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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}} <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.
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:
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>
\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":
\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":
\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 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>
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>
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...)