# How many mAh = 1 farad? Answered

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The question is very easy to understand, the assumptions behind it are not.
1) You seem to assume the capacitor will be charged at a constant 1 Amp, I can't find any evidence for this as the charge time can be 1 min using a full car battery to 40 min on a USB charger, a constant 1 Amp makes the maths easier so lets accept it.

2) If you are charging at a constant current of 1 Amp the voltage increase each second will be the same. Your calculation implies the voltage increased for example by 3 V in the 3rd second so you included the 1st and 2nd voltage increase to get 3W, but it should only be 1V for each second. But your Watts calculations are not important as I show below.

3) I don't think Watts can be added as you have done.
If a heater with 500, 200 &100 W settings is used for 2 hours at 500W, 5 hours at 200W & 10 hours at 100W we can't say it is a 500 + 200 +100 W = 800W heater.
You could argue that, as you're calculating the Watts for 1 second then that is equal to energy (measured in Joules), so you are adding energy not power, BUT...

4) Why do you assume the voltage rises by 1 Volt each second?
Farads x Volts = Amps x seconds
20 x Volts = 1 x seconds (assuming 1 amp constant current)
or 20 x Volts = seconds (voltage rise is constant with constant current)
to rise 1 Volt takes 20 seconds (not 1 second)
to rise 2 Volts takes 20 x 2 = 40 seconds
to rise 16 Volts takes 20 x 16 = 320 seconds

5) The energy stored in the capacitor =(Farads x Volts x Volts) / 2 Joules
=(20 x 16 x 16)/2 Joules
= 5120 Joules
Watts x seconds = Joules
or seconds =Joules / Watts
seconds = 5120 / 136
seconds = ~ 38 seconds (~ means about)
In ideal conditions a 20F capacitor could provide 136 Watts (your answer) for 38 seconds.
In 3) above the heater uses energy from the mains electricity supply so does not run out. The capacitor is an energy store so runs out of energy & needs recharging.

Hope the above helps, I've enjoyed the challenge, but "Are you right or Not"
You decide.

Statement or question?

As a statement it is true, as we need the capacitor’s internal resistance to calculate the Amps.

If it is a question & the leads are shorted together with a very thick lump of metal (i.e. zero Ohms).

A thin wire could glow red hot and give a nasty burn.

Assuming the capacitor’s internal resistance will be between 40mOhms & 300mOhms.

Ohms law gives us Volts= Amps X Ohms

Rearranging Amps = Volts / Ohms

If internal resistance = 40 mOhms (0.04Ohm) Amps = 3V7 / 0.04 = ~92 Amps

If internal resistance = 300 mOhms (0.3Ohm) Amps = 3V7 / 0.3 = ~ 12 Amps

Most capacitors will be somewhere in-between 40 mOhm & 300 mOhm

BUT only for a short time.

The voltage will be ~1V2 (~63% of original) in (resistance x capacitance) seconds

0.04 Ohms internal resistance.

0.04 Ohms x 3000F = ~120 seconds

& AMPS=1v2 /0.04 = ~30 Amps

0.3 Ohms internal resistance.

0.3 Ohms x 3000F = ~900 seconds

& AMPS=1v2 /0.3 = ~4 Amps

The charged capacitor’s energy content =.(Farad * Volts * Volts)/2 Joules.

= (3000 * 3.7 * 3.7)/2 Joules

= ~42000 Joules or 42 kJ about the same as 2 grams of chocolate.

16V 20F Ultracapacitor Engine Battery Starter Booster Car Super Capacitor

If i read everything right about capacitors, this capacitor would charge to 16 Volts, in 16 seconds and draw :

1 AMP 1 VOLT ( Sec 01 ) 1 Watt VxA = P?
1 AMP 2 Volts ( Sec 02 ) 2 Watts
1 3 3
1 4 4
1 5 5

xx xx xx xx xx

1 16 16 Watts

16 +15 +14 + 13 .... + 1 = 136 Watts.

Am I right or NOT?

I think we've got to make some assumptions here about what exactly is being asked. The question is: How many mAh = 1 Farad? If this means how many farad are needed to store energy equivalent to a mAh, then the answer is "it depends on the voltage". Of course, we can make a very good guess at what voltage is being implied here - 5V, the standard for phones and many many other consumer devices which use Lithium ion based batteries.

Then the question also becomes "it depends" because capacitors don't hold constant voltage, but we're looking at a theoretical maximum here. Plus, lithium ion batteries also aren't constant in voltage output and do not even output 5V. So we can assume some sort of ideal theoretical voltage regulation here, 100% efficiency with no overhead.

The energy of a fully charged capacitor at a certain voltage is ( 1/2 * CV^2 ). A 1F capacitor at 5V has the energy 12.5 J. 1 mAh at 5V is 18 J. And so with our above assumptions:

At 5V, to store the energy equivalent of 1mAh , you need a 1.44F capacitor.

For reference, the iPhone 7 supposedly has a 1960mAh battery while the iPhone 7 Plus has a 2900mAh battery. If you wanted to store the iPhone 7 plus 2900mAh equivalent you'd need a minimum of 4176 F in capacitance - assuming perfect voltage regulation and energy transfer as above.

That's a lot of those little caps, huh?

I see all of these comments that seem to be people having potshots at the intelligence of the other people making comments... Not only that, but seem to be trying to inflate their own ego...

So my question, what is the reason for not being able to connect a cap to a power meter, measure from the maximum voltage of your circuit to the minimum voltage of your circuit, and see what the rated mAh is... Seems so dang simple to me, but perhaps I am overlooking something...

Well first of all, what is a power meter? A multimeter can measure voltage and current.. Secondly, the power at the beginning is not maintained.

You could connect your charged capacitor to a small load say (10K ohm) and measure both the current and voltage periodically, until the capacitor is drained to 0V. Then you could work out the power it delivers at each point in time by multiplying voltage and current (P=IV). But what we see is that power decreases over the duration of the experiment. You see, the voltage falls as the capacitor loses charge, so the current and power also fall.

A power meter is a device that measure power... I have one sitting here that I built. (I am not being sarcastic, or snark in any way, so please do not take offense, I am not very skilled at communicating and am often mistook for being negative or sarcastic when I am not... Just wanted to put that out there). A multimeter is able to measure various parameters that make up power, but it doesn't actually measure power, but if you were able to monitor the current and voltage, you could surely ascertain how power (or whatever other measurements you were interested in) by doing some basic math with those values...

My power meter is able to display/log current, voltage, Ah, Wh and surely anything else needed with some basic conversion/math... As such, by connecting the Capacitor to a load, surely I could obtain the mAh, along with any other applicable metrics...

On top of that, all portable power sources (that is to say, batteries) that I am aware of have a voltage that varies with their capacity, or rather, their state of charge, so why is measuring the capacity of these in any way different from measuring a Capacitor? Again, unless there is some significant difference that I am completely unaware of, I don't see the problem with finding the mAh of any Capacitor, and as mentioned, all you would need to do is run the Capacitor from the nominal voltage, down to the minimum voltage of your circuit, that way you are not measuring the energy storage that is of no use to your circuitry.

Again, I am not being sarcastic, or rude, or anything of the sort, I am just sharing information as I understand it, and given my, admittedly, relatively limited (by comparison to an engineer or such) knowledge, I do fully accept that there may be something that completely negates everything that I have just stated... :-)

Thank you for the communications, and be well!

~R

How many mAh = 1 farad?

Yeah this is a doozy of question, but I think it is the result of a slight misunderstanding: basically, although the farad and mAh units both describe capacitance, you can’t charge a capacitor with 1 Farad, but you can charge it with 1mAh! Also, capacitors and batteries behave differently. Read on if you are curious for more.

mAhours are the number of milliamperes that can be sustained for 1 hour, and this unit tends to be used to describe batteries, which have a fairly flat discharge curve, meaning they don’t lose much voltage until they are exhausted.

Farads describe capacitors, which increase their voltage linearly as you charge them up and also decrease their voltage linearly as they release charge (assuming constant amperage!). 1 Farad describes a capacitor that gets to 1V when a constant charge flow of 1amp is pushed into it for a period of 1 second. If you did it for 2 seconds you would have stored 2 ampSeconds of charge but the voltage would be at 2V - and you could keep pushing AmpSeconds in until you hit the capacitor’s maximum voltage rating (at which point the dielectric would break down). As you discharge at 1 amp, each second results in a drop of 1V (for a 1F capacitor).

Now, if your capacitor is losing voltage, it’s also losing power rapidly (P=V^2/R) and current, so it’s hard to compare capacitors with batteries when supplying circuits with power.

Unless you have some electronics to regulate the voltage, you won’t be able to supply your milliamps at a constant rate with a capacitor. If you do use a regulator, you will be fine, except you are wasting a lot of power at the beginning of the discharge (Imagine a capacitor charged to 10V with a 3V regulator: that’s initially 7V across your regulator and 70% of your power wasted on the regulator). This probably needs to be taken into consideration if you are thinking of replacing a battery with a cap.

I’ll try to use water as an analogy: “1 Farad = how many mAh?” is a bit like saying “if I have a tank that fills to a height of 1m when I put in 1L of water, how many mL per second will I be able to leak and maintain this for exactly 1 hour. (You can answer it, but the hole in the bottom of the tank actually leaks more if the height of the water [voltage] is higher and leaks less as the tank gets empty, complicating things)

I’m sure what you want to know is “how big a capacitor do I need to replace a certain battery?” Well that depends on a lot of things: the circuit, the type of regulator, the voltage you are charging your capacitor to (limited by its rating). I will say that in practice, it is usually surprising how big a capacitor you need for even the smallest applications.

Try this simulation and let it run for 2 min then experiment by changing the values. https://circuits.io/circuits/5256559-capacitor-supplying-regulator-with-load

In my opinion, different battery with the same mAh (coulomb) have different farad(capacitance) assuming chemical reaction charge/discharge rate is involved or not. It also depends on both + - plates and their surrounding medium inside the battery. There are a lot factors involved so there is no simple answer to that but it's interesting to figure it out.

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.

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?

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).

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.

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.

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.

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

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.

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.

Steve

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