value of the time constant CR such that the triangular waveform at the output has a 20-V peak-to-peak amplitude.a)0.25msb)0.50msc)2.5msd)5.0msCorrect answer is option 'B'. Can you explain this answer?
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