How many mAh = 1 farad?

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AsimB612 days ago

The answer is very simple...

A 1-farad (1,000 mF or 1,000,000uF) capacitor can store one coulomb (coo-lomb) of charge at 1 volt. A coulomb is 6.25e18 (6.25 * 10^18, or 6.25 billion billion) electrons. One amp (1,000 mA) represents a rate of electron flow of 1 coulomb of electrons per second, so a 1-farad capacitor can hold 1 amp-second of electrons at 1 volt.

So yes. you need very very high capacity if you want to replace a battery, search for Supercapacitors.

But you will also need a voltage regulator.

Bodragon.1 month ago

I think what the questioner is really asking here, (and this is something I have pondered myself). Not: does mAh have an equivalent value in farads, which, if I dare be so bold to venture, I would say yes, I think an estimate could be calculated. Surely, mAh represents the total amount of power in any given battery, (or cell). Isn't the Farad scale a similar kind of measure? Cannot both be converted into Joules, and then a direct comparison made, or am I missing something more fundamental?

JeremiahA143 months ago

Laughing out loud, basically what you need is a completely separate circuit that would include voltage regulation, and the required energy storage... simply said, if you need a specific voltage and draw, and you need it to last for a specific time length (i.e. 3.7v 2000mAh), you would have to design a special circuit. A capacitor will not function as a direct replacement for a battery because the voltage changes too much from empty to full (i.e. 0v-2.4v 10F).

Slowpoke5 years ago
None. Your question is probably related to how many mAh is available in 1 Farad as compared to a battery. To answer this you need to know the voltage the capacitor is charged to & what your current draw will be. Also you need to know what is the maximum & minimum voltage is suitable. Generally you cannot compare a large value capacitor with a battery. Kelseymh has supplied the necessary info to calculate the available energy in a capacitor.

thx for the help

SuartanaD11 months ago

If I need a 20AH at 36 Volts, how large capasitor do I need to replace?

mgosiewski1 year ago

I 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.

caarntedd5 years ago

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:


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.


* 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.


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?


How many kilderkins in a MeV?

KaroR1 year ago

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.

Lowurey2 years ago

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

kelseymh5 years ago
None. "mAh" is a unit of charge, equal to 3.6 coulombs (A=C/s, h=3600 s). The farad is the SI unit of capacitance, equal to coulombs/volt. That should be enough information for you to answer the question for the particular battery you have in your hand.
Oh yeah? how many kilderkins in a furlong?
7654321 (author)  caarntedd5 years ago
Actually, 99,486,918.8 kilderkins = 1 furlong^3
Well technically none - a kilderkin is an old English measure of volume (beer or Ale)

2 Kilderkins = 1 Barrel
1 1/2 Barrels = 1 Hogshead
How many k baud per square fathom?
orksecurity5 years ago
How many elephants make one afternoon?
7654321 (author)  orksecurity5 years ago
3 if it's sunny
Without complex electronics, its impossible to "replace" one, because the capacitor voltage drops as soon as you start to pull charge from it, so it won't be 1.2 V....

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

Other answerers  have said "none", but I think the correct answer to this question is "mu", or "empty set", or maybe "syntax error" because the question itself does not make sense as asked.

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. 

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

Anyway you should meditate on these facts and formulas, and then come back when you've got a question that makes sense.
7654321 (author)  Jack A Lopez5 years ago
But they are both measures of capacitance, and I'm pretty sure you could measure how long a capacitor can supply a given load, which means you could find out how many mA it can supply in an hour, then you could look at how many F it is and tell how many mAh= F. I know they arent the same, I was asking for a close number of about how much.
No, The FARAD is the ONLY measurement of capacitance. You can't equate mA to F, because the discharge rate depends on the voltage on the capacitor.,

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.

7654321 (author)  steveastrouk5 years ago
oh, so you're saying that a capacitor charged to 10V could supply current longer than one of the same capacitance charged to 5V?
Yes. The energy stored = CxV^2/2
7654321 (author)  steveastrouk5 years ago
So... If I wanted to replace a 1.2V 2600mAh battery with a capacitor charged to 1.2V, how many farads are we talking?
Without complex electronics, its impossible to "replace" one, because the capacitor voltage drops as soon as you start to pull charge from it, so it won't be 1.2 V....

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

They are not both a measure of capacitance. A farad is capacitance and a mAH is a measure of "capacity", not capacitance. Slightly different sound but technically very different.
No, that is not correct. You should (a) read my answer, and (b) do some research, for example on Wikipedia. Ampere-hours are a unit of charge (coulombs). Farads are a unit of capacitance, which is charge per voltage. Get it right.
"syntax error" - quality! It's so many years since I heard that from a computer that I can't remember (and it may have been a ZX Spectrum...