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How about a superduper capacitor substitute for a battery?

Let us hypothetically have a dual-AA battery powered lamp that runs for 30 minutes on 3 volts. And lets us replace the two AAs with some superduper capacitors.

Now, lets make some slight adjustments. Super caps usually sell 2.5 volts caps. So lets have two AA cells that output 2.5 volts and some super caps that have a max rating of 2.5 volts.

So lets pretend that in my left hand, I have two ni-mh AA cells and a light bulb. The voltage is 2.5 volts. It will run for 30 minutes.

Lets now pretend I want to do the same, but with super caps.

How many 2.5 volt capacitors do I need at n-farads? (assuming there are no single billion-farad caps). Bare in mind that today's super capacitors are no longer the size of oil drums. And I know the lamp will dim as the cap runs down.

Real rough. Found some nimh AAs that are 1300mAhr. So 30 minutes would 2.6A load (you won't get the rated amphours discharging them that quick)

1 RC time constant is 63% discharged.
R = 2.5V / 2.6A = 0.96Ohms
Time in seconds is 30min * 60S /min = 2160S
T = RC so C = T/R
C = 21605S / 0.96Ohms = 22505F

30 supercaps each 100F in parallel might get in the area of 30 minutes.
(60 supercaps if 30% discharged is dim enough.)

(Been a ll oo nn gg time. Anybody, please check my arithmatic))

Cheers,

Just spotted a 3500F super-super-cap.

One or two of those should do too.

(a question - is this an incandescent light? because the white LEDs I've seen need 3.8V)

I can't see where the OP expressed the current demand of said bulb. Is it a small LED? Incandescent?

It was indirect. He said he expected AAs to last 30 minutes.

If it is rated 1300mAHr, that is 1.3A for 1 hour. So 2.6A for half an hour. (Though at high current loads you don't get the rated time.)

A lot of LEDs, a LED headlight, or an incandescent does not matter for the question; "How much capacitance to equal two LiMh AAs in series?"

Really, in order to compare in an even fashion, you will need to convert to joules. It's been so long since I've had to do that, that I'll have to look it up again. Until I do, here's hoping someone will chime in.

OK, I see that now, even though we know 2.6 amps for 30 minutes from a AA is pushing reality.
Time would be 30 mins X 60 secs/min = 1800 seconds.

C = T/R, so with a time of 1800, and an R of 0.96 ohm- 1800/.96 = 1875 Farads

So, who has that capacity for us to try it?

Thank you! (Wonder what I pushed on the calculator???)

Power X Time is energy, so 1.3 AH X 2.5 volts X 1800 seconds = 5850 joules (or watt-seconds) (From the battery)

Energy stored in a capacitor is 1/2CVsquared, so, 1/2 X 1875 Farads X 2.5volts squared = 5859 joules (or watt-seconds) (From the capacitor)
Close enough!

Sounds very doable.
Just came in from Digikey: 45F 5.6 volt cap. It's the size of a 'n' cell, but slightly smaller.
I charged it to about 4 volts and it powered an LED for a while. I didn't time it. The cap started heating up fast.

Hey, why does it heat up like that? I connected a 1 watt white LED. Caps should discharge rapidly, right? Or do super caps follow different rules?

And, yes, I would love to replace battery packs, if possible, if practical.

Super caps have a higher internal resistance than filter caps.

That is lucky because you don't mention current limiting, and you didn't "blow" the LED. LEDs are not incandescent bulbs. With bulbs an increase in voltage increases current, bulb gets hotter which increases resistance which regulates the current for small changes. Resistance and fixed voltage drop in LEDs is pretty constant, so changes in voltage make big changes current.

Parallel caps will increase time and keep caps cooler, but risk "blowing" the LED(s). Perhaps a buck/boost circuit between caps and LED(s) to regulate current. Will also reduce dimming as caps discharge.

((if this was economical way to replace batteries it would be done already (IMO)))

thanks