Step Response of an RC Circuit

To find the step response of an RC circuit, the following is required:

  • The initial capacitor voltage \(v(0)\).

    Since the capacitor voltage cannot change instantaneously,

    \(\boxed{v\left(0^-\right) = v\left(0^+\right) = v\left(0\right) = V_0}\)

    where \(v\left(0^-\right)\) is the voltage across the capacitor just before switching and \(v\left(0^+\right)\) is its voltage immediately after switching.

  • The final (steady state) capacitor voltage \(v(\infty)\).

    Use the fact that the capacitor acts like an open circuit to DC at steady state.

  • The time constant \(\tau\).

    The time constant \(\tau\) for an RC circuit is

    \(\boxed{\tau = RC}\)

    where \(R\) is the thevenin resistance at the capacitor terminals and \(C\) is the equivalent capacitance.

When the DC source of an RC circuit is suddenly applied, the voltage or current source can be modeled as a step function, and the response is known as a step response.

Applying KCL

\(\displaystyle C\frac{dv}{dt} + \frac{v - V_s\,u\left(t\right)}{R} = 0\)

\(\displaystyle \frac{dv}{dt} + \frac{v - V_s\,u\left(t\right)}{RC} = 0\)

\(\displaystyle \frac{dv}{dt} + \frac{v}{RC} = \frac{V_s}{RC} u\left(t\right)\)

For \(t > 0\)

\(\displaystyle \frac{dv}{dt} + \frac{v}{RC} = \frac{V_s}{RC}\)

\(\displaystyle \frac{dv}{dt} = -\frac{v - V_s}{RC}\)

\(\displaystyle \frac{dv}{v - V_s} = -\frac{dt}{RC}\)

Integrating both sides and introducing the initial conditions,

\(\displaystyle \int_{V_0}^{v\left(t\right)}\frac{dv}{v - V_s} = -\int_{0}^{t}\frac{dt}{RC}\)

\(\displaystyle \left.\ln{\left(v - V_s\right)}\right|_{V_0}^{v\left(t\right)} = -\left.\frac{t}{RC}\right|_{0}^{t}\)

\(\displaystyle \ln{\left[v\left(t\right) - V_s\right]} - \ln{\left(V_0 - V_s\right)} = -\frac{t}{RC}\)

\(\displaystyle \ln{\frac{v - V_s}{V_0 - V_s}} = -\frac{t}{RC}\)

\(\displaystyle \frac{v - V_s}{V_0 - V_s} = e^{\displaystyle -\frac{t}{RC}}\)

\(\displaystyle v = V_s + \left(V_0 - V_s\right)\,e^{\displaystyle -\frac{t}{RC}}\)

\(\boxed{v\left(t\right) = V_s + \left(V_0 - V_s\right)\,e^{\displaystyle -t/\tau}, \quad t > 0}\)

Thus,

\(\boxed{v\left(t\right) = \begin{cases}V_0\,, & t < 0 \\ V_s + \left(V_0 - V_s\right)\,e^{\displaystyle -t/\tau}\,, & t > 0\end{cases}}\)

where time constant \(\tau = RC\)

This is known as the complete response of the RC circuit to a sudden application of a DC source.

The complete response may be written as

\(\boxed{v\left(t\right) = v\left(\infty\right) + \left[v\left(0\right) - v\left(\infty\right)\right]\,e^{\displaystyle -t/\tau}, \quad t > 0}\)

where time constant \(\tau = RC\), \(v\left(0\right)\) is the initial voltage at \(t = 0^+\) and \(v\left(\infty\right)\) is the final or steady-state value.

This equation can be used in source-free RC circuit.

If the switch changes position at time \(t = t_0\) instead of at \(t = 0\), there is a time delay in the response.

\(\boxed{v\left(t\right) = v\left(\infty\right) + \left[v\left(t_0\right) - v\left(\infty\right)\right]\,e^{\displaystyle -\left(t - t_0\right)/\tau}, \quad t > t_0}\)

where time constant \(\tau = RC\) and \(v\left(t_0\right)\) is the initial value at \(t = t_{0}^{+}\).

This equation can be used in source-free RC circuit.

Links to this page
#first-order-circuit #circuit