# Loss in transferring charge from C to L through R and diode

Discussion in 'General Electronics Chat' started by jgubby, Sep 3, 2010.

1. ### jgubby Thread Starter New Member

Sep 3, 2010
4
0
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

File size:
7 KB
Views:
11
2. ### mossman Member

Aug 26, 2010
131
3
What's the inductor for?

3. ### jgubby Thread Starter New Member

Sep 3, 2010
4
0
If a capacitor is charged straight from a voltage source, the highest efficiency that can be achieved is 50% due to the mismatch in voltage. The idea with this is, when the current is at its peak the voltage source is disconnected, and the inductor free-wheels back into the voltage source to recover the stored energy.

James

4. ### Ghar Active Member

Mar 8, 2010
655
73
I think you're stuck with your equations.

The simple fact is that the majority of electronics equations are ugly and this is especially the case when it comes to reactive elements and energy.

Not much else you can do to this aside from approximations once you get an equation out of it.

Edit:
Also, one interesting thing I discovered a few months back is that if you charge the capacitor in steps through a resistor your efficiency is much better than if you apply the entire source at once.

The reason is that when you charge up by dV you always lose 0.5C(dV)^2 in the resistor but the change in stored energy is 0.5C[(Vo + dV)^2 - Vo^2]

So, with 1F if you apply 1V you'll lose 0.5J and store 0.5J, that's 50%.
If you apply 0.5V, you lose 0.25J and store 0.25J, then if you apply 1V you'll lose another 0.25J but the stored energy increased to 1J, giving you 0.5J lost and 1J stored for 66% efficiency.

Naturally a ramp is like a huge amount of steps, giving you a good efficiency.
And of course a ramp voltage on a capacitor is like a current source... sort of like dumping current from an inductor.

Last edited: Sep 3, 2010
5. ### jgubby Thread Starter New Member

Sep 3, 2010
4
0
Ghar

Im increasingly of that view, unfortunately. I think my resistance to an approximation (for low values of t, the exponential could be approximated (1-Rt/2L) is that the losses are themselves small, and by approximating the expression I am essentially approximating the losses away!

James

6. ### Ghar Active Member

Mar 8, 2010
655
73
If you were consistently switching you can model this as you would a switching converter, since this is exactly what a boost converter does.

There are 'averaged switch' models and stuff like that, giving you some fairly simple equations for efficiency.

The problem is that those models usually assume 'small ripple' and things like that, meaning you can't be significantly changing the voltage on the capacitor.

At least that's how I remember it, haven't done it in a few years. You might look into it, the book Fundamentals of Power Electronics does it really well. The author has some slides here: