Hello all
I have a circuit that is a voltage source charging a capacitor through an inductor (with series resistance) and a diode. We know the Q factor of a simple RLC series circuit is:
\(\frac{1}{R}\sqrt{L \over C}\)
But I can't think of an elegant way to express this for the case where there is a diode. I'm happy to model the diode as a constant voltage drop Vd.
The current waveform when the 'switch is closed' is:
\(i(t) = \frac{V_{ss} - V_d}{\omega_n L} e^{-\frac{R}{2L}t} sin(\omega_n t)\)
But to calculate the total loss from this is a rather horrible integral (i(t) squared times R, plus i(t) times Vd!) and isn't much practical use when it comes out.
Any ideas? Any simplifications I could make? Ideally I'd like the proportion of energy lost as a ratio.
James
I have a circuit that is a voltage source charging a capacitor through an inductor (with series resistance) and a diode. We know the Q factor of a simple RLC series circuit is:
\(\frac{1}{R}\sqrt{L \over C}\)
But I can't think of an elegant way to express this for the case where there is a diode. I'm happy to model the diode as a constant voltage drop Vd.
The current waveform when the 'switch is closed' is:
\(i(t) = \frac{V_{ss} - V_d}{\omega_n L} e^{-\frac{R}{2L}t} sin(\omega_n t)\)
But to calculate the total loss from this is a rather horrible integral (i(t) squared times R, plus i(t) times Vd!) and isn't much practical use when it comes out.
Any ideas? Any simplifications I could make? Ideally I'd like the proportion of energy lost as a ratio.
James
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