rlc circuit

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

f_c = {1 \over 2 \pi \sqrt{L C}} 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}} 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.

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} \, 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) Rearranging the equation will result in the following second order differential equation:

  +{R \over L}  + {1 \over {LC}} I(t) = {1 \over L}

The ZIR (Zero Input Response) solution

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

  +{R \over L}  + {1 \over {LC}} I(t) = 0

with the initial conditions for the inductor current, I_L(0), and the capacitor voltage V_C(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) \, The second one is obtained employing KVL again:

V_R(0)+V_L(0)+V_C(0)=0 \,

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

\Rightarrow I'(0)={1 \over L}\left\right 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 In case of an electrical circuit ω_k > 0 and therefore, there are three possible cases:

Over damping

\alpha>1 \Rightarrow R^2 C>4L \, In this case, the characteristic polynomial's solutions are both negative real numbers. This is called "over damping":

Two negative real roots, the solutions are:

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

Critical damping

\alpha=1 \Rightarrow R^2 C=4L \, In this case, the characteristic polynomial's solutions are identical negative real numbers. This is called "critical damping":

The two roots are identical (

\lambda_1=\lambda_2=\lambda

), the solutions are:

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

Under damping

\alpha<1 \Rightarrow R^2 C<4L \, In this case, the characteristic polynomial's solutions are complex conjugate and have negative real part. This is called "under damping":

Two conjugate roots (

\lambda_1 = \bar \lambda_2 = \alpha + i\beta

), the solutions are:

I(t)=e^{\alpha t} \leftsin(\beta t) + b cos(\beta t) \right

The ZSR (Zero State Response) solution

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

\left\{\begin{matrix} +{R \over L} + {1 \over {LC}} I(t) = {1 \over L} \\ I(0^{-})=I'(0^{-})=0 \end{matrix}\right. 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} +{R \over L} + {1 \over {LC}} I_u(t) = 0 \\ I(0^{+})=0 \qquad I'(0^{+})={1 \over L} \end{matrix}\right. Assuming λ₁ and λ₂ are the roots of

P(R)= R^2+2\xi \omega_k R + \omega_k^2 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)}} \leftt}-e^{\lambda_2 t} \right

\Rightarrow I_{\delta}(t)={1 \over {L(\lambda_1-\lambda_2)}} \lefte^{\lambda_1 t}-\lambda_2 e^{\lambda_2 t} \right

Critical Damping

The two roots are identical (

\lambda_1=\lambda_2=\lambda

), the solution is:

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

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

Under Damping

Two conjugate roots (

\lambda_1 = \bar \lambda_2 = \alpha + i\beta

), the solution is:

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

\Rightarrow I_{\delta}(t)={1 \over {\beta L}} e^{\alpha t} \leftsin(\beta t) + \beta cos(\beta t) \right (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:

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

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

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

Taking the magnitude of the above equation:

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

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

\omega_o = \frac{1}{\sqrt{L C}}

Rlc Circuit