Series RLC Circuit Response to a Step Voltage

Table of Contents

Use of Laplace transforms to study the response of an RLC circuit to a step voltage. Formulas for the current and all the voltages are developed and numerical examples are presented along with their detailed solutions.
An online calculator for step response of a series RLC circuit may be used check calculations done manually.

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Fomulas for the Current and Voltages in a Series RLC Circuit in Response to a Step Voltage

Problem
Find expressions for the current \( i \) and voltages across the capacitor \( C \) , the inductors \( L \) and the resitor \( R \) as functions of time in the ciruit below given that the voltage source \( v_i = V_0 \; u(t) \), where \( V_0\) is a constant and \( u(t) \) is the unit step function . The initial current at \( t = 0 \) is equal to zero.
series RLC transient circuit analysis
Solution to the Above Problem
Use Kirchhoff's law of voltages to write
\( v_i - v_R - v_L - v_C = 0 \)       (I)
Use Ohm's law to write
\( v_R = R \; i \)
Relationship between voltage and charging current of a capacitor
\( \displaystyle v_C = \dfrac{1}{C} \; \int i dt \)
Relationship between voltage and charging current of an inductor
\( \displaystyle v_L = L \; \dfrac{d i}{dt} \)
Substitute \( v_R \) , \( v_L \) and \( v_C \) by their expressions in equation (I)
\( \displaystyle v_i - R i - L \dfrac{d i}{dt} - \dfrac{1}{C} \int i dt = 0 \)
Take the Laplace transform of both sides of the above equation
\( \displaystyle \mathscr{L}\{ v_i - R i - L \dfrac{d i}{dt} - \dfrac{1}{C} \int i dt \} = \mathscr{L}\{ 0 \} \)
Use the property of linearity of Laplace transform and aslo the fact that \( \mathscr{L}\{ 0 \} = 0 \) to rewrite the above as
\( \displaystyle \mathscr{L}\{ v_i \} - R \mathscr{L}\{ i \} - L \mathscr{L} \left\{ \dfrac{d i}{dt} \right\} - \dfrac{1}{C} \mathscr{L} \left\{ \int i dt \right\} = 0 \)
Since \( v_i(t) = V_0 \; u(t) \) where \( V_0 \) is a constant and \( u(t) \) is the unit step function, \( \mathscr{L}\{ v_i \} = \dfrac{V_0}{s} \)
Let \( \mathscr{L}\{ i\} = I(s) \)
Use the property of the derivative and integral (see formulas and properties of Laplace transform) to write
\( \mathscr{L} \left\{ \dfrac{d i}{dt} \right\} = s I(s) - i(0) = s I(s) \) since the initial current is equal to zero \( i(0) = 0 \)
\( \displaystyle \mathscr{L} \left\{ \int i dt \right\} = \dfrac{I(s)}{s} \)
After substitution, our equation becomes
\( \dfrac{V_0}{s} - R \; I(s) - L \; s \; I(s) - \dfrac{I(s)}{C s} = 0 \)
NOTE that we have transformed our initial differential equation from the \( t \) (time) domain to the \( s \) domain.
Multiply all terms in the above equation by \( s \) and simplify
\( V_0 - R \; s \; I(s) - L \; s^2 \; I(s) - \dfrac{I(s)}{C} = 0 \)
Factor \( I(s) \) out and rewrite the above equation as
\( I(s) (L \; s^2 + R \; s +\dfrac{1}{C}) = V_0 \)
Solve the above for \( I(s) \) and rewrite as follows
\( I(s) = \dfrac{V_0}{L} \times \dfrac{1}{s^2 + \dfrac{R}{L} s + \dfrac{1}{L C} } \)
Complete the square in the denominator
\( I(s) = \dfrac{V_0}{L} \times \dfrac{1}{ \left(s + \dfrac{R}{2 L} \right)^2 + \dfrac{1}{L C} - \left(\dfrac{R}{2 L}\right)^2 } \)
Let \( \alpha = \dfrac{R}{2L} \)
and rewrite the above equation as
\( I(s) = \dfrac{V_0}{L} \times \dfrac{1}{ \left(s + \alpha \right)^2 + \dfrac{1}{L C} - \left(\dfrac{R}{2 L}\right)^2 } \)



We now consider 3 cases depending on the sign of the expression \( \dfrac{1}{L C} - \left(\dfrac{R}{2 L}\right)^2 \)
Case 1: \( \dfrac{1}{L C} \gt \left(\dfrac{R}{2 L}\right)^2 \) : The circuit is underdamped

Let \( \omega = \sqrt {\dfrac{1}{L C} - \left(\dfrac{R}{2 L}\right)^2} \) and rewrite \( I(s) \) as
\( I(s) = \dfrac{V_0}{\omega L} \times \dfrac{\omega}{ \left(s + \alpha \right)^2 + \omega^2 } \)
Use formulas and properties of Laplace transform to find the inverse Laplace transform of \( I(s) \) as
For \( t \ge 0 \) , \( v_i (t) = V_0 \) and we have the following
\( i(t) = \dfrac{V_0}{\omega L} \; \sin (\omega t) \; e^{-\alpha t} \)
\( v_R(t) = R \; i(t) = \dfrac{R V_0}{\omega L} \sin (\omega t) e^{-\alpha t} \)
\( \quad \quad = V_0 \dfrac{2 \alpha}{\omega } \sin (\omega t) e^{-\alpha t} \)
\( v_L(t) = L \; \dfrac{d i}{dt} = V_0 \left\{ \cos (\omega t)- \dfrac{\alpha}{\omega} \sin (\omega t) \right\} e^{-\alpha t} \)
\( v_C(t) = v_i(t) - v_R(t) - v_L(t) \)
\( \quad \quad = V_0 - V_0 \dfrac{2 \alpha}{\omega } \sin (\omega t) e^{-\alpha t} - V_0 \left\{ \cos (\omega t)- \dfrac{\alpha}{\omega} \sin (\omega t) \right\} e^{-\alpha t} \)
\( \quad \quad = V_0 - V_0 \left\{ \cos (\omega t) + \dfrac{\alpha}{\omega} \sin (\omega t) \right\} e^{-\alpha t} \)

Numerical Applications - Example 1 - Underdamped Circuit
Let \( V_0 = 1 \; V\) , \( R = 10 \; \Omega \) , \( L = 0.4 \; H \) and \( C = 50 \;\mu F \)
\( \dfrac{1}{L C} = 50000\)
\( \left(\dfrac{R}{2 L}\right)^2 = 156.25 \)
Hence \( \dfrac{1}{L C} \gt \left(\dfrac{R}{2 L}\right)^2 \) ; the circuit is underdamped
\( \alpha = \dfrac{R}{2L} = \dfrac{10}{2 \times 0.4} = 12.50 \)
\( \omega = \sqrt {\dfrac{1}{L C} - \left(\dfrac{R}{2 L}\right)^2} = \sqrt {\dfrac{1}{0.4 \times 50 \times 10^{-6}} - \left(\dfrac{10}{2 \times 0.4}\right)^2} = 223.26 \)
\( i(t) = \dfrac{1}{223.26 \times 0.4} \; \sin (223.26 t) \; e^{-12.5 t} \)
Simplify
\( i(t) = 0.011 \; \sin (223.26 t) \; e^{-12.5 t} \)
The voltages may be calculated as follows
\( v_R(t) = V_0 \dfrac{2 \alpha}{\omega } \sin (\omega t) e^{-\alpha t} \)
\( \quad \quad = 0.11198 \; \sin (223.26 t) \; e^{-12.5 t} \)
\( v_L(t) = V_0 \left\{ \cos (\omega t)- \dfrac{\alpha}{\omega} \sin (\omega t) \right\} e^{-\alpha t} \)
\( \quad \quad = \left\{ \cos \left(223.26t\right) - 0.0559875 \sin (223.26t ) \right\}e^{-12.5t} \)
\( v_C(t) = V_0 - V_0 \left\{ \cos (\omega t) + \dfrac{\alpha}{\omega} \sin (\omega t) \right\} e^{-\alpha t} \)
\( \quad \quad = 1 - \left\{ \cos (223.26t) + 0.055988 \sin (223.26t) \right\} e^{-12.5t} \)
You may use the calculator for step response of a series RLC circuit to check all the above calculations.



Case 2: \( \dfrac{1}{L C} \lt \left(\dfrac{R}{2 L}\right)^2 \) : The circuit is overdamped

Let \( \beta = \sqrt { \left(\dfrac{R}{2 L}\right)^2 - \dfrac{1}{L C} } \) and rewrite \( I(s) \) as
\( I(s) = \dfrac{V_0}{\beta L} \times \dfrac{\beta}{ \left(s + \alpha \right)^2 - \beta^2 } \)
Use formulas and properties of laplace transform to find the inverse Laplace transform of \( I(s) \) as
For \( t \ge 0 \) , \( v_i (t) = V_0 \) and we have the following
\( i(t) = \dfrac{V_0}{\beta L} \; \sinh (\beta t) \; e^{-\alpha t} \)
\( \quad \quad = \dfrac{V_0}{\beta L} \; \left\{ \dfrac{e^{\beta t} - e^{\beta t}} {2} \right\} \; e^{-\alpha t} \)
\( \quad \quad = \dfrac{V_0}{2\beta L} \; \left\{ e^{ (\beta - \alpha) t} - e^{ ( - \beta - \alpha) t} \right\} \)
\( v_R(t) = R \; i(t) = \dfrac{R V_0}{2\beta L} \; \left\{ e^{ (\beta - \alpha) t} - e^{ ( - \beta - \alpha) t} \right\} \)
\( \quad \quad = V_0 \dfrac{\alpha}{\beta } \; \left\{ e^{ (\beta - \alpha) t} - e^{ ( - \beta - \alpha) t} \right\} \)
\( v_L(t) = L \; \dfrac{d i}{dt} = \dfrac{V_0}{2\beta} \left\{ (\beta - \alpha) e^{ (\beta - \alpha) t} + (\beta + \alpha) e^{ ( - \beta - \alpha) t} \right\} \)
\( v_C(t) = v_i(t) - v_R(t) - v_L(t) \)
\( \quad \quad = V_0 - V_0 \dfrac{\alpha}{\beta } \; \left\{ e^{ (\beta - \alpha) t} - e^{ ( - \beta - \alpha) t} \right\} - \dfrac{V_0}{2\beta} \left\{ (\beta - \alpha) e^{ (\beta - \alpha) t} + (\beta + \alpha) e^{ ( - \beta - \alpha) t} \right\} \)
\( \quad \quad = V_0 - V_0 \left\{ \dfrac{\beta + \alpha}{2 \beta} e^{(\beta - \alpha) t} + \dfrac{\beta - \alpha}{2 \beta} e^{(-\beta - \alpha) t} \right\} \)

Numerical Applications - Example 2 - Overdamped Circuit
Let \( V_0 = 1 \; V\) , \( R = 200 \; \Omega \) , \( L = 0.4 \; H \) and \( C = 50 \;\mu F \)
\( \dfrac{1}{L C} = 50000\)
\( \left(\dfrac{R}{2 L}\right)^2 = 62500 \)
Hence \( \dfrac{1}{L C} \lt \left(\dfrac{R}{2 L}\right)^2 \) ; the circuit is overdamped
\( \alpha = \dfrac{R}{2L} = \dfrac{200}{2 \times 0.4} = 250 \)
\( \beta = \sqrt { \left(\dfrac{R}{2 L} - \dfrac{1}{L C}\right)^2} = \sqrt {\dfrac{1}{0.4 \times 50 \times 10^{-6}} - \left(\dfrac{200}{2 \times 0.4}\right)^2} = 111.80339 \)
\( i(t) = \dfrac{V_0}{2\beta L} \; \left\{ e^{ (\beta - \alpha) t} - e^{ ( - \beta - \alpha) t} \right\} \)
\( \quad \quad = 0.01118 \; \left\{ e^{ -138.2 t} - e^{ -361.8 t} \right\} \)
\( v_R(t) = V_0 \dfrac{\alpha}{\beta } \; \left\{ e^{ (\beta - \alpha) t} - e^{ ( - \beta - \alpha) t} \right\} \)
\(\quad \quad = 2.23613 \; \left\{ e^{ -138.2 t} - e^{ -361.8 t} \right\} \)

\( v_L (t) = \dfrac{V_0}{2\beta} \left\{ (\beta - \alpha) e^{ (\beta - \alpha) t} + (\beta + \alpha) e^{ ( - \beta - \alpha) t} \right\} \)
\( \quad \quad = -0.617754 e^{ -138.2 t} + 1.617246 e^{ -361.8 t} \)

\( v_C(t) = V_0 - V_0 \left\{ \dfrac{\beta + \alpha}{2 \beta} e^{(\beta - \alpha) t} + \dfrac{\beta - \alpha}{2 \beta} e^{(-\beta - \alpha) t} \right\} \)
\( \quad \quad = 1 - 1.61806 e^{ -138.2 t} + 0.61806 e^{-361.8 t} \)
You may use the calculator for step response of a series RLC circuit to check all the above calculations.



Case 3: \( \dfrac{1}{L C} = \left(\dfrac{R}{2 L}\right)^2 \) : The circuit is critically damped

\( I(s) \) simplifies to
\( I(s) = \dfrac{1}{ \left(s + \alpha \right)^2} \dfrac{V_0}{L} \)
Use formulas and properties of laplace transform to find the inverse Laplace transform of \( I(s) \) as
For \( t \ge 0 \) , \( v_i (t) = V_0 \) and we have the following
\( i(t) = \dfrac{V_0}{ L} \; t \; e^{-\alpha t} \)
\( v_R(t) = R \; i(t) = \dfrac{R V_0}{ L} \; t \; e^{-\alpha t} \)
\( \quad \quad = 2 V_0 \alpha \; t \; e^{-\alpha t} \)
\( v_L(t) = L \; \dfrac{d i}{dt} = V_0 \left( 1 - \alpha t \right) e^{-at} \)
\( v_C(t) = v_i(t) - v_R(t) - v_L(t) \)
\( \quad \quad = V_0 - 2 V_0 \alpha \; t \; e^{-\alpha t} - V_0 e^{-at} \left( 1 - \alpha t \right) \)
\( \quad \quad = V_0 - V_0(1+\alpha t)e^{-\alpha t} \)

Numerical Applications - Example 3 - Critically damped Circuit
Let \( V_0 = 1 \; V\) , \( R = 100 \; \Omega \) , \( L = 0.4 \; H \) and \( C = 160 \;\mu F \)
\( \dfrac{1}{L C} = 15625\)
\( \left(\dfrac{R}{2 L}\right)^2 = 15625 \)
Hence \( \dfrac{1}{L C} = \left(\dfrac{R}{2 L}\right)^2 \) ; the circuit is critically damped
\( \alpha = \dfrac{R}{2L} = \dfrac{100}{2 \times 0.4} = 125 \)
\( i(t) = \dfrac{V_0}{ L} \; t \; e^{-\alpha t} \)
\( i(t) = 2.5 \; t \; e^{- 125 t} \)
\( v_R(t) = 2 V_0 \alpha \; t \; e^{-\alpha t} \)
\(\quad \quad = 250 e^{ - 125 t} \)
\( v_L(t) = L \; \dfrac{d i}{dt} = V_0 ( 1 - \alpha t ) e^{-at} \)
\( \quad \quad = (1 - 125) e^{ - 125 t} \)
\(v_C (t) = V_0 - V_0(1+\alpha t)e^{-\alpha t} \)
\( \quad \quad = 1 - (1 + 125) e^{ - 125 t} \)
You may use the calculator for step response of a series RLC circuit to check all the above calculations.



More References and Links

Solve Differential Equations Using Laplace Transform
Step Response of a Series RLC Circuit
Dirac Delta and Unit Heaviside Step Functions - Examples with Solutions