Diode Switched RC Load:
An ideal diode with RC Load
Ideal means reverse recovery time (\(t_{rr}\)) and forward voltage drop (\(V_{D}\)) are negligible \[t_{rr} = 0 ~\text{and} ~ V_D = 0\]
The source voltage (\(V_s\)) is a dc constant voltage
Switch \(S_1\) closed at \(t=0\), the charging current \(i\) flows
\[\begin{aligned} & V_s=v_R+v_c=v_R+\frac{1}{C} \int_{t_0}^t i d t+v_c(t=0) \\ & v_R=R i \end{aligned}\]
With initial condition \(v_c(t=0)=0\), gives the charging current \(i\) as \[i(t)=\frac{V_s}{R} e^{-t / R C}\]
The capacitor voltage \(v_c\) is \[v_c(t)=\frac{1}{C} \int_0^t i d t=V_s\left(1-e^{-t / R C}\right)=V_s\left(1-e^{-t / \tau}\right)\]
\(\tau=R C\) is the time constant of an \(R C\) load.
The rate of change of the capacitor voltage is \[\frac{d v_c}{d t}=\frac{V_s}{R C} e^{-t / R C}\]
The initial rate of change of the capacitor voltage (at \(t=0\) ) is \[\left.\frac{d v_c}{d t}\right|_{t=0}=\frac{V_s}{R C}\]
The current \(i\) is unidirectional and does not tend to change its polarity, the diode has no effect on circuit operation.
Problem:
A diode circuit with \(R=44 \Omega\) and \(C=0.1 \mu \mathrm{F}\) load has initial capacitor voltage, \(V_{c 0}=V_c(t=0)=220 \mathrm{~V}\). If switch \(S_1\) is closed at \(t=0\), determine
the peak diode current
the energy dissipated in the resistor \(R\)
the capacitor voltage at \(t=2 \mu \mathrm{s}\).
Solution:
The peak diode current \(I_p\) is \[I_P=\frac{V_{c 0}}{R}=\frac{220}{44}=5 \mathrm{~A}\]
The energy \(W\) dissipated is \[W=0.5 C V_{c 0}^2=0.5 \times 0.1 \times 10^{-6} \times 220^2=0.00242 \mathrm{~J}=2.42 \mathrm{~mJ}\]
For \(R C=44 \times 0.1 \mu=4.4 \mu \mathrm{s}\) and \(t=t_1=2 \mu \mathrm{s}\), the capacitor voltage is \[v_c(t=2 \mu \mathrm{s})=V_{c 0} e^{-t / R C}=220 \times e^{-2 / 4.4}=139.64 \mathrm{~V}\]