# Full wave rectifier capacitance calculation

For making 100 amperes, 50 volts full wave rectifier, how do I calculate the circuit capacitance to avoid the ripple voltage? I mean which size of capacitor should I use?

• Note that charging the capacitor puts an additional current load on the input, which depending on where your power comes from may not be desired so you have to find a compromise by adding a resistor. – PlasmaHH Oct 14 '14 at 14:31
• In addition to the answers showing how to determine the capacitance, you would need to look at the ripple current rating of your capacitors as well, and also how your AC supply is going to cope with this kind of load. Those are large currents and voltages. If your question is more than theoretical then I suggest getting professional advice. – nekomatic Oct 14 '14 at 14:32

Here's the circuit I believe you refer to: -

And here's the formula: -

It's an approximate formula because it assumes the discharge of the capacitor between recharges is linear (it's actually exponential) but is reasonable for ripples up to 10%.

So for 100 amps with a ripple voltage of (say) 5Vp-p at 60Hz, the capacitance is 0.333 farads.

However, given your power requirement (5kW) I'd be strongly considering a 3 phase transformer and 3-ph bridge rectifier to make life easier.

You can't avoid the ripple, but you can make it small enough that you can regulate it out.

For the reservoir capacitance:

$$C=\frac{I\ t}{\Delta V}$$

Where:

C is the capacitance in farads,

I is the DC load current in amperes,

$t$ is the period of the full-wave rectified waveform, in seconds, and

$\Delta V$ is the allowable ripple across the load, in volts.

In your case, if you're working with 50Hz mains and you can stand, say, 1 volt of ripple, then

$$C=\frac{100A\times 0.01s}{1V} = {1F}$$

• It seemed strange to me that this formula doesn't have any term that brings in the peak to peak or rms voltage of the input waveform, but presumably as those go to zero C should go to zero. I googled it briefly and this formula is correct, but only in the limit where the ripple voltage is small compared to the input voltage. – Doug McClean Oct 14 '14 at 17:56