Electric Circuits & Networks · Lecture 20

Step Response of RC Circuits

Transient Analysis

Dr. Mithun Mondal Engineering Devotion Electric Circuits & Networks
Concept Demonstration (JoVE)
Demonstrative Video
SECTION 01

Demonstrative Video

SECTION 02

RC Circuit with Source

SECTION 01

Step Response of an RC Circuit

  • When dc source (voltage or current) is suddenly applied to an RC circuit , the source can be modelled as a step function, and the response is known as a step response.

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  • \[v(0^-)=v(0^+) = V_0\]
    Since the voltage of a capacitor cannot change instantaneously,
  • \[C \frac{d v}{d t}+\frac{v-V_{s} u(t)}{R}=0 \quad \Rightarrow~ \frac{d v}{d t}+\frac{v}{R C}=\frac{V_{s}}{R C} u(t)\]
    \(t>0\)\(v\)Applying KCL,
\[\begin{gathered} \frac{d v}{d t}+\frac{v}{R C}=\frac{V_{s}}{R C} u(t) \\ \frac{d v}{d t}+\frac{v}{R C}=\frac{V_{s}}{R C} \\ \frac{d v}{d t}=-\frac{v-V_{s}}{R C} \\ \frac{d v}{v-V_{s}}=-\frac{d t}{R C} \end{gathered}\]
SECTION 01

Complete or Total Response of an RC Circuit

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\[\begin{aligned} &v(t)= \begin{cases}0, & t<0 \\ V_{s}\left(1-e^{-t / \tau}\right), & t>0\end{cases} \\ &v(t)=V_{s}\left(1-e^{-t / \tau}\right) u(t) \\ &i(t)=C \frac{d v}{d t}=\frac{C}{\tau} V_{s} e^{-t / \tau}, \quad \tau=R C, \quad t>0 \\ &i(t)=\frac{V_{s}}{R} e^{-t / \tau} u(t) \end{aligned}\]
\[\begin{aligned} \text { Complete response }&=\underset{\text { stored energy }}{\text { natural response }}+\underset{\text { independent source }}{\text { forced response }}\\ v = v_n+v_f & \quad v_n = V_0\cdot e^{-t/\tau} \quad v_f = V_s(1-e^{-t/\tau})\\ \text { Complete response }&=\underset{\text { transient }}{\text { temporary part }}+\underset{\text { steady-state }}{\text { permanent part }}\\ v = v_t+v_{ss} & \quad v_t = (V_0-V_s)e^{-t/\tau} \quad v_{ss} = V_s \end{aligned}\]

  • Transient response is the circuit’s temporary response that will die out with time.

  • Steady-state response is the behaviour of the circuit a long time after an external excitation is applied.

SECTION 01

Problem Solving Steps

\[v(t) = v(\infty) + [v(0)-v(\infty)]e^{-t/\tau}\]

  • Determine the initial capacitor voltage \(v(0)\)

  • Determine the final capacitor voltage \(v(\infty)\)

  • Determine the time-constant \(\tau\)

\[v(t)=v(\infty)+\left[v\left(t_{0}\right)-v(\infty)\right] e^{-\left(t-t_{0}\right) / \tau}\]
, there is a time delay in the response so equation becomes, instead of at Note : if the switch changes position at time

Solved Problem

The switch has been in position \(A\) for a long time. At \(t=0\), the switch moves to \(B\). Determine \(v(t)\) for \(t>0\) and calculate its value at \(t=1 \mathrm{~s}\) and \(4 \mathrm{~s}\).

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\[\begin{gathered} v\left(0^{-}\right)=\frac{5}{5+3}(24)=15 \mathrm{~V} \\ v(0)=v\left(0^{-}\right)=v\left(0^{+}\right)=15 \mathrm{~V} \\ \tau=R_{\mathrm{Th}} C=4 \times 10^{3} \times 0.5 \times 10^{-3}=2 \mathrm{~s} \\ v(t)=v(\infty)+[v(0)-v(\infty)] e^{-t / \tau} \\ =30+(15-30) e^{-t / 2}=\left(30-15 e^{-0.5 t}\right) \mathrm{V} \\ \text { At } t=1, \quad v(1)=30-15 e^{-0.5}=20.9 \mathrm{~V}\\ \text { At } t=4, \quad v(4)=30-15 e^{-2}=27.97 \mathrm{~V} \end{gathered}\]