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Hi, I have a very basic electronics question. I built a DC switch that has 4 70A 2 pole contactors the basic idea is: a 2 position rotary switch lets me switch between 2) 24V battery banks, the contactors are wired to allow selection between bank "A" and Bank "B" each consisting of 2) 24V batteries (for a total of 4 batteries). The outputs are wired in common so that either bank "A" is directly connected, or bank "B" is, as 2) 24V sources. I put a large motor run capacitor across each output pair to attempt to eliminate the short period of time the output drops to 0V during switching since there are some digital loads that are resetting during the switch. I need to hold the voltage up during the 25/50 ms the contacts are open. The load could be as much as 1400 watts so I'll need (potentially) to supply 60 amps for 50 ms during the switch - any advice? TIA, - James

Two ways to solve it.

For a capacitor with a constant C value, you need to extract from it: $${ \Delta}Q= 60 A {\times} 0.05 s = 3 Coulomb$$

The capacitor will necessary drop its voltage from 24 V to ... ? (24 V because we can assume the cap will start while fully charged). Assuming that you allow a voltage drop of 1 Volt after these 50 ms, taking the standard law for a cap:

Q = C V

or

$${ \Delta}Q = C {\Delta}V$$

we solve:
3 [Coulomb] = C [Farad] * 1 [Volt].

C=3 Farad

The second way it through the energy:

1440 W * 0.05 s = 72 Joule

Energy from the Capacitor:

E = 0.5 C V²

Start with

E=0.5 C 24²

End with

E=0.5 C 23²

for a difference of 72:

72 = 0.5 (24² - 23²) C

C= 3.8 Farad

(larger, since the computed 1440 W = 24 V * 60A is based on a constant 24V, which is over-evaluated).

In both cases, that predicts a LARGE capacitor (and many super-caps do not match the assumptions leading to Q = C V and even less E = 0.5 C V², and have a relatively high internal resistance which will consume energy too). I would have use a safety factor of 3 if that would not have led to capacitor of 10F.

You can play with a different final acceptable voltage than here the assumed 23 V.

Oops, and super cap do not support 24V, in general.

I'm ignoring your question of course, but why are you switching banks in the first place? Are they removeable?

@vanderghast has the right idea with the math as well as his conclusions.

Why not considering that the two batteries be in parallel for 50 ms: Flipping the switch connect on the second battery but only close (through a transistor) the first one after 50 ms ? No cap required.