practical supercapacitor?

Not going to happen. Energy stored is proportional to volume of device. Gasoline has a very high energy / cu inch ratio. Batteries have much less. Super caps? Less again I suspect.

Yes, a car battery.

I can buy supercaps from distributers for a few dollars each in the less than 10F range. While that doesn't compete with batteries for size or cost, they do charge much faster, which is what I really need.

Homer J Simps>

How much energy is stored in 10 F at 3.5 V?

Energy, U = 0.5 * C * V * V where V is in volts, C in farads, and U is in Joules. 122.5 Joules is a *LOT* of energy!

How long will that run a 240 VAC inverter?

Actually for a 3.5v cap it would be 61.25 Farads, which is enough to power a

60 watt bulb for 1 second. 1 Farad = 1wattsecond.

Look at the ESR of the supercaps you're interested in. At that price, they are most likely CMOS backup caps and are only good for a few milliamps current draw. Even the large 12v 10F caps used in car audio applications would still be useless as a power source, not to mention much more expensive than a battery.

Time to look at battery packs, or extension cords...

Chris

There are some new super or ultra capacitors that are a cross between a lead acid battery and a large capacitor. They are finding applications on cross-over duty in UPS systems to sustain large electrical drain until the generators kick in - without the maintenance hassle of lead acid batteries, and electric vehicles to provide bursts of energy. Charge is measured in amperes/joules versus time.

Search on electrochemical capacitors or ultra caps. They are already in production. See:

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If 122.5J is a *LOT* of energy how come my domestic electricity supply meter reads in units of 3,600,000J? or the gas tank in my car holds about

1,800,000,000J?

The capacitor only stores 61J anyway.

Hmm a gas tank or nearly 30 million 10F 3.5v supercaps - tough call.

A 1.8 billion joule cap would blow your ass off the planet same as premium, what's the point?

1 F = 1 amp at 1 volt for 1 second

10F & 3.5 V will power let's say a 100mA device at 3.5 V for 100 seconds.

If I had the ability to grab a decent amount of electricity in a few seconds, I would be willing to sacrifice on the device that use it.

And of course, I would use more than 10F. Probably more like 200F.

No, I don't think so. You don't get 3.5V the whole time; if the cap voltage was the same at the end as at the beginning, then you wouldn't have drawn any power. In fact, the voltage starts dropping as soon as you start drawing current; and it drops at dV/dt = i/C = 10mV/second in this example. If you draw constant current, at the end of 100 seconds, you're down to

2.5V.

But in the real world, you can't draw constant current regardless of voltage, without throwing a lot of it away. Even with a switch-mode supply, which could efficiently maintain a fixed output voltage, you'd have to draw more and more current as the cap voltage dropped. Power is power; if the V goes down, the I has to go up.

The better way of thinking about the problem is in terms of energy; as Christopher pointed out (although he confused F and J), 10F charged to 3.5V is about 61J, which is 61 watt-seconds. Your 220F capacitor is still only

22 times that; great, you can light a 60W bulb for 22 seconds.

But again, that's assuming that you can actually get that energy out of the capacitor in a useful way. In practice, you'll be lucky to get half that, because you always end up leaving a bit of voltage behind, because of heating in the capacitor's internal resistance, and because of inefficiencies in the switching power supply (the wider the supply voltage range it has to deal with, the worse the average efficiency).

Wikipedia has a decent article on supercapacitors, touching on this use:

not really.

no it won't the voltage reduces as the capacitor discharges.

Bye. Jasen

How do you get 1 wattsecond powering a 60 watt bulb for a second? wouldn't it take a wattminute (or 60 wattseconds)? The watt second would power a 60 watt bulb for .017 seconds (1/60)

Carbon aerogel caps aren't ridiculously expensive (4.7F Time to look at battery packs, or extension cords...

Agreed. While you could make something with capacitors which would charge in seconds and deliver hundreds, even thousands, of watts you'd be paying a lot of cash for the luxury of 15-second charging compared to the 15-minute charging available with (some) Ni-Cad or Ni-MH cells. Enough to pay for a couple of spare packs and a decent charger.

Bear in mind the power supply you'd need to charge the capacitors quickly too. If you want your pack to, say, run a 120W inverter for ten minutes and charge in 10s you'd need a 120*10*(60/10) = 7.2kW charger - and that's only the average over the 10s, initial draw would be higher. I suppose your charger could be another, even larger, capacitor bank :)

Tim

This is interesting. If I wanted to experiment with using a voltage converter to maintain a steady voltage as long as possible, could you recommend an IC or hybrid circuit for that? In Wikipedia they say such things exist. My target voltage is enough to run some bright LEDs, e.g. 3 volts.

Thanks.

I believe they are called switching regulators...

The MAXxxx, MCxxxxx and LMxxxx step-up regulators look promising (available thru Farnel or any other component distributer). Some will work down to input voltages of around 2.3V (output current is limited to around 0.5A or less). Some examples are LM2577M, LM2587-12, MC34063, etc. NOTE - the ones mentioned above are rated for an output voltage of 12V.

Don't bother with the 12-240V inverter. Even at low output currents (or no load connected at all !) the inverter will suck the capacitors dry and stop working when the voltage drops from 12V down below about

10V, almost instantly.

What type of supercaps do you have, as it sounds like you have the CMOS memory backup type (small coin-cell sized high internal impedance (i.e.- 100 or more ohms) supercaps designed for low discharge current (low mA or less)) with a voltage rating of 2.5V (or slightly higher, depending on supplier, etc). As an example (assuming standard 4.7F 2.5V supercaps) if you wanted to get 12V out you would need 5 in series, giving 0.94F at 12.5V (remember capacitance in series reduces).

Additional problems you may run into are - voltage balancing issues when running so many caps in series.

Supercaps have there uses. The low impedance types are very useful for providing fast discharge and charge currents (due to there high power storage properties compared to batteries) or for proping up a high impedance batteries voltage during high discharge load peaks, but as for energy storage, they are very poor when compared to batteries (one of those tiny Alkaline 12V car garage door remote batteries will do a much better job in most cases. Not to expensive either !).

Exactly what are you planing to do (sounds like a miniature power supply for a LED application)?

Don't forget the series resistor for charging!

I'm not sure why you'd want my advice on what regulator to use, given that my advice, like that of most everyone else who's responded, is "don't use a capacitor in this application." But it sounds like you're more interested in doing this your way than in doing it the sensible way. Within that constraint, my advice would be this:

Don't think in terms of "target voltage." LEDs aren't interested in constant voltage; they want constant current. But anyway, the problem that you're working on, powering a collection of white LEDs (at constant brightness) from a battery that is gradually being depleted, is a very common one; it's what all the laptop and cellphone and white-LED-flashlight manufacturers have to solve. So, you want to solve it the same way they do: with a white LED driver IC.

If you Google "white LED driver" you'll find a plethora of options. Personally, I'd avoid Maxim, since their chips are notoriously hard to find in reliable supply, and focus on more readily available manufacturers such as TI, National, Fairchild, and Linear.

Or, you could just take apart an LED flashlight. If it's powered by two

1.5V batteries, it'll work just as well from a 3V (or a bit more) capacitor charge.