How to figure out the maximum throughput on a capacitor?

How can I figure out what the maximum throughput of a capacitor is?

(I have a 20V DC power source that is very unstable. When you start drawing from it, the voltage briefly drops. I'm using some circuitry that can't handle this drop. This is why I need to smooth it out using a capacitor, and need a capacitor that can handle a continuous throughput of 30A at 20V.)

• Define "throughput of a capacitor"; it's a phrase I've not heard before and, I suspect, a notion that isn't well defined. – Alfred Centauri Apr 24 '14 at 22:06
• The impedance of the load is? The maximum allowable voltage drop is? Precisely how unstable is "very unstable"? – Ignacio Vazquez-Abrams Apr 24 '14 at 22:07
• @AlfredCentauri I'm not an electrical engineer, so I don't know the terms. The capacitor is going to be used to smooth out the voltage on the 20V power supply. There's going to be a 30A draw from the power supply. If I'm not wrong here, this means the capacitor will have to supply 30A for the time that the power supply has a voltage drop. Again, I'm no electrical engineer. But if the capacitor is being constantly charged and discharged (0.01F), how can I know how much it will handle before getting too hot? – Friend of Kim Apr 24 '14 at 22:11
• @IgnacioVazquez-Abrams I don't know how unstable it is as I have no instrument to measure it. – Friend of Kim Apr 24 '14 at 22:11
• @AlfredCentauri To simplify the comment above: If the capacitor is charged, then discharged at a rate of 30A, then charged again forever. How does one know how many amps can be drawn this way without overloading the capacitor? – Friend of Kim Apr 24 '14 at 22:13

If you can tolerate X% spike or dip from charging a load cap, then the cap on the source must be 1/X times bigger. Like 10x bigger. Bad idea.

So if load cap is 0.01F (huge) and assuming this cap has a very very low series resistance {ESR spec} the rate of voltage charging will be limited by your supply current.

> Formula=> Time to reach ~ 63% difference voltage change =T = Volts/Amps * Farads [sec] = "time constant"

Thus T = 20V/10A *0.01F = 0.02 seconds However if load is already using 8A leaving 2A spare , it will take 5x longer = 100ms and if 63~ % is nit high enough , you need to wait another couple time constants, to get to 95% or 3T= 300ms in this hypothetical scenario.

To reduce this down x10 in time you need a current 10x bigger or 100A, which would over heat the cap. (Fizz boom)

S0 the cap needs to tolerate 200Watts for 20ms.

Not good idea, To increase current and reduce dip duration.

Better idea is use an inrush current limiter (metal oxide ICL .. cheap ) rated for X Amps and wait as long as it takes to safely charge up this supercap.

Need specs on part to check.

May I ask why you need to make a rapid charge?

• I have a charger that needs a steady 17V supply. It is going to charge some lipo batteries. The PSU using is a 12V battery charger. The problem is when the lipo charger increases the charge current, the voltage from the PSU drops and the lipo charger gets problems. – Friend of Kim Apr 25 '14 at 6:30
• I don't have any advanced measurement devices, and found out about this by testing a 12V battery as the PSU, which worked fine. This is why I need a way to smooth the voltage from the PSU. – Friend of Kim Apr 25 '14 at 6:32