RLC circuit

From Academic Kids

An RLC circuit is a kind of electrical circuit composed of a resistor (R), an inductor (L), and a capacitor (C). See RC circuit for the simpler case. A voltage source is also implied. It is called a second-order circuit or second-order filter as any voltage or current in the circuit is the solution to a second-order differential equation. Since the circuit components are assumed ideal and are linear, an RLC circuit is an example of an electrical harmonic oscillator.

The resonant or center frequency of such a circuit (in hertz) is:


f_c = {1 \over 2 \pi \sqrt{L C}} <math>

It is a form of bandpass or bandcut filter, and the Q factor is


Q = {f_c \over BW} = {2 \pi f_c L \over R} = {1 \over \sqrt{R^2 C / L}} <math>

There are two common configurations of RLC circuits: series and parallel.


Series RLC Circuit

In this circuit, the three components are in series with the voltage source. An RLC series circuit has a resonant frequency and is often used as a model for analysing electronic circuits, such as calculating impedance.

Missing image
Image:RLC series circuit.png

Where the notations in the figure above are:

  • V - the voltage of the power source (measured in volts V)
  • I - the current in the circuit (measured in amperes A)
  • R - the resistance of the resistor (measured in ohms = V/A)
  • L - the inductance of the inductor (measured in henries = Vs/A)
  • C - the capacitance of the capacitor (measured in farads = C/V = As/V)

Given the parameters V, R, L, and C, the solution for the current (I) using Kirchoff's voltage law (or KVL) is:


{V_R+V_L+V_C=V} \, <math>

For a time-changing voltage V(t), this becomes


RI(t) + L { {dI} \over {dt}} + {1 \over C} \int_{-\infty}^{t} I(\tau)\, d\tau = V(t) <math>

Rearranging the equation will result in the following second order differential equation:


{{d^2 I} \over {dt^2}} +{R \over L} {{dI} \over {dt}} + {1 \over {LC}} I(t) = {1 \over L} {{dV} \over {dt}} <math>

The ZIR (Zero Input Response) solution

Nullifying the input (voltage sources) we get the equation:


{{d^2 I} \over {dt^2}} +{R \over L} {{dI} \over {dt}} + {1 \over {LC}} I(t) = 0 <math>

with the initial conditions for the inductor current, IL(0), and the capacitor voltage VC(0). However, in order to solve the equation properly, the initial conditions needed are I(0) and I'(0).

The first one we already have since the current in the main branch is also the current in the inductor, therefore


I(0)=I_L(0) \, <math>

The second one is obtained employing KVL again:


V_R(0)+V_L(0)+V_C(0)=0 \, <math>


\Rightarrow I(0)R+I'(0)L+V_C(0)=0 \, <math>


\Rightarrow I'(0)={1 \over L}\left[-V_C(0)-I(0)R \right] <math>

We have now a homogeneous second order differential equation with two initial conditions. Usually second order differential equations are written as:


I''+2\xi \omega_k I' + \omega_k^2 I = 0 <math>

In case of an electrical circuit ωk > 0 and therefore, there are three possible cases:

Over damping


\alpha>1 \Rightarrow R^2 C>4L \, <math>

In this case, the characteristic polynomial's solutions are both negative real numbers. This is called "over damping":

Missing image

Two negative real roots, the solutions are:


I(t)=a e^{\lambda_1 t} + b e^{\lambda_2 t} <math>

Critical damping


\alpha=1 \Rightarrow R^2 C=4L \, <math>

In this case, the characteristic polynomial's solutions are identical negative real numbers. This is called "critical damping":

Missing image

The two roots are identical (<math> \lambda_1=\lambda_2=\lambda <math>), the solutions are:

<math>I(t)=(a+bt) e^{\lambda t}


Under damping


\alpha<1 \Rightarrow R^2 C<4L \, <math>

In this case, the characteristic polynomial's solutions are complex conjugate and have negative real part. This is called "under damping":


Two conjugate roots (<math>\lambda_1 = \bar \lambda_2 = \alpha + i\beta<math>), the solutions are:


I(t)=e^{\alpha t} \left[ a \sin(\beta t) + b \cos(\beta t) \right] <math>

The ZSR (Zero State Response) solution

This time we nullify the initials conditions and stay with the following equation:


\left\{\begin{matrix} {{d^2 I} \over {dt^2}} +{R \over L} {{dI} \over {dt}} + {1 \over {LC}} I(t) = {1 \over L}{{dV} \over {dt}} \\ I(0^{-})=I'(0^{-})=0 \end{matrix}\right. <math>

Separate solution for every possible function for V(t) is impossible, however, there is a way to find a formula for I(t) using convolution. In order to do that, we need a solution for a basic input - the Dirac delta function.

In order to find the solution more easily we will start solving for the Heaviside step function and then using the fact our circuit is a linear system, its derivative will be the solution for the delta function.

The equation will be therefore, for t>0:


\left\{\begin{matrix} {{d^2 I_u} \over {dt^2}} +{R \over L} {{dI_u} \over {dt}} + {1 \over {LC}} I_u(t) = 0 \\ I(0^{+})=0 \qquad I'(0^{+})={1 \over L} \end{matrix}\right. <math>

Assuming λ1 and λ2 are the roots of


P(R)= R^2+2\xi \omega_k R + \omega_k^2 <math>

then as in the ZIR solution, we have 3 cases here:

Over Damping

Two negative real roots, the solution is:


I_u(t)={1 \over {L(\lambda_1-\lambda_2)}} \left[ e^{\lambda_1 t}-e^{\lambda_2 t} \right] <math>


\Rightarrow I_{\delta}(t)={1 \over {L(\lambda_1-\lambda_2)}} \left[ \lambda_1 e^{\lambda_1 t}-\lambda_2 e^{\lambda_2 t} \right] <math>

Critical Damping

The two roots are identical (<math> \lambda_1=\lambda_2=\lambda <math>), the solution is:


I_u(t)={1 \over L} t e^{\lambda t} <math>


\Rightarrow I_{\delta}(t)={1 \over L} (\lambda t+1) e^{\lambda t} <math>

Under Damping

Two conjugate roots (<math>\lambda_1 = \bar \lambda_2 = \alpha + i\beta<math>), the solution is:


I_u(t)={1 \over {\beta L}} e^{\alpha t} \sin(\beta t) <math>


\Rightarrow I_{\delta}(t)={1 \over {\beta L}} e^{\alpha t} \left[ \alpha \sin(\beta t) + \beta \cos(\beta t) \right] <math>

(to be continued...)

Sinusoidal steady state analysis

The series RLC can be analyzed in the frequency domain using complex impedance relations. If the voltage source above produces a pure sine wave with amplitude V and angular frequency ω, KVL can be applied:

<math>V = I \left ( R + j \omega L + \frac{1}{j \omega C} \right ) <math>

Where I is the complex current through all components. Solving for I:

<math>I = \frac{V}{ R + j \omega L + \frac{1}{j \omega C} } <math>

Taking the magnitude of the above equation:

<math>I_{mag} = \frac{V}{\sqrt{ R^2 + \left ( \omega L - \frac{1}{\omega C} \right )^2 }} <math>

If we choose trivial values where R = 1, C = 1, L = 1, and V = 1, then the graph of magnitude of current as a function of ω is:

Missing image

Note that there is a peak at ω = 1. This is known as the resonant frequency. Solving for this value, we find:

<math>\omega_o = \frac{1}{\sqrt{L C}} <math>

Parallel RLC Circuit

Missing image
Image:RLC parallel circuit.png

A much more elegant way of recovering the circuit properties of an RLC circuit is through the use of nondimensionalization.

For a parallel configuration of the same components, where Φ is the magnetic flux in the system

<math> C \frac{d^2 \Phi}{dt^2} + \frac{1}{R} \frac{d \Phi}{dt} + \frac{1}{L} \Phi = I_0 \cos(\omega t) \Rightarrow \frac{d^2 \chi}{d \tau^2} + 2 \zeta \frac{d \chi}{d\tau} + \chi = \cos(\Omega \tau) <math>

with substitutions

<math>\Phi = \chi x_c, \ t = \tau t_c, \ x_c = L I_0, \ t_c = \sqrt{LC}, \ 2 \zeta = \frac{1}{R} \sqrt{\frac{L}{C}}, \ \Omega = \omega t_c . <math>

The first variable corresponds to the maximum magnetic flux stored in the circuit. The second corresponds to the period of resonant oscillations in the circuit.

The expressions for the linewidth in the series and parallel configuration are inverses of each other. This is particularily useful for determining whether a series or parallel configuration is to be used for a particular circuit design. However, in circuit analysis, usually the reciprocal of the latter two variables are used to characterize the system instead. They are known as the resonant frequency and the Q factor respectively.

See also

de:Schwingkreis fr:Circuit RLC it:Circuito RLC zh:RLC电路


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