Design an op-amp differentiator that will differentiate an input signal with $f_{\max}=100 Hz$. Draw the output waveform for sine wave of 1 V peak at 100 Hz applied to the differentiator. Also repeat it for square wave input.

Select $f_a=f_{\max}=100 Hz=\dfrac{1}{2πR_f C_1 }$

Let $C_1=0.1 μF$

Then $R_f=\dfrac{1}{2π(10^2)(10^{-7})}=15.9 kΩ$

Now choose $f_b=10f_a \\ =1 kHz \\ =\dfrac{1}{2πR_1 C_1}$

Therefore, $R_1=\dfrac{1}{2π(10^3)(10^{-7})}=1.59 kΩ$

Since $R_f C_f=R_1 C_1$

We get,

$$C_f=\dfrac{1.59 \times 10^3 \times 10^{-7}}{15.9 \times10^3 }=0.01 μF$$

Since, $v_i=1\sin 2π(100)t$

$v_o= -R_f C_1 \dfrac{dv_i}{dt} \\ = -(15.9 kΩ)(0.1 μF)\dfrac{d}{dt}[(1V)\sin(2π)(10^2)t] \\ = -(15.9 kΩ)(0.1 μF)(2 π) (10^2)\cos[(2π)(10^2)t] \\ = -0.999 \cos[(2π)(10^2)t] \\ = -1 \cos[(2π)(10^2)t]$

The input and output waveforms are shown in Fig.(a)

For a square wave input, say 1V peak and 1 KHz, the output waveform will consist of positive and negative spikes of magnitude $V_{sat}$, which is approximately 13V for ± 15V op-amp power supply. During the time periods for which input is constant at ± 1V, the differentiated output will be zero. However, when input transits between t 1V levels, the slope of the input is infinite for an ideal square wave. The output, therefore, gets clipped to about ±13V for a ± 15V op-amp power supply as shown in Fig (b).