Differentiating operational amplifier

Question

Answer 1

Step 1:

The amplitude of the triangular pulse is 5 V.

The average value of pulse is 0.

The frequency is .

The time period of the pulse is .

Substitute .

.

The time period of the triangular pulse is 0.001 sec.

An op-amp differentiator with triangular waveform as input, the output waveform is a rectangular waveform.

Hence, for one cycle of triangular pulse as input, we have two cycles of rectangular waveform.

Draw the triangular pulse with time period of 0.0005 sec:

Find the input voltage using the above triangular waveform.

Observe the graph:

Two points in the interval are and .

Find the voltage in .

Slope = .

Point-slope form of the equation is .

Slope = 20000 and the point is .

Voltage in is  V.

The output voltage of the differentiator is .

Step 2:

Similarly for .

Two points in the interval are and .

Slope = .

Slope = and the point is .

Voltage in is .

The output voltage of the differentiator is .

Peak to peak voltage

Peak to peak voltage is 8 V.

Solution:

Peak to peak voltage is 8 V.