Thu Sep 29, 2011 11:59 am

Physics time: The voltage over a capacitor is linearly dependent on the charge and reciprocal on the capacitance, or the other way around, Q=V*C, charge equals voltage times capacitance. Charge is measured in coulombs, capacitance is measured in farads. A 1F capacitor charged to 1 coulomb will measure 1V. From this you can see how much energy is stored in a given capacitor at a given voltage.

Let's say you can use voltages between 3V and 5V to power your circuit and you're using a 10F capacitor (or ten 1F capacitors) charged to 5V to store energy. At 5V, there's a charge of 50C, at 3V the remaining charge is 30C. The average voltage is 4V and you got Q equals 20 coulombs out of the capacitor. You know that power is voltage times current, P=V*I. Work (or energy) is power times duration, W=P*deltat. Thus W=V*I*deltat. The current is charge divided by duration, so W=V*Q. In our example, you got 4V*20C=80J out of the capacitor. One joule is one Ws (wattsecond), so if the circuit consumes 2W, then you can power it this way for 40 seconds.

The energy stored in a capacitor is usually given as W=0.5*C*V^2. To see that our calculation is correct, set the low voltage to 0V: Charge 50C, average 2.5V, so W=2.5V*50C=125J. That's the same as W=0.5*10F*(5V)^2=125J. In practice you can't use voltages below a certain threshold.

Let's say you buy two 2600F 2.5V supercaps and put them in series to get to 5V. Again, let's say you can use 3V to 5V. That's W=0.5*(5V+3V)*(5V-3V)*2600F=20800J. At 2W, that would be 20800Ws/2W=10400s or about 2 hours and 50 minutes, if the capacitor could hold a charge that long. Of course you'd have to pay more than twice the price of a Raspberry Pi Model B just for the capacitors. Four rechargeable AA batteries store about twice as much energy.