RC Step Response

Charging or discharging a capacitor through a resistor.

Basically, we have some voltage sitting on a capacitor and we close the switch at time $t=0$. We have:

$$\begin{array}{rl}{I}_C& =-C\frac{dV(t)}{dt}\\ I_R& =\frac{V(t)}{R}\end{array}$$

Since these currents are the same, we put them together:

$$\begin{array}{rl}-I_C& =I_R\\ \frac{V(t)}{R}& =-C\frac{dV(t)}{dt}\end{array}$$

The equation above is called a first order system, which means there are no oscillations and we will see exponential decay.

Solving for $V(t)$, we have:

$$\begin{array}{rl}-\frac{1}{RC}V(t)dt& =\frac{1}{V(t)}dV(t)\\ -\frac{1}{RC}{\int }_0^{t_f}dt& ={\int }_{V(t=0)}^{V(t=t_f)}\frac{1}{V(t)}dV(t)\\ -\frac{t}{RC}& =\ln (V(t))-\ln (V_i)\end{array}$$

So:

$$\begin{array}{rl}V(t)& =e^{\ln (V_i)}{e}^{\frac{-t}{RC}}\\ & =V_i{e}^{-\frac{t}{RC}}\end{array}$$

We define a time constant $\tau =RC$, such that the voltage discharge is:

$$V(t)=V_i{e}^{-\frac{t}{\tau }}$$

For charging we would have:

$$V(t)=V_i(1-e^{-\frac{t}{\tau }})$$