Capacitor sizing for Time off Delay Formula?

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Hello all,
Iv been searching the net for an answer to a problem im having and iv
found this great forum and saw a thread that came close to what I was after but just not right. So I thought that i would start a new one asking the question.
Im trying to calculate a capacitor size for a time off delay circuit. Im still trying to find out the wattage of the motors that im trying to delay so i cant post these up.
I am working with a 12v dc power source and im trying to keep these motors going for at least 40 sec. All this info is useless with out the wattage i know but the formula should still be the same.
So the real question is do anyone know the formula for capasitor sizing in a 12v dc system. If so that would be greatly appreciated.
Thanking you in advance.
Ben
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dieagothecat_at_hotmail_dot snipped-for-privacy@foo.com (DiegotheCat) wrote:

Is that 40 sec or 40 msec? 40 seconds will take a very large capacitor unless you are talking a very small motor. Normal "big" capacitors don't store much energy at all.

Energy in a capacitor is 1/2 C V^2
If you want a 1 watt motor to turn for 40 sec, that's 40 watt seconds of energy or 40 joules.
At 12 volts that's:
40 = 1/2 C 12^2
so C is .55 Farad
Or 550,000 uF - a very large capacitor.
I'm not completely sure I got all the units above correct BTW.
And this only gives you the lower limit of capacitor size because voltage drops as the charge drains so the motor will slow down and stop at some point long before all the energy is actually used. So you will need a much larger capacitor (10x the size) if you actually want to keep driving the motor near the same speed without the help of a regulator.
And of course, the wattage rating of the motor is not the same thing as the actual watts being used for a given application. You need to measure the current and voltage to calculate wattage while the motor is running in your application under the actual load you want to keep driving for 40 seconds.
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snipped-for-privacy@kcwc.com (Curt Welch) wrote:

I realized there's a fairly easy way to do the calculation to deal with the fact the voltage is dropping while the charge drains from the capacitor. You just need to to first determine how much voltage drop is acceptable in your 40 second time frame. If the answer is that the voltage can drop from 12, to 10 volts, but no further, then the capacitor will need 10 volts of charge after 40 seconds. You can calculate how much energy is left, and how much was needed at the start, and directly determine the capacitor size that way.
So, again with the same numbers as the above example:
40 j = start - end
40 = 1/2 C 12^2 - 1/2 C 10^2 40*2/(12^2-10^2) = C = 1.8 Farads, or 1,800,000 uF
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Are you really trying to put enough of a charge on a capacitor to keep the motors turning for 40 seconds? I don't know what your actual project is, but this seems very unlikely as the best way to accomplish it. Why not put a 40 second timer on a relay?
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