[SOLVED] Formula for Hold up capacitor

DC/DC converter:
Input voltage 72V
Output voltage 3.3V and current 22 A.
Output accepted 3V (10%-discharge)

Can you please tell me a generic formula to calculate the Hold up capacitor?

i=C*(dv/dt).

But you should already know that.

But to hold a charge for 1ms,The capacitor value is coming around 73mF.
C = I x (dt/dV) = 22 * (1ms/3.3-3) = 73mF.
It is too high value and the Package is too big.
Is there any other suggestion to reduce the capacitor value for the same configuration.

You can't fight the laws of physics. But I'm not sure I quite understand your application. Aren't you supplying continuous current from the input? Assuming 100% efficiency, you should be able to maintain a constant output voltage with 1Amp input; in other words, all the output current doesn't have to come from the cap, does it?

Thanks for your reply.

Our application is N+1 redundant power supply.It has two identical power module,each has a capability of providing complete power.
The load will take the power from the first power module.When this module fails,then it will switch to second power module.But the switching time is in terms of millisecond.To maintain continuous output we have to compensate this switching time.So how to do this?

If you can't reduce the switching time, you will need a large value of capacitance. It doesn't have to be big though. For example, you could use 12 of these 10000uF caps. That would cost about $20USD and fit into a space 5cm * 6.5cm.

BTW, don't forget to take ESR into account in your calculations. That's why I suggested 12 instead of 8.

Yeah this is very good idea.
But why i shouldn't take exact number.12 not 8.
Can you please tell me how ESR affects.

From your calculation, you need 73000uF, so eight 10000yF caps should be enough if the caps are perfect, but they're not.

Electrolytic capacitors have some series resistance (ESR). The caps I linked have ESR = about 0.06Ω. So with 12 in parallel, you effectively get 12000uF in series with 0.005Ω.

As soon as you start to draw 22A current from the caps, the voltage will drop by 22A * 0.005Ω = 0.11V. So you start with 3.19V instead of 3.3V.

Then, after 1mS, the voltage has dropped by V = 22A * 1mS / 120000uF = 183mV. So the actual voltage after 1mS = 3.19V - 0.183V = 3.007V.

Actually, that's too close to your minimum, especially because the cap tolerance is only +-20%. It would be better to use 15 or 16 of them, I think, to be safe. (I'm choosing numbers that are easy to lay out neatly in rows and columns e.g. 3*5 or 4*4).

Another thing to look for in the capacitor spec is the ripple current rating. These ones are OK, they can handle more than 2A each (I forget exactly).

There's come advantages to using lots of small caps in parallel instead of one big one. Firstly, it results in lower inductance (and maybe lower resistance too). Also, if one cap explodes, the whole system doesn't die - 14 caps works almost as well as 15.

I often works out cheaper too. I didn't check but it might cost less if you use more capacitors of lower value, or fewer capacitors of higher value.

Should I use hold up capacitor in the input side of the DC - DC module or in the Output side of the module.

You don't really care about the voltage on the input side, it's the OUTPUT you want to keep constant, right? So, put the cap on the output.