# How many mAh = 1 farad?

caarntedd3 years ago
None.
6 months 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:

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.

6 months ago

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.

6 months ago

How many kilderkins in a MeV?

kelseymh3 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.
3 years ago
Oh yeah? how many kilderkins in a furlong?
7654321 (author)  caarntedd3 years ago
Actually, 99,486,918.8 kilderkins = 1 furlong^3
3 years ago
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
3 years ago
How many k baud per square fathom?
orksecurity3 years ago
How many elephants make one afternoon?
7654321 (author)  orksecurity3 years ago
3 if it's sunny
3 years ago
2.999999999999999999999999999999999999999999999999999999999999999999
steveastrouk3 years ago
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

Steve
Jack A Lopez3 years ago
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.
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.
7654321 (author)  Jack A Lopez3 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.
3 years ago
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.

Steve
7654321 (author)  steveastrouk3 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?
3 years ago
Yes. The energy stored = CxV^2/2
7654321 (author)  steveastrouk3 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?
3 years ago
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

Steve
3 years ago
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.
3 years ago
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.
3 years ago
"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...

L
Slowpoke3 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.