Capacitor Power Calculation. Am I doing this correctly?

I want to use some solar panel and super caps to power a “low voltage circuit”.

I want to calculate the effective “mAh” the caps would store at full charge.

Seeming as Q=CV
and Q=It

We can assume CV=It

Therefore I = CV / t

So having 2 100F 2.7V caps in series gives a 5.4V cap with 50F capacity:

Assuming t = 3600seconds (an hour):

I = (50*5.4) / 3600 = 0.075Amps = 75mAh

Now, the “usable voltage range” would be say a minimum of 3.3V before brownout occurs

So 5.4 - 3.3 = 2.1V

So I should use:

I = (50F * 2.1V) / 3600s = 29mAh

So over a day, assuming the caps can be fully charged over the day, I can pull about 1mA of current on average?

One nrmally uses :

energy stored = 1/2 c * v * v. Joules. Not useful here unless you use a switchmode psu.

And voltage / time characteristic as I = C dv/dt.

so if you have 50 F and current of 1mA, the rate of voltage drop dv/dt wiil be will be 0.001/50 => 20 exp - 6 v/s.

And to drop 2.1 volts will take 2.1 / (20exp-6 ) = > 105,000 seconds or 29.2 hours.

Your sums are correct, if unconventionally expressed.

Allan

We can assume CV=It

No, but C*(change in)V = I*t is a useful approximation.