How long will a 1 Farad capacitor last under load? 12 volt/.72amp?

Question

I'd like to run a fan (12 volt .72 amps) from a charged 1 Farad Cap but don't know how to calculate the time it will last! So far my crude math yields 190 days! Yeah right! I'd prefer the math so I can see what a 5 F would do etc... Plus I'd learn a bit! Thanks!

I've always been Told 1 F is a TON of power! And 1F is now easy to find...they have 3000F caps!!!
http://en.wikipedia.org/wiki/Ultracapacitor
Maybe a smaller fan would work? Shooting for 3 or 4 hours of fan use...

Answer 1

The trouble is that your fan would slow down all the time. As capacitors discharge the voltage goes down too.

Energy stored is 0.5*CV^2 so you have 76J.

The fan uses 12*.72 =8.6J

Even if you could run a boost regulator to avoid the voltage drop you would only run the fan for 76/8.6 = 8.8 seconds.

Answer 2

I think you're out of luck. Let's say you find a 1 F cap; I've only seen 5 volt versions but let's say you found a 12 volt one. Now, charge it to 12 volts since q = C V, you have 1 F at 12 volts so you have 12 coulombs. The voltage will drop on the cap as you draw charge but let's do an optimistic calculation and say it stays at 12 volts. Now, you're drawing .72 amps. You would bleed the cap in 12 coulombs/.72 amps; you'd get about 15 seconds best best case. Realistically it won't work like this because of a dropping voltage.

Answer 3

The voltage will decay exponentially. The load resistance of the fan is 12/0.72 = 16.7 Ohms. The R*C time constant is therefore, 16.7 seconds.
That is the time it will take for the 12v across the capacitor to decay to 4.4 volts.
In 33.3 seconds, the voltage will be 1.7 volts.
In 50.0 seconds the voltage will be 0.6 volts
In 66.6 seconds the voltage will be 0.24 volts
In 83.3 seconds the voltage will be 0.12 volts

There are integrated circuit boost converters that accept low-battery input voltages down to about 0.8 volts and they are about 80% efficient, so you might get 30 seconds operation from a fully charged capacitor.

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Answer 4

Yea, your math is off a bit, the link in "sources" below gives a pretty good explanation of farad. Assuming the capacitor is charged to 12v, it might run the fan for 10seconds or so, remember the voltage is constantly falling during that time.

Answer 5

It depends what you mean by "last".
As Peter K. says, the voltage drops, in a non-linear fashion, and the answer to your question depends on what is the "cut-off-voltage" you require, or the minimum voltage to keep something still working on that charge.
I suggest you look at
http://www.play-hookey.com/dc_theory/rc_circuits.html
or more heavy reading (bottom of page)
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capdis.html

Answer 6

I have never heard of it taking long like that ever. All you do is get the test light and put one of the leads to the power wire and one of the other leads on the positive side of the batter. The light should go on and then dim quickly. Thats how i was told by people and the directions. and to discharge it you take both leads from the test light to the positive and negative side the ligh should light up and dim.