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| I currently have a 24v circuit that pull only about 1/4 amp. Lets say I want to be able to run this circuit for 10 minutes off of a supercapacitor. I have not found any formulas that seem to link amp/hour with capacitance. Any suggested equations or places to start looking? Thanks! | |
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| Try using the time constant C = R x C - I suspect your requirements are FAR above what a super capacitor can provide though. | |
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| The equation for a constant current is C= (I x t)/V where I is the current, t is the time in seconds, and V is the amount of voltage drop you can tolerate before the circuit quits. For example for 1/4 amp, 10 minutes, and a 4V drop (24V to 20V) the required capacitance is 37.5 Farads, a rather large capacitor. | |
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| And a high voltage one too, most supercaps are low voltage. | |
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| United Chemi-con makes some very large capacitors. Some are used in cars. After a car crash you may not have 12 volts to make the electric windows work. They make capacitors that hold enough charge to run the window motors once. Why not use a battery? | |
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| Years ago I had some 5F low voltage capacitors. The charge and discharge curves are not like a capacitor but like a battery. I think they are actually batteries. The UCC capacitors are real capacitors. I know. You need to look at the current rating on the caps. You may not get them charged in 10 seconds. | |
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| Might have a high ESR too. They're generally designed for clock backup. | |
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| I've got over 1000 supercaps in stock at the moment The temptation to hook them all up in parallel rather than selling them is immense ..... | |
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