Diode Switched LC Load:
\[\begin{aligned} V_s & =L\frac{di}{dt}+\frac{1}{C}\int_{t_0}^tidt+v_c(t=0) \\ &\text{With initial conditions }i(t=0)=0~\mathrm{and}~v_{c}(t=0)=0 \\ i\left(t\right)&=V_{s}\sqrt{\frac{C}{L}}\sin\omega_{0}t\\ &=I_{p}\sin\omega_{0}t \qquad \omega_0 = 1/\sqrt{LC}\\ \end{aligned}\]
\[\begin{aligned} I_p & =V_s\sqrt{\frac{c}{L}}\\ \frac{di}{dt} & =\frac{V_s}{L}\cos\omega_0t \\ \left.\frac{di}{dt}\right|_{t=0} & =\frac{V_s}{L} \\ v_c(t) & =\frac{1}{C}\int_0^tidt=V_s(1-\cos\omega_0t)\\ \end{aligned}\] At \(t=t_1=\pi\sqrt{LC}\) the diode current \(i\) falls to zero and the capacitor is charged to 2\(V_s\).
In an LC circuit without resistance (no energy loss), energy transfer occurs between C and L due to oscillating current.
Diode in series with the switch, prevents negative current flow
Without Diode, the LC circuit would oscillate indefinitely.
Electronic switches (BJT, MOSFET, IGBT) typically block reverse current flow.
The switch and diode simulate electronic switch behavior.
Capacitor output can be connected to similar circuits with switches and diodes to obtain multiples of the DC supply voltage for high-voltage applications like pulse power and superconducting systems.
Solved Problem:
A diode circuit with an \(LC\) load has initial capacitor voltage,\(V_c(t=0)=-V_{c0}=V_0-220V;\) capacitance, \(C=20~ \mu F\); and inductance, \(L=80~\mu H\). If switch \(S_{1}\) is closed at \(t=0\), determine
the peak current through the diode,
\[\begin{aligned} &L\frac{di}{dt}+\frac{1}{C}\int_{t_{0}}^{t}idt+v_{c}(t=0)=0 \\ &i(t)=V_{c0}\sqrt{\frac{C}{L}}\sin\omega_{0}t \\ &\omega_{0}=1/\sqrt{LC}=10^{6}/\sqrt{20\times80}=25{,}000\mathrm{rad/s}. \\ &I_{p}=V_{c0}\sqrt{\frac{C}{L}}=220\sqrt{\frac{20}{80}}=110\mathrm{A} \end{aligned}\]
the conduction time of the diode \[t_1=\pi\sqrt{LC}=\pi\sqrt{20\times80}=125.66\mu s\]
the final steady-state capacitor voltage.
\[\begin{aligned} v_{c}(t) &=\frac{1}{C}\int_{0}^{t}idt-V_{c0}=-V_{c0}\cos\omega_{0}t\\ \mathrm{For}~t &=t_{1}=125.66\mathrm{\mu s},v_{c}(t=t_{1})\\ &=-220\cos\pi=220\mathrm{V}. \end{aligned}\]