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I am calculating the output of a capacitor bank in A/s using Q = C·V, but I need to get the raw amperage. Can I do I = Q/t (in seconds)?

ocrdu
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Teslingy
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  • Depends on the parasitic resistances and inductances of the bank itself, the interconnecting wiring, and whatever the load is. Amps/s is a current ramp rate. – vir Sep 01 '22 at 17:32
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    \$I = \dfrac{dq}{dt}\$ hence, asking about amps per second is meaningless it seems. – Andy aka Sep 01 '22 at 17:33
  • "amps per second" is not usually a useful measurement unless you really do want to know what the *rate of change* of current is. But whatever it is you're wanting, you're not going to calculate it without knowing the ESR of your capacitors or specifying the load that's drawing current from the capacitors. – brhans Sep 01 '22 at 18:33

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If in fact you are trying to establish how quickly the voltage on your capacitor bank depletes when subject to a particular load current try this: -

$$Q = C\cdot V$$ $$I_{LOAD} = \dfrac{dq}{dt} = C\cdot \dfrac{dv}{dt}$$

So, if the load current is 1 amp and the capacitance is 1 farad, you would expect to see the capacitor terminal voltage fall at 1 volts per second.

Feels more like the asker wants to know how much current they'll be able to get out of a charged capacitor bank. – Scott Seidman

That depends on the effective series resistance of the capacitor (ESR). A capacitor with a low ESR can supply many tens to many thousands of amps (depending on the model). The effective series inductance can play a role too. Normally, for an ultracap it's going to be about 100 μH and this can cause problems when trying to take high load currents rapidly.

Andy aka
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