Hello All,

I am trying to charge two 400F 2.7V supercaps in series as fast as possible and I am having trouble finding a charge circuit for this result.

Any help would be greatly appreciated.

Thanks,

-Cory

 

How fast is, "fast"?
2000 amps for .54 seconds?
10,000 amps for .108 seconds?

 

Our goal is 30 seconds to have the caps at 5 volts.

What equation would that be?

 

33&1/3 amps

IT = CV
amps times seconds = Capacitance times voltage
(amp-seconds of charge) = (capacity times voltage)
33&1/3 times 30 seconds = 200F times 5 volts

 

Ok great thanks! So we would now be looking at a step down transformer of some sort; or a buck and boost circuit?

We have a dc adaptor for our source that is 12 volts and 2 amps.
We have thought that a step down transformer would be able to step our source down to 5 volts and then boost our current, but probably not to 33.3 Amps?

Any ideas there?

 

Your best bet would be to use a low voltage high current transformer and an automatic disconnect that shuts off the charging power when it reaches the voltage you want.

 

You would need about a 5 volt transformer rated at 600 watts with 3 rectifiers in series to get the voltage low enough without a regulator. You have to ask yourself, "How much money do I have for this?"

See the second drawing on the left for the factors you will use.

 

Your best bet would be to use a low voltage high current transformer and an automatic disconnect that shuts off the charging power when it reaches the voltage you want.

A 5 volt detector on the output, opto-coupled to a triac might work to avoid exploding the capacitors, or just use up the extra voltage with extra rectifiers in series with each other.

 

Will the batteries be fully discharged every time?

If the 200F cap powers a "5v device" that device may stop working when the voltage is about 3v, so cap charge cycle only needs to be from 3v to 5v, not 0v to 5v?

And the caps won;t have 200F of capacity, if they only power your device from 5v down to 3v they only have useful capacity of 80F.

Just something to think about. Caps are not batteries!

 

We have seen several other people go through this discovery process in the last year or two.

200F @ 2 v difference is 6.7 amp minutes.
Do you want to buy a 600 watt transformer for that kind of performance?

 

The project has been simplified now;

We just have 5 amps to work with for charging the caps (since 33.3 amps is a lot for an undergrad to be handling).

That would yield a 200 second charging time which isn't bad at all.

The supply sources in the lab range from 0-6v @ 2.5 amps, or 6-20v @ .5 amps.

We have a dc wall socket adaptor that puts out 12 volts @ 2 amps.

We can use what ever source would be viable (or buy a new one).

 

The total energy stored in a capacitor is 0.5 x C x V^2 (one-half C V-squared). Your two caps are in series, so the effective value is 200 F and you are charging from 0 V to 5 V, so the total energy is 2500 ws (watt-seconds). This shows you the minimum size of the charger system neded: 2500 watts for one second, 250 watts for 10 seconds, or 83.3 watts for 30 seconds. Your DC source is 24 watts. Assuming a 90% efficient DC/DC converter, you're down to 21.6 W delivered so the charge time can not be any faster than 116 seconds.

#12's equation above (and my example above) is for a constant-current charger, the smallest and most efficient approach in terms of time because there is no decrease in the rate of charging as the capacitor approaches full charge.

Something to consider is the peak current capability of the capacitors (maybe called ripple current on the datasheet). At 33 amps for many seconds, an internal connection might blow like a fuse.

ak

 

Check out the app circuit, center of page 20.

ak