Capacitors as powersources

[A3rd.Zero]

I am interested in using a capacitor as a powersource, much like a battery. I am however, a noob in terms of my understanding of electronics. If I have therefore a 50F capacitor that runs at 3.6V and I have a current draw of 3V an 30mA, what is the equation for figuring out how long this capacitor will supply power?

I know that Capacitors discharge over a curve, is it possible to use a voltage regulator to keep the discharge reasonably flat? If used will a voltage regulator consume much power just by existing in a circuit?

Thank you for your help. I am sure that many of my questions sound silly to many of you.

Milo

[Bill Fitzpatrick]

http://www.tpub.com/neets/book2/3d.htm

[A3rd.Zero]

Thank You!!!!!!!

[Sch3mat1c]

Hm, 50F at 30mA eh? What is your cutoff voltage?

For a resistive load the equation is: V = Vo * e^(-t/RC), where Vo ("vee naught") is the initial voltage (3.6V), V is the voltage at time t, R and C are the obvious circuit constants and e is the natural base (2.7 something). Some algebra can solve for t rather than V. For small changes in voltage the resistor looks like a constant current sink and the equation reduces to dV/dt = CdQ/dt. But I didn't get enough sleep to figure the rest so...l8r...

Tim

[A3rd.Zero]

This is all good info but it has generated another question. I have figured out the requirements for one led with resistor run off a 50F capacitor and it looks at least experimental, but if I add led's in parallel then I will be drawing more current. How do I figure current draw into this equation? Thanks for your help.

milo

[Sch3mat1c]

For LEDs, the dynamic resistance the cap sees (as a power supply vs. load situation) will be equivalent to the total series current-limiting resistor values. So say you have four red LEDs with 100 ohm current limiting resistors: 100/4 = 25 ohms would be used for R in the formula.

I hope you aren't using blue or white LEDs on a 3.6V rated capacitor, as to have adequate voltage overhead to limit current (with the resistors) you'll need more like 5V. 6 to 8V would be even better as far as intensity stability is concerned... it'll drop off quite quickly as you can only use the voltage between when the LEDs turn on, and the max. voltage you are using. (Blue LEDs might turn on at 3V; using 4V on the cap gives 1V of mildly usable range, all of 25J. Life might be 6 minutes? Just guessing...

Tim

[DUG]

Or you can use a constant current draw as a model.

Basic formulae:

C=Q/V

i=Q/t

(I'm leaving out the delta's on purpose)

The first one can be re-written as Q=C*V
The second one can be re-written as Q=i*t

Q=Q

then C*V=i*t

or re-written as t=C*V/i

in words: the time it takes a 35F cap to drop 1 V at a current of 1 A is 35 seconds.

for 0.1 Amp 350 seconds
for 0.01 amp 3500 seconds
etc

Also to charge that capacitor to 3.5V with 1 Amp would take over 100 seconds

I use this (and some trig) to calculate ripple on power supply filter capacitors.

[Sch3mat1c]

Ah, that's what I needed... thanks

Tim (still tired, and going to bed soon)