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active| newest | oldestI need to be more precise:

* theoretical battery of 1000mAh and 1,2V: at the beginning has 1,2V at output, and keeps that 1,2V at output for as long as you do not reach 1000mAh (for instance discharging it with 1000mA in 1 hour, or 1mA in 1000 hours). After that it immediately dies and the output is 0V.

* theoretical condenser starts with 2V and immediately after the current starts to flow out the voltage drops linearly and reaches 0V after whole charge had been drained.

In practice: the battery is not ideal, the voltage starts at 1,4V or similiar, drops a bit during lifetime of the battery, also the circuit is never ideal and usually how many mA it drains depends on input voltage. Also the number of mAh the battery has depends on the current drained out. Usually batteries have more mAh when you drain small current and less mAh when you drain higher current .... up to some point when tendency reverses and with very small currents the mAh of the battery will be lower than rated.

So the real question is: how large capacitor do I need so that my circuit works the same time as it would when having 1000mAh battery? This makes perfectly sense and my equation gives quite good approximation for small currents.

The 0,2777 coefficient is taken directly from the capacitor's current to voltage ratio:

C=Q/V. See the 'capacitor' in Wikipedia - all equations are there.

Remember: V drops with time (condenser discharges)

In electronics there is nothing like ideal theoretical circuit and some shortcuts in thinking are perfectly justifieble and ...

...long story short: the above question makes sense.

There are many theoretically oriented people who want to be more exact than necessary. The person asking wants simply to replace the battery with charged capacitor. The answer is:

1F=0,2777mAh/V

so: if you have 1,2V battery with 0,27mAh capacity, than you can replace it with the capacitor of 1F charged to let's say 2V. Your circuit should cope with it. After consuming 0,27mAh from the capacitor it will get discharged down to 1V which should be also ok for your circuit.

So:

* check the minimum and maximum voltage for your circuit. Take the difference between those two,

* multiply it by 0,2777 per each needed mAh

and you will get needed capacity in F.

And remember to start with fully charged capacitor up to the maximum possible voltage.

It would be nice to know how you calculated the 0.2777 constant. :)

sorry one correction:

*check the minimum and maximum voltage, take the difference

* divde 0,277 (not multiply) by the min/max voltage difference

you will get number of F.

Hi,

I'm trying to replace a 2.4v 500mah nicd battery with an equivalent or higher energy dense capacitor. What rating of voltage and capacitance in farad do it need?

Thanks

How many kilderkins in a MeV?

So why complicate everything so much. Just get a Capacitor of said Fards conected to a synchronous boost converter pull needed voltage from device and see how long it lasts.

if you use a resister to lower the voltage out of the capacitor will it have a higher f or amp hour?

2 Kilderkins = 1 Barrel

1 1/2 Barrels = 1 Hogshead

If we say that we do have perfect electronics, 100% efficient, then since Q=CV , C=Q/V and we have 2.6 C/ 1.2 V = 2F

Steve

You might ask the question, "How many donkeys are equal to one school bus... approximately?", and then wonder why no one knows what the monkey you're talking about.

There is some thing unstated, some thing donkeys and school buses have in common, some thing which will make the question make sense. Is it that donkeys and school buses are both heavy? Is it that donkeys and school buses can both be used to transport people? Knowing the quality which is common to both donkeys and school busses, the thing

that you care about, is important.Regarding milliampere*hours and farads, I suspect the thing you are interested in is energy storage.

For this you should know batteries and capacitors store energy in somewhat different ways. For a battery, the energy stored is approximately its voltage multiplied by its current*time capacity, e.g. volts times ampere*hours.

http://en.wikipedia.org/wiki/Battery_%28electricity%29#Battery_capacity_and_discharging

For a capacitor, stored energy is equal to one half the square of the voltage multiplied by the capacitance of the capacitor, in farads.

http://en.wikipedia.org/wiki/Capacitor#Energy_storage

Anyway you should meditate on these facts and formulas, and then come back when you've got a question that makes sense.

If you say, that the charged capacitor C, with voltage V has a charge on it, Q=CV, you know how many Couloumbs of charge are stored. Now, if 1 Couloub flows past a point in a second, we say a current of 1A is flowing. Likewise 0.001 C/S = 1mA.

Steve

If we say that we do have perfect electronics, 100% efficient, then since Q=CV , C=Q/V and we have 2.6 C/ 1.2 V = 2F

Steve

charge(coulombs). Farads are a unit ofcapacitance, which is charge per voltage. Get it right.L