Hello,

I have a small question that I would like some help with.
Normal electrolytic capacitors, how much power can they generally handle?
For example if I want a capacitance of 330uF and the power to be delivered is around 25W. Is it possible with normal capacitors or would I need super capacitors?

Thank you!

 

If you are talking about how much energy can be stored in a capacitor (sorta like a battery), then you need to know four things:

1. The capacitance value
2. The load current
3. the peak capacitor voltage before the load starts drawing current
4. the minimum capacitor voltage when it is too low to power the load.

As you draw energy from a capacitor, its terminal voltage decreases. This means that the voltage presented to whatever the capacitor is powering is changing constantly, downward. So you have to know the starting voltage *and* the lowest useful voltage. Once you have these numbers, use this equation:

Total energy = 1/2 x C x V^2

One-half-C-Vsquared is the total energy in a capacitor. A 330 uF cap charged up to 100 V has 1.65 watt-seconds of total energy. If you discharge the capacitor all the way down to 0 V, it will power a 1 watt load for 1.65 seconds. If your load will run only down to 25 V, then you use the equation again.

TE (100 V) = .5 x .00033 x 10000 =
TE (25 V) = .5 x .00033 x 625 =

Subtract those two numbers and you have the total energy available to your load in watt seconds. With a 25 W load, that works out to 62 milliseconds.

ak

 

@AnalogKid Thank you for your reply. I am familiar with this formula, but now my actual question would be whether the capacitor can actually withstand the current drawn or not. The surge drawn by the load is 0.45A for about 75ms. The minimum voltage is 46V and max is 57V. Would the normal 330uF electrolytic capacitor work or do I need a super capacitor?

 

The capacitor datasheet will list its max ripple current, usually in rms. From this you can derive a safe peak current. Non-repetitive or widely separated peaks can be higher, but I don't have a rule of thumb for that kind of undocumented extension. You also can use the cap's ESR to calculate internal resistive heating.

ak

 

I get 2170 uf for the answer. Available to purchase for about $2 @www.mouser.com
With a load of about 120 ohms, the ESR would not be a problem.

Anybody that knows the answer is welcome to correct the number I derived from my power supply formula √2 C Erip(p-p) F = I

 

11 V delta-V (57 V down to 46 V), at 0.45 A, for 75 ms:

Total energy method for 2200 uF > 16.6 W, which is 33.6% less than 25 W.

Constant current method for 0.45 A > 15.3 V delta-V, which is 39% greater than 11 v.

Therefore, 2200 uF is not enough, by approximately 30-40%.

2200 uF x 1.4 = 3080 uF

Constant current method yields 3069 uF. Use a 3300 uF, 100 V capacitor.

ak