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I'm thinking, is it possible to make a battery out of a capacitors? If yes, how many 10,000 µf capacitors will be equivalent to a 100 mAh battery?

If possible how do I calculate it to get the values myself?

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    \$\begingroup\$ What voltage are they charged to and how much lower can that voltage become under load without preventing the target circuit load failing? \$\endgroup\$
    – Andy aka
    Commented Aug 18, 2023 at 10:41
  • \$\begingroup\$ It might be of interest that the first thing to be called an "electrical battery" was made out of Leyden Jars, which were an early form of capacitor: en.wikipedia.org/wiki/History_of_the_battery \$\endgroup\$ Commented Aug 19, 2023 at 17:24
  • \$\begingroup\$ Toolbox or something can't do this? uf to mah, provided the voltage? \$\endgroup\$
    – Mazura
    Commented Aug 19, 2023 at 17:50
  • \$\begingroup\$ Firstly capacitor self-discharge is incomparably higher compared to a chemical cell. Secondly "battery" means "one thing many times": see "anti-aircraft battery". \$\endgroup\$
    – Vorac
    Commented Sep 3, 2023 at 18:00

3 Answers 3

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A cells's energy storage capacity depends on its voltage. Given the cell's charge capacity \$Q\$ (in amp-hours, Ah), and average cell voltage \$V\$ (which will be somewhere between the fully-charged and fully depleted voltages) the energy \$E_{CELL}\$ (in Joules, J) of a fully charged cell will be roughly:

$$ E_{CELL} = Q \times 3600 \times V $$

By contrast, the energy \$E_{CAP}\$ stored in a capacitance \$C\$ is given by:

$$ E_{CAP} = \frac{1}{2}CV^2 $$

The main behavioural difference between cells and capacitors is that a cell's voltage remains more or less constant over the entire charging or discharging cycle, whereas a capacitor's voltage will ramp up from 0V during charging, never ceasing to rise until either you stop charging it, or it breaks. Similarly if you keep drawing current from a charged capacitor, its voltage will keep falling until it reaches 0V (and continue to fall, even, if current keeps flowing in the same direction).

This means that a cell's voltage is well defined, but the voltage across a capacitor can be arbitrarily high. This makes capacitors difficult to use as power sources for electrical equipment, because most loads require a fixed voltage to operate properly. Therefore you'd usually have to employ a voltage regulator of some kind (such as a DC-DC converter) to produce a steady, controlled voltage from the ever-decreasing voltage across the capacitor as it discharges.

Let's say you have a 3.7V LiPo cell, with a capacity of 100mAh. The 3.7V is "nominal", giving you some idea of the average voltage you can expect from it over a useful discharge cycle; it might start at 4.2V (fully charged) and end up at 3.2V at the point where you should stop discharging it, for example.

The energy stored in this fully-charged cell, which is available to be delivered over time to whatever load is connected to it, will be:

$$ \begin{aligned} E_{CELL} &= Q \times 3600 \times V \\ \\ &= 0.1Ah \times 3600 \times 3.7V \\ \\ &= 1300J \\ \\ \end{aligned} $$

That's a bit optimistic, since the charging process isn't 100% efficient, but since we are only estimating here, that's close enough.

To find out what capacitance we must use to store that same amount of energy, we first need to specify at what voltage we would declare the capacitor to be "fully charged". The upper limit is of course the capacitor's own voltage rating. Let's assume we are using 50V rated capacitors, but we don't want to operate them at full voltage, which would reduce their longevity, so we'll say that 40V is the maximum voltage.

We are aiming for 1300J of energy stored, when the capacitor has 40V across it:

$$ \begin{aligned} E_{CAP} &= \frac{1}{2}CV^2 \\ \\ 1300J &= \frac{1}{2}\times C \times (40V)^2 \\ \\ C &= \frac{2 \times 1300J}{(40V)^2} \\ \\ &= 1.6F \end{aligned} $$

To achieve this using 10,000μF capacitors, you would need this many capacitors:

$$ N = \frac{1.6F}{10000\mu F} = 160 $$


Update

One of the commenters below, user Elec1, has criticised my statement that

$$ E_{CELL} = Q \times 3600 \times V $$

The criticism is justified, since this relationship is not what you would find in textbooks, and the factor of 3600 is ambiguous. I'll try to clear that up here.

If all the units are SI, then the relationship should be written:

$$ E_{CELL} = Q \times V $$

The cell's charge \$Q\$ in that latter expression is assumed to have units of Coulombs (amp-seconds), the standard SI unit of charge. The product \$Q \times V\$ (with \$Q\$ expressed in amp-seconds) yields energy in units watt-seconds (Ws), which is the same as Joules.

However, the charge quantity \$Q\$ we are working with has the units amp-hours (Ah), and the product \$Q \times V\$ would have units watt-hours (Wh), not Joules. Since we are comparing with capacitor energy in Joules, we require cell energy also to be expressed in Joules, requiring conversion according to the relationship \$1\text{Ah} = 3600\text{As}\$. The number 3600 is 60×60, the number of seconds in an hour.

Perhaps it would have been more appropriate if I had separated the calculation into a conversion from Ah to As, using unambiguous variable names, and then arrived at the final energy calculation:

$$ \begin{aligned} Q_{[Ah]} &= 100\text{mAh} \\ \\ Q_{[As]} &= 3600 \times Q_{[Ah]} \\ \\ V &= 3.7\text{V} \\ \\ E_{CELL} &= Q_{[As]} \times V \\ \\ &= 3600 \times Q_{[Ah]} \times V \\ \\ &= 3600 \times 100\text{mAh} \times 3.7\text{V} \\ \\ &= 1300\text{J} \\ \\ \end{aligned} $$

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    \$\begingroup\$ I would like to add that we can't "fully discharge" capacitors, and C = 2 * Ecell / (Ustart^2 - Uend^2) indeed. The difference is negligible for a 50V capacitor that is discharged from 40V to 5V, but is notable if we have a 5.5V supercap that is discharged from 5V to 3.3V. \$\endgroup\$
    – ReAl
    Commented Aug 18, 2023 at 7:27
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    \$\begingroup\$ @ReAl I have a step-up regulator devboard from LT that can step up voltages as low as 20 mV, but I doubt it's very efficient at those levels! \$\endgroup\$
    – pipe
    Commented Aug 18, 2023 at 12:44
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    \$\begingroup\$ @u2n: The battery will also have discharging losses and will probably be used with a buck/boost converter. \$\endgroup\$
    – Michael
    Commented Aug 18, 2023 at 16:46
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    \$\begingroup\$ @Simons Fitch The correct formula is ECELL=Q×V and not ECELL=Q×3600×V. The latter form is a is a sloppy use of formulas and sometimes leads to the crash of Mars probes. A popular method with practitioners, because one believes to know the units of the result, if one uses only the "correct" numerical values. The correct way is to use arbitrary units and to carry them along consistently in the calculation process. \$\endgroup\$
    – Elec1
    Commented Aug 19, 2023 at 5:43
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    \$\begingroup\$ @Elec1 I agree with you, that was sloppy. I've appended a half-baked correction. \$\endgroup\$ Commented Aug 20, 2023 at 9:06
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If you take a battery that is a single-cell Li-ion and considered fully charged at 4.2V and discharged at 2.9V, we can calculate how many 10,000uF capacitors it would take to directly replace a battery without added circuitry.

Assume a constant 100mA discharge rate, the voltage change will be dv/dt = 1.3V/3600 seconds.

So C = I/(dv/dt) = 0.1*3600/1.3 = 277F. So you would need 277F/0.01F = 27,700 capacitors.

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    \$\begingroup\$ I think this answer is much closer to the spirit of the original question than the accepted one. It does answer "how many capacitors will be equivalent to a battery". Not how many capacitors PLUS some hypothetical conversion circuit. Still, I would base calculations on the energy, to remove discharge current from the picture. \$\endgroup\$
    – Maple
    Commented Aug 18, 2023 at 21:38
  • \$\begingroup\$ @Maple. Thanks. The discharge current won't matter in this case, it will cancel out since the capacitors don't care (much) about discharge current. Eg. 10 hours at 10mA. would be 0.01*36000/1.3. \$\endgroup\$ Commented Aug 18, 2023 at 22:08
  • \$\begingroup\$ It maybe depends on what you're doing with the current. If you're linear regulating it so the load draws a constant current (or fixed amount of total charge for its workload), this is appropriate. If you're using a switching regulator, you care about the amount of total energy you get out of the caps or battery, the integral of V*I. IDK if it makes much difference, but I think a battery stays higher voltage over more of the total charge (in coulombs) you pull from it, only dropping quickly toward 2.9V near the end. (So the discharge stopping point you pick matters a lot for cap vs. bat) \$\endgroup\$ Commented Aug 19, 2023 at 10:49
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It is possible but most of the time it makes no sense to do it.

You can use two very common equations:

Q = I * t and Q = C * U

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