1 Farad caps in series as batteries

[sean k]

I picked up some 5.5V 1 Farad batteries and I'm going to set them up in series for a potential of 11V with a couple of 100k resistors in parallel as bleeder resistors and by my calculation I'll have a 200mA battery.

If I remember right 1 volt stored at 1 amp equals 1 Farad therefore 10V (1 x 10=10)stored at 2 Farads = 200mA (2/10 =.2).

The idea is that, with even 4 to 6 0f these batteries, one will be able to just plug in 9V for a few seconds to fill them up then they'll be of use for, hopefully an hour or two in the average stompbox, then you'll just plug in the 9V again for less time than it takes to change a battery. Am I completely off the mark or will this idea work?

[solderman]

Hi
Without thinking to much, I think you are confusing effect with energy here. You are forgetting the letter "h" in mAh as effect over time. If you're calculations were right i am positive that you have solved the energy need for this planet for all time to come . You will need the same amount of energy (mAh) to play for an hour regardless of power source. Higher effect (mA) when loading the caps will give shorter time (h) to reach the same potential energy (mAh).   

[sean k]

Hey solderman, I've kinda been aware of the time thing but wondered how it fitted in and now that you've said that I'm wondering where I'll find the answers other than to charge up the caps and run a metermeasuring current into a load. I know they are batteries of a kind, caps, because the tube amp runs on as the caps discharge. Also you get sag, volts go down, in the tube amp when the caps are too small and the current load is high. Theres also the caps in cameras that charge up to run the flash. I'll go back and do some reading...

[alanlan]

It sounds like a crazy idea!!!

Seriously though, a number of points:

2 X 1F caps in series is 0.5F not 2F.  If you put them in parallel then you would have 2F but of course it would be rated at 5.5V maximum.

The equation you need is Q (charge in Coulombs) = C (capacitance in Farads) X V (Volts).

So, if you charge the 2F cap to 5.5V (which happens to be the maximum before damage may be sustained), you have a charge of 11C.

Current (A) is the rate of change of charge or, Q / time.  So, if you drained a constant current of 1mA from the capacitor, all the charge would be gone (in theory) in 11000 seconds.  But as the charge on the capacitor is reduced, then so must the voltage (look at the equation again).  So the voltage would ramp down linearly as the current flows.  

If you load a charged capacitor with a resistive load (rather than a constant current load), the voltage will decrease exponentially i.e. it will reduce more quickly initially and level off towards zero.

I don't mean to dampen your enthusiasm and maybe I'm missing something here!

[sean k]

No problem, I'm learning and thats a good thing
I'll get two more and have each two in series in parallel so I'll have 1F capable of about 10V and then I'll charge them up and use then on a booster and see how long they go. 8 of these caps takes the same space as a 9V battery and given the price of 9V rechargeables and the charger... I don't know I'll see what happens. Thanks for the info chaps.

[petemoore]

RG was talking about "Supercaps' IIRC.
  I like the idea...especially if it could wait for a few hours then do a whole say 4 hour gig.
  Zap your cap and not have to worry about opening up the pedal case next week to get the discharged, new battery out.

[R.G.]

Q = C*V, or C = Q/V

You can also write this as
C = (dv/dt)/(dV/dt) Since C is constant with time. Mostly at least.

dq/dt is the same thing as current I, so C = I/(dV/dt) where dV/dt is "the change of voltage per unit time".

Finally,
dv/dt = I/C, which tells us that the voltage runs down at a rate of I/C.The form we need this in is dt = Cdv/I, or "the time it takes to drain a capacitor of C capacitance is C times the difference in the starting and finishing voltages divided by the current you pull."

A "9V" battery is considered dead at 7V by the battery makers. If you run an LED in your effect at 10ma (which is pretty common) and the effect is 1ma (a reasonable guess for simple effects) then I = 11ma.

The time between 9V and 7V at 11ma for a 0.5F cap is dt = dv*C/I = 2V*0.5F/0.02 = (1/2)*10E2 V-F/A = 50seconds.

If I didn't drop a decimal somewhere, two 1F/4.5V supercaps in series will run an effect using 11ma for 50 seconds before reaching the same voltage at which a battery would be considered drained.

[sean k]

Thanks R.G.,
                   So that means that 2 in series paralleled with another two in series equals about 100 seconds with the LED. So if we take out the LED we can safely times that by 10 for 1000 seconds which is about 16 minutes.
What I'm thinking now is that one rechargable battery or powersupply could then be hooked up to all the FX and be used to charge up the individual caps when they start sounding nasty. It'd save money in batteries and space in individual power supplies.