# How to calculate the capacitances for a voltage doubler

#### mahela007

Joined Jul 25, 2008
45
Hello everyone,
I'm trying to build a voltage doubler circuits. I understand how they work and also know the basics of how capacitors work. Could you show me how to calculate the appropriate capacitances for the voltage doubler circuit given the input and output voltages and the current draw?

#### Hi-Z

Joined Jul 31, 2011
158
Interesting question, and before I tried to answer I looked up the voltage doubler in Wikipedia, because I had in mind what I call the "Cockcroft-Walton" doubler/multiplier and I can never remember the circuit topology (I have to work it out from scratch).

http://en.wikipedia.org/wiki/Voltage_doubler

Well, what a surprise - there are 2 types of circuit, and my favourite circuit is should actually be called a Greinacher doubler, because he invented it before Cockcroft and Walton invented their multiplier.

Anyway, the other circuit is a bridge circuit, and I'll deal with this first because I think it's easier to understand. As I see it, you end up with a slightly-depressed output with a sawtooth ripple at twice the mains frequency. The amount of voltage depression (i.e. reduction in peak output voltage relative to the no-load doubling) will, by my reckoning, be half the ripple voltage.

The magnitude of the voltage depression and ripple is of course due to the effect of a load current, and will be given by:

Vr = t*I/C

where Vr is the ripple voltage, I the load current and t the half-period.

So, for a load current of 1mA, each C being 1uF and t = 1/100 second, we'd get a droop of 5V and a ripple of 10V.

If you find the above a bit glib, I'd urge you to try drawing out the ripple waveforms in order to get a better understanding. Please feel free to criticise, because I might have made an error!

Meanwhile I'll have a think about the other circuit...

• mahela007

#### mahela007

Joined Jul 25, 2008
45
The magnitude of the voltage depression and ripple is of course due to the effect of a load current, and will be given by:

Vr = t*I/C

where Vr is the ripple voltage, I the load current and t the half-period.

So, for a load current of 1mA, each C being 1uF and t = 1/100 second, we'd get a droop of 5V and a ripple of 10V.
Thanks for the reply. I think I understand the bit I've quoted above. Basically, the equation you have provided is derived from the definition of capacitance isn't it? So ripple voltage = (loss of charge from cap) / capacitance.

But what do you mean by 'half period'? Is it 'half of the period of the input AC voltage"?
How did you arrive at the conclusion that the voltage drop will be half the ripple voltage?

#### Hi-Z

Joined Jul 31, 2011
158
Actually, I made a mistake - the decay period is a full cycle, so t would be the period of the input waveform, not half the period. So, in the example I gave, which referred to 50Hz mains, t would actually be 1/50 second, and the resulting ripple would be 20V, the "droop" 10V. Ripple frequency is at 100Hz though.

You have to view the way the circuit works: each capacitor takes turns to be be fully charged via its associated diode. This charging process occurs for a small fraction of the half-cycle in which the diode conducts (the diode conducts only when the input voltage exceeds the voltage across the slightly-discharged capacitor). When one capacitor is being charged, the other is halfway through the process of being discharged.

Of course, the output voltage is the sum of the voltages across the capacitors, and you'll see that the peak of the ripple corresponds to whenever one of the capacitors is being charged. Because the other capacitor is halfway through its discharge at this point in time, the output voltage is less than it would have been had there been no load current - hence the "droop".

As I say, it's something which is best sketched on a piece of paper. If its still a bit unclear, let me know and I'll maybe post a sketch.

#### mahela007

Joined Jul 25, 2008
45
Ok.. I'll ponder this on my own for a while and post back in a couple of hours. thanks again..

#### freefireant

Joined Mar 15, 2016
11
It's been a couple of hours