Q. Draw the differentiator circuit. Explain its principle of operation with necessary waveforms ?

A circuit in which the output voltage is directly proportional to the derivative of the input signal is called differentiator circuit i.e. v0=d/dt(input). A differentiating circuit can be arranged by a simple RC series circuit. Here the values of R and C are selected in such a way that the time constant (RC) of the circuit should be very small than the time period of the input wave and reactance of the capacitor is very large as compared to the value of the resistor. Thus the current flowing in the differentiator circuit is to a large extent decided by the capacitive reactance; the current will be leading the input voltage by 90 degrees approximately. Due to this current, the voltage developed across the resistor also leads the input voltage by 90 degrees.

If the input is a sinusoidal signal and given by

            v_i = V sin(ωt)

Then the output will be leading this voltage by 90 degrees.

           v_o = Vsin (ωt+90)= V cos(ωt).

Thus we find that the output is a differentiated form of the input signal. This can be proved mathematically.

       The output voltage V_O = i*R

But since XC>>R, the current (i)is dependent upon capacitor alone. Therefore, the rate of change of charge across the capacitor gives the current

             i = dQ/dt

but the charge Q across the capacitor C is given as

             Q = C*VC

Where VC is the voltage across the capacitor

           i = dQ/dt = d/dt(C*VC) =C* d(vt)/dt

            v_o = i* R = CR*d/dt(vt)

When a d_c voltage is applied to the differentiating circuit, its output will be zero since the derivative of a constant is zero. For a square wave input, the differentiator output will be sharp narrow pulses. When the input to the differentiator circuit is a triangular wave, the output will be a square or rectangular wave. For a sine wave input, the output of the differentiator will be a cosine wave.