# Can I replace a 3.8V 2900mAh battery with capacitors?

I have a 3.8V 2900mAh battery and I was thinking if it's possible to replace it with a bunch of capacitors

Here more specs about the battery

This is for a mobile phone and was planning to use capacitors instead of batteries since I use the phone as a modem and it always powered by its charger (Not sure how much power is given to the battery by the phone while charging) and I prefer to power the capacitors by what phone gives to the battery if its possible

• Why are you asking? What do you expect to achieve? Commented Apr 23, 2020 at 20:31
• You could try a super capacitor. If you plan to keep the capacitors outside the cellphone, then you can try to use one 2.7V super capacitor and step up the voltage to 3.9V or use two super caps in series to get 5.4V and step down and get 3.9V. But when charging the capacitor will never be fully charged as the max voltage for a Li-ion battery is 4.2V when charging. Commented Apr 23, 2020 at 20:37
• Use the formula: $U=\frac{1}{2} CV^2$ to calculate the energy stored in a charge capacitor. Then compare that to what a battery stores. Realize that you will need to discharge the capacitor all the way to zero volts to get all the energy out. Yes it is possible but not practical except for certain special applications. If you do not care that the amount of energy is much smaller then sure, what you propose might work. Commented Apr 23, 2020 at 20:41
• If the phone is always plugged into the wall... why does this matter? Also, phone batteries have more than just + and - terminals. There's some smarts going on in them. You stick caps (or even some unknown battery) in the phone, I'd expect the phone to reject them. LiPo batteries tend to blow up when mistreated so phones have alotta protections built in. Commented Apr 23, 2020 at 20:59
• I don't think there is usually much "smarts" in the battery, just a protection circuit and a temperature sensor. Commented Apr 23, 2020 at 21:29

So the goal here is to trick the phone into working without a battery? not to provide useful runtime.

I suspect you will have to do a few things.

First I suspect you will have to pre-charge the capacitor to a voltage that is reasonable for a lithium ion battery. Deeply discharged lithium ion batteries can be dangerous if later recharged, so lithium battery systems will generally have an under-voltage lockout.

Secondly the capacitor will have to be big enough to prevent over or under-voltage swings. Just how big may require some experimentation. IIRC the charge circuits in smartphones are often insufficient to cover peak power demand with the battery being expected to take up the slack.

Thirdly many phone battery packs will have some kind of temperature sensor in them. I suspect you may need to find a way to fake the response from said sensor (in many cases I suspect this will just be a resistor of appropriate value).

• Thats the thing this phone doesn't have any of those health pins Commented Apr 23, 2020 at 21:37
• Well that's good news for you then, you only have items 1 and 2 on the list to worry about. Commented Apr 23, 2020 at 21:38

For a short time, a capacitor can behave as battery because it stores energy. This can be useful for keeping a device on when you need to change batteries or if you want to have something turn on temporarily. However, because capacitor voltage and current decrease as capacitor discharges across a load, it wouldn’t be a practical long term replacement for a real battery. Also to even have a capacitor last for a long time would require very large capacitors and a light load. This is why batteries can be rechargeable is a useful aspect.

• Batteries cannot provide infinite load. Using a battery will gradually reduce its voltage. Commented Aug 4, 2020 at 19:40

Yes, you can replace a battery with a capacitor. The energy densities are much lower with capacitors, so the phone will have a very limited power on time, unless you use a lot of capacitors.

The voltage of a capacitor falls exponentially also so unless you have a DC DC converter to boost the voltage, you'll get even less time.

You need to know how much current the phone is taking and for how long. Just to give you an idea, a 2900maH is about 10000 Coulombs. To store the equivalent energy you'll need at least a 4V 2500F capacitor (or capacitors in parallel).

That last calculation is assuming that all the energy in the capacitors can be used, which it can't. And there will also be inefficiencies in the voltage regulator or converter to get a usable voltage to the phone. So you'll need an even greater amount of capacitance.

• Well it doesn't matter if the phone shutdown instantly once I remove the charger. also how I manage that spec? Commented Apr 23, 2020 at 21:34

The capacity of the capacitors can be back-calculated. Let's do some math.

The capacity of the battery is 2900mAh. Let's say that the phone will run for 6 hours on this. Roughly the phone is then consuming 480mA. Also, a typical Li-ion battery goes from 4.2 ish volts down to 3.4 with a hard cutoff at 3.0v. Armed with these facts suppose we want the phone to run for 10 minutes on only capacitors.

First, we need the total charge that the phone would consume. 0.480 * 10(60) = 288 Current is coulombs per second. Multiplying current by seconds cancels our the seconds leaving the total coulombs consumed.

Next, we use Q = C * V. If the capacitance remains the same we can mess with the charge levels to see what voltage we get. We can then build a system of equations.

4.2 * C = Q

3.0 * C = Q - 288

The first equation calculates our total charge at 4.2 volts. The second equation states that with the capacitance remaining the same, by the time the voltage reaches 3.0 volts the capacitor should have lost 288 coulombs.

The final answer is 240 Farads, which is huge. A minute of runtime cuts that value by 10. 24 Farads is more achievable. On Digikey currently, a 5F super cap with a 5v rating runs just over \$5. The rating of -0% +100% is terrible but it runs in your favor. You would need 5 of those to get roughly 1 minute of run time. These were very rough calculations, the run time is probably worse. Good luck, you can play with the numbers to see what is acceptable.