10,000 Farad Supercapacitor Amp Hours?

Thread Starter


Joined Jul 15, 2014
I watched a Youtube video yesterday where (Robert Murray Smith)showed he had built a 10,000 Farad super-capacitor,I think he said it was 3.7 volts and 38,000 joules? What Id like to know is what the amp hours would be with those specs? Ive tried to work it out using ohms law but how to calculate the equation escapes me? Ive looked up lots of articles on ohms law but they never explain adequately (to me at least) how they come to the conclusion?


Joined Oct 2, 2009
The charge on the capacitor is given by the formula, Q coulombs = C farads x V volts

Q = C x V

Current I amps = Q coulombs / t seconds

That would give about 10 Ah.


Joined Aug 1, 2013
Amp-hours, as used in describing battery capacity, does not apply to capacitors. A battery has a relatively flat discharge curve. That is, with a constant load on the battery there is a time period when the output voltage changes very little. Then as the battery is running out of charge the output voltage drops quickly. In very round numbers, a battery's amp-hour rating describes how long the output will be in that slowly changing flat spot.

In a similar situation, a capacitor does not have that flat spot. Depending on the type of load, the discharge curve usually is either a downward sloped straight line or exponential curve. As above, you can calculate the total energy in a cap and relate it to the total energy in a battery, but the discharge characteristics are so different that "amp-hours" does not apply meaningfully to a cap.

What you can do is determine the minimum capacitor voltage you need to run your circuit and work from there. For example, if you have 5 V on the cap and you are running a circuit that will function down to 3 V, like most CMOS logic chips, and you know the current the circuit draws, you can calculate how long it will take the cap to discharge from 5 V to 3 V. Again, the equation changes depending on the load characteristics, but in this case the cap can be thought of as functioning as a battery.



Joined Oct 2, 2009
You are both correct.

The voltage and current from a capacitor with a fixed resistive load will follow a decreasing exponential curve.
Another simple calculation is time-constant tau (seconds) = RC

Suppose you apply a 3.7Ω load resistance across the capacitor

time (seconds) to fall to 37% of the original voltage = R x C = 37,000 seconds = 10 hours

A rough estimate of "useful" amp-hours might be something like 1-2Ah.