The energy stored in a capacitor is equal to 1/2 of the Capacitance value in farads multiplied by the square of the voltage that the capacitor is charged to.
As an example, if you wanted to power an led through a resistor from your 6.5 volt 1 farad capacitor, it will of course actually draw more power at the higher voltage than at the lower voltage end of the range, but for this example I will simplify and say that it draws .1 watt through the range of voltage that it will operate through, say 6.5 volts to 3.5 volts, then, you will have .5*1 Farad*((6.5 volts)^2-(3.5 volts)^2)= 15 Joules of energy. Now a watt is equal to a rate of energy usage of one Joule per second, so if we are using .1 watt and have 15 joules of energy that we are using over this range, then 15 Joules divided by .1 Joules per Second = 150 Seconds, which is the amount of time it will take for your capacitor to reach the 3.5 volt level.
In reality the resistor and LED would start out bright, and then dim as the voltage of the capacitor dropped.
For a capacitor where the capacitance is known, you can calculate the amount of energy stored just by knowing the voltage. Conversly, you can determine the capacitance if you can withdraw a known quantity of energy and measure the starting and ending voltages.
The larger a capacitor bank is, the flatter the voltage curve becomes as a function of the energy removed from the capacitor.
If the capacitor in the example above had been 5 farads instead of one, the voltage after 150 seconds would have been ((.5*5 farads*(6.5 volts)^2) - 15 Joules)=.5*5 farads*(final voltage)^2, or (105.625-15)/2.5)^.5 = 6.02 volts. About 6 volts, so the voltage doesn't change nearly as much as for the 1 Farad capacitor. It would take it exactly 5 times as long to discharge to 3.5 volts as well. For a very large capacitor, or a very small load, a capacitor can mimic a battery in that the voltage can remain within 10 or 15% of the starting value. You would not be using the capacitor very efficiently, however, because if you discharge a capacitor to even as low as half of its original voltage, 1/4 of the original energy remains in the capacitor. Most electronic devices cannot handle that much of a voltage swing, so when capacitors are used as batteries special circuits are used to convert the energy taken from the capacitor under varying states of charge and supply the energy at the voltage required for the end use.
The energy stored in a capacitor per unit of weight of the capacitor is much lower than that of most batteries. The cost per joule of useable energy is much higher than that for batteries in general. Because of this, you won't see capacitors as the main power supply for your cell phone, cordless drill, or electric car anytime soon. The power density is greater than that for many batteries meaning that they can discharge more amps per unit of mass for a short period than batteries can, so they see some use in conjunction for batteries for high current uses as in regenerative breaking in hybrid vehicles and in burst transmissions in some cell phones. The potential for improvements exists in ultracapacitor technology, and it appears likely that the energy density of production models will continue to increase for the forseeable future. The fact that they can cycle many more times than any battery technology out there that I am aware of plus the potential for further improvements makes me very optimistic about the future of these devices.
Rich Hagen
