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I want to power a laptop on supercapacitors. How would i do it with 2.5-2.7 v caps? what capacitance will i need? Answered

battery: 10.8 v, 47 whr
the charger says 18.5 v output, 3.5 A

Laptop: Presario f500, Turion 64 Mk-36 2.0 ghz (800 mhz or less for most tasks) 2 gb ram, 100 gb hdd, ubuntu 9.10. i think its supposed to last aroun 2 1/2 hours on a charge, but the battery's really old so I can't know for certain. i want to switch to capacitors so i don't have to buy a new battery every couple of years.


Lemonie's right, but for the hell of it, here's some rough maths.

Assume the thing pulls 25 Watts, and it takes 18V from the supply and a current of 25/18 amps. Assume you're right, and it'd last 2.5 hours, or 9000 seconds.

Current = Charge/time, 1A is one Couloumb per second  - we need 25/18 C / sec, so for 9000 seconds we need 25 x 9000/18 Couloumb - 12500 Couloumb

For a capacitor, Q = CV, 12500 = C x 18 .....rearrangel for C = 694 Farads.

694 Farads at 18V is some supercap bank.


You'd need 20 of these

10 in series, in parallel with another 10.
Then you'd need to build a buck/boost inverter, to keep the output stable as the caps discharge, or an inverted output inverter would work.


For something like that, You'd practically need a room to keep them in.

Thanks for doing the maths, hard day: I had to sum it up like that.


It'd be cheaper and easier to replace the battery.


But he's got these 300 throw away cameras just sitting there and he don't know what to do with them.

Make a "nuclear" flash-gun, is the obvious project (to me)