Diode Switched RLC Load:
\[\begin{aligned} & L\frac{di}{dt}+Ri+\frac{1}{C}\int idt+v_c(t=0)=V_s \\ & \frac{d^2i}{dt^2}+\frac{R}{L}\frac{di}{dt}+\frac{i}{LC}=0 \Rightarrow s^2+\frac{R}{L}s+\frac{1}{LC}=0 ~~ \Leftarrow \textbf{characteristic equation}\\ & s_{1,2}=-\dfrac{R}{2L}\pm\sqrt{\left(\dfrac{R}{2L}\right)^2-\dfrac{1}{LC}}\\ &\alpha=\frac{R}{2L} ~~\Leftarrow \textbf{Damping factor} \qquad \omega_0=\frac{1}{\sqrt{LC}}~~\Leftarrow \textbf{Resonant frequency}\\ & s_{1,2}=-\alpha\pm\sqrt{\alpha^2-\omega_0^2}~~ \Leftarrow \textbf{Roots of the characteristic equation}\\ \end{aligned}\]
If \(\alpha=\omega_0\), the roots are equal, \(s_1=s_2\), and the circuit is called critically damped. The solution takes the form \[i(t)=\left(A_1+A_2 t\right) e^{s_1 t}\]
If \(\alpha>\omega_0\), the roots are real and the circuit is said to be overdamped. The solution takes the form \[i(t)=A_1 e^{s_1 t}+A_2 e^{s_2 t}\]
If \(\alpha<\omega_0\), the roots are complex and the circuit is said to be underdamped. The roots are \[s_{1,2}=-\alpha \pm j \omega_r\] where \(\omega_r\) is called the ringing frequency (or damped resonant frequency) and \(\omega_r=\sqrt{\omega_0^2-\alpha^2}\). The solution takes the form \[i(t)=e^{-\alpha t}\left(A_1 \cos \omega_r t+A_2 \sin \omega_r t\right)\] which is a damped or decaying sinusoidal.
A switched underdamped \(RLC\) circuit is used to convert a dc supply voltage into an ac voltage at the damped resonant frequency.
The constants \(A_1\) and \(A_{2}\) can be determined from the initial conditions of the circuit.
Solving for these two constants requires two boundary equations at \(i(t=0)\) and \(di/dt(t=0)\) .
The ratio of \(\alpha/\omega_0\) is commonly known as the damping ratio, \(\delta=R/2\sqrt{C/L.}\)
Power electronic circuits are generally underdamped such that the circuit current becomes near sinusoidal, to cause a nearly sinusoidal ac output or to turn off a power semiconductor device.
For critical and underdamped conditions, the current \(i(t)\) will not oscillate and there is no need for the diode.
Equations are the general forms for the solution of any second-order differential equations. The particular form of the solution will depend on the values of \(R, L,\) and \(C.\)
Solved Problem:
The second-order \(RLC\) circuit has the dc source voltage \(V_s=220\mathbb{V}\), \(L= 2\)mH, \(C=0.05~\mu F\), and \(R=160\Omega\). The initial value of the capacitor voltage is \(v_c(t=0)=V_{c0}=0\) and inductor current \(i( t= 0) = 0.\) If switch \(S_1\) is closed at \(t=0\), determine:
an expression for the current \(i(t)\)
\[\begin{aligned} \alpha &=R / 2 L=160 \times 10^3 /(2 \times 2)=40,000 \mathrm{rad} / \mathrm{s} \\ \omega_0 & =1 / \sqrt{L C}=10^5 \mathrm{rad} / \mathrm{s}\\ \omega_r & =\sqrt{10^{10}-16 \times 10^8}=91,652 \mathrm{rad} / \mathrm{s} \\ \alpha&<\omega_0 \Leftarrow ~\text{under-damped circuit }\\ i(t) & =e^{-\alpha t}\left(A_1 \cos \omega_r t+A_2 \sin \omega_r t\right) \end{aligned}\]
At \(t=0, i(t=0)=0\) and this gives \(A_1=0\). The solution becomes \[i(t)=e^{-\alpha t} A_2 \sin \omega_r t\]
\[\begin{aligned} \frac{d i}{d t} & =\omega_r \cos \omega_r t A_2 e^{-\alpha t}-\alpha \sin \omega_r t A_2 e^{-\alpha}\\ \left.\frac{d i}{d t}\right|_{t=0}&=\omega_r A_2=\frac{V_s}{L}\\ A_2 & =\frac{V_s}{\omega_r L}=\frac{220 \times 1,000}{91,652 \times 2}=1.2 \mathrm{~A}\\ i(t) & =1.2 \sin (91,652 t) e^{-40,000 t} \mathrm{~A} \end{aligned}\]
the conduction time of diode.
The conduction time \(t_1\) of the diode is obtained when \(i=0\). That is, \[\omega_{\mathrm{r}} t_1=\pi \quad \text { or } \quad t_1=\frac{\pi}{91,652}=34.27 \mu \mathrm{s}\]