Ultracapacitors as batteries

For several years now, I have been toying with the idea of using capacitors, instead of batteries, to store power. During that time, technology has moved on, and "super-capacitors" or "ultra capacitors" have become available. These have a rating of many Farads, and, I thought, should be feasible for the home experimenter to have a play with.

Capacitors have several advantges over rechargeable batteries. They can be charged quickly, hold their charge well, and don't suffer from many of the problems associated with rechargeable batteries.

However, I wanted to be able to do some calculations without going to the expense of buying some just to try out. I wanted to know what I should be able to power, and how long it would be likely to run for. I thought that if I could run a small portable radio for an hour or so at a reasonable cost, it might be worth a try.

Having spent many frustrating sessions on the internet, and looking in books for this information to no avail, I put the idea to one side. I have recently come across this information, and decided to put it here in the hope that others might find it useful.

Please be warned - high value capacitors can be dangerous. They can discharge very quickly - through you! You may suffer a dangerous (even lethal) electric shock from these, even at one or two volts. They do not behave like batteries, where you can safely touch the terminals. Do not attempt to use them without being sure that you fully understand what you are doing. Proceed at your own risk. I can't think how to make this warning any clearer!

If that hasn't put you off, here is the information!

We can use the following formula to convert Farads to Amp hours

(Vmin + Vmax)/2 * F / 3600 = Ah

Vmin & Vmax is the upper and lower voltage. This is the voltage range that your device has to work in. As capacitors "run down" and need charging, their voltage drops, so we have to allow for that. If we assume that we want to power a 3 volt radio, we can guess that it should work happily between 3.5 volts and 2.5 volts. That's what the first part of the calculation allows for;
In our example, (Vmin + Vmax)/2 would be (3.5v + 2.5v) / 2 equals 3 volts.

In the rest of the calculation, F - Farads and

3600 - coulomb, a unit of electrical measure. One amp hour = 3,600 C

Ah - Amp hours ( 1000 mAh = 1Ah)

So we now have 3 * F / 3600 = Ah

Let's say we can get hold of a 3.5 volt 100 F super-capacitor. Using 3 * 100 / 3600 = Ah,

we could expect it to power a device drawing 0.083 amps, or 83 milliamps, for one hour.

What if this wasn't enough? What if we wanted a 3 volt capacitor to power a 3 volt device which drew 250mA for one hour?

We can use the formula "current in amps * 3600 / voltage = capacitance needed in Farads" to find out how big the capacitor needs to be. In this example

0.25 * 3600 / 3 = 300 F

The capacitor would need to be 3v 300F to power a 3 volt device drawing 250mA for 1 hour. Ideally, you should have a capacitor of slightly more than 3 volts to allow for the voltage drop mentioned above.

Information is available to explain how capacitors can be wired in parallel and in series, to get more voltage and capacity (Farads) than from a single capacitor.

Please bear in mind that the above information is intended to help hobbyists and experimenters. Use the information at your own risk! I would be very interested to hear if you find this useful, spot any errors, or if you have done any experiments with ultra caps, or have any suggestions.

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Comments 14 comments

AYAN 4 years ago

hi my names anna and i love to do technology design


Mark 1 4 years ago from UK Author

Hi Ayan, thanks for your comment - I do hope you found this useful


Chris 4 years ago

Hi,

i have pair up 2 350F capacitor and tested it, this is the result.

note that we do not have a high end volt/amp calculator.

result are appox. the load in question is an led lamp.

5.06V 2:55 - 13Xma drain

2:59 : 130ma drain

3:02 : 120ma drain

3:07 : 110ma drain

3:17 : ~9Xma drain

3:20 : ~ 8Xma drain

4.07V 3:25 : 80ma drain

in 30min a 350F capacitor drain at about 100mah and decrease by a volt.

there is no way a 300f capacitor could deliver 250mah for an hour from 4v to 3v even if you use a volt converter to acquire more amp with steady 3v.

there is a calculation.

F = (a/s) / V

F * V = a/s

so 350F * 5v = 1750 a/s

so 350F = 0.4864ah

however this is not accurate as ah decrease as voltage decrease

and as such not possible to really calculate ah from farad.

what we have calculated even if we use 4v show 0.3888ah

what our result show is 50mah from 5v to 4v.

i have yet to find a real way to calculate this and it is confusing, lol.

i hope this this information find you useful.

if we were to take this :

so 350F * 5v = 1750 a/s

so 350F = 0.4864ah / 10 = 48.6mah

we need 35kf to deliver 4.8ah @ 5v

and that is 200x 350f capacitor.

each weight about 68G

which mean we need about 13.6KG of UC to match just about 4ah li-ion battery of about 100g on my desk.

not sure if we want to use ultra capacitor to replace lead acid even if it is capable of 1m charge cycle as compare to about 500- 1k CC for lead acid and li-ion.

would love to hear what you think of the result.


chris 4 years ago

also the 0.4864ah should be the combine from 0v -5v of amp.

@ 4v 0.0972 ah

@ 5v 0.1216 ah

so as we have tested @5v it is indeed giving out about 50mah to 60mah.


Mark 1 4 years ago from UK Author

Hi Chris,

thanks very much for taking the time to share your results - really interesting!

I wonder if you could clarify the columns slightly;

I assume

5.06V 2:55 - 13Xma drain

means 5.06 volts at a time reading of 2 hours 55 minutes (is this right?), and what does the X stand for please?


John 3 years ago

Hello Mark, a late post but here it is1 I use two 1 f , 5V caps, I charge in parallel from RF energy from a strong broadcast station. when charged (abt 24hr) to 5 volts I can operate a efficient audio amp for nearly 30 minutes to amplify a crystal set audio output (I cannot tolerate headphones) cul jn85@comcast.net


Mark 1 3 years ago from UK Author

Hi John, thanks for that, that's really interesting. I managed to power a small LCD clock from the signal from a local medium wave station by rectifying the RF, but haven't tried charging capacitors from it. I'm impressed that you can run an audio amp for nearly 30 minutes this way. Thanks again


John 3 years ago

You can't get a shock from 1 or 2 volts, no matter how high the capacity is, because the resistance of your body is too high. Look up ohms law. The only dangerous aspect is the high current, if you short it, because wires can overheat and even explode.


Olivier Butler 2 years ago

This is a fantastic post, thank you so much for it. I hope if you find more accurate or revised formulae you will update it :)

For all you capacitor lovers, I recently stumbled across this site, it has a Japanese sister company, and they are Lithium Ion capacitors, that sit somewhere between batteries a capacitors, having very high energy density (more even than a regular battery), yet hold their charge for years and still work after many thousands of charge cycles. They also can't explode, woop woop. They don't seem to be for sale, but I've emailed them and hope to hear back. http://www.jsrmicro.be/en/lic_laminated_cells


Mark 1 2 years ago from UK Author

Thanks very much Oliver, glad you liked it. Thanks for the information as well, I han't come across Lithium Ion capacitors. Please post again if you get a reply or if you come across any other info


Mark 1 2 years ago from UK Author

Thanks John, that's very reassuring! That explains why a friend of mine is still alive :-) I did warn him not to touch one, and felt puzzled when he did - and felt nothing. Bit of a risky way for him to check it out though...


Matthew 2 years ago

Hi, just wanted to point out your formula for converting Farads to Amp hours doesn't seem to be quite right.

Using your formula and assuming a 3600 Farad capacitor:

With a device that functions between 1 and 2 volts your formula would suggest:

(1+2)/2 * 3600 / 3600 = 1.5Ah

Whereas a device that functions between 0.1 and 2 volts your formula would suggest:

(.1+2)/2 * 3600 / 3600 = 1.05Ah

So although the second device works over a larger voltage range it can draw less power? That doesn't make sense so lets break it down a bit:

1 Farad = 1 Coulomb per Volt

1 Coulomb = 1 Amp for 1 Second or 1 Amp-Second

If we want to express that in Amp-Hours rather than Amp-Seconds we must divide it by 3600. That gives us 0.000277 (7 recurring).

We can then convert that into milli Amp-Hours.

1 Coulomb is therefore equivalent to 0.278 mAh

Finally therefore 1 Farad = 0.278 mAh per Volt

To get the useful voltage we subtract minimum usable voltage from maximum usable voltage.

Finally we arrive at this formula:

(Vmax - Vmin) * F * 0.278 = x (mAh)

or

(Vmax - Vmin) * F * 0.000278 = x (Ah)

That is: Useful Voltage multiplied by Capacitance multiplied by our previously calculated constant.

So applying that formula to the above examples:

(2-1) * 3600 * 0.000278 = 1 Ah

and

(2-0.1) * 3600 * 0.000278 = 1.9 Ah

Which makes a lot more sense.


Mark 1 2 years ago from UK Author

Hi Matthew,

thanks very much for taking the time to post that, much appreciated.


Mark 1 5 months ago from UK Author

Thank you, glad you found it useful

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