In an RC differentiator, the capacitor

RC Differentiator:

• The input voltage is given to the capacitor and the output is taken at the resistor.
• At low frequencies, the reactance of the capacitor X_C = ∞. So, the capacitor blocks DC voltage or behaves like an open circuit.
• At high frequencies, the reactance of the capacitor X_C = 0. So, the capacitor allows the varying signals or behaves like a short circuit.

Capacitor current is given by:

\({i_C} = C\frac{{d{V_{in\left( t \right)}}}}{{dt}}\)

The capacitor voltage is given by:

\({V_C} = \frac{1}{C}\smallint {i_c}\left( t \right)dt\)

The amount of charge across the capacitor plates is:

Q = C × V_C

Analysis:

Consider the input signal is unit step signal i.e., u(t)

The KVL in the loop gives:

\({V_{in}}\left( s \right) = \frac{{I\left( s \right)}}{{sC}} + I\left( s \right)R\)

\(\frac{1}{s} = I\left( s \right)\left[ {\frac{1}{{sC}} + R} \right]\)

\(I\left( s \right) = \frac{{sC}}{{s\left( {1 + RsC} \right)}}\)

\(I\left( s \right) = \frac{1}{{R\left( {s + \frac{1}{{RC}}} \right)}}\)

\(i\left( t \right) = \frac{1}{R}{e^{ - \frac{t}{{RC}}}} = \frac{1}{R}{e^{ - \frac{t}{τ}}}\)

The capacitor charge is:

\({Q_C} = \smallint i\left( t \right)dt = \smallint \frac{1}{R}{e^{ - \frac{t}{τ }}}\)

\({Q_C} = \frac{1}{R}\frac{{{e^{ - \frac{t}{τ }}}}}{{ - \frac{1}{τ }}} = - \frac{τ }{R}{e^{ - \frac{t}{τ }}}\)

We observe that the charge across the capacitor is depending on the time constant τ.

RC differentiator output waveforms for the different time constants are shown below:

From the above, it is clear that the capacitor charge exponentially at a rate depending on the RC time constant.