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# capacitor discharge time

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#### aashlesha

##### Newbie level 1
I want to run a dc motor by a capacitor bank.The motor needs 24volts and 1 amps. I am using capacitor bank of 0.2F charged upto 60volts. but I am unable to find for how much time the motor will ...plz help

Assuming an ideal (buck-boost) voltage converter and complete capacitor discharge, you can simply use capacitor energy storage formula.
E = 0.5*CU²

I am unable to find for how much time the motor will ...plz help
Not enough information.
Motor at full power, or motor to a complete stop ?

If at full power, you can work out the stored capacitor energy at 60 volts, and the stored energy at 24 volts.

Stored energy at 60 volts = 0.5 x 0.2F x 60v x 60v = 360 Joules
Stored energy at 24 volts = 0.5 x 0.2F x 24v x 24v = 57.6 Joules.

So you can pull out 360 - 57.6 = 302.4 Joules before the capacitor voltage falls to 24v.

That is 302.4 watt seconds, and your motor draws 24 watts (24v x 1amp).

If you use an efficient buck regulator to regulate the motor voltage to 24v, it should run for about 12.6 seconds at full speed then pretty much die after that.

Probably more like ten seconds in practice I would say

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