# Power from a capacitor

I need to discharge a lot a energy in a short time from a capacitor, I don't have the numbers yet, but my question is if the power of a capacitor is limited for some others parameters of this.

Theoretically the energy stored depends of the capacity of the capacitor and the time of discharge can be adjusted with a resistance in order to change it. With this parameters I can obtain the power that the capacitor will give.

I'm trying to "power" a tungsten filament, like a light bulb, in order to transfer the energy for a chemical reaction, also I want to transfer a few Joules in a few nanoseconds.

Should I consider another parameters for this problem?

• You also need to pay attention to ESR (equivalent series resistance), since at high power (current) levels, a lot of your energy will get dissipated there. – Dave Tweed Jan 5 '15 at 20:40
• Actually energy stored in a capacitor only depends on capacitance and the voltage it has $E=\frac 1 2 C V^2$. Power=E/t depends on time, i.e. you can get a lot of power during a short time, or less power during longer time. – Roger C. Jan 5 '15 at 20:45
• You MUST say what you are wanting to "power". Using a resistor to limit energy transfer rates is usually an extremely bad idea when dealing with "a lot of energy". Telling us what you are trying to achieve will help a lot. – Russell McMahon Jan 5 '15 at 21:35
• Real capacitors also have dielectric absorption losses (dissipation factor) as well as reduction in capacitance with applied working voltage. Be sure to read the manufacturer's data sheet when choosing a capacitor for this kind of application. (high-voltage? rapid discharge?) – MarkU Jan 5 '15 at 22:25
• How much is a lot of energy? How long is a short time - seconds or nanoseconds? – pjc50 Jan 6 '15 at 0:42

"I want to transfer a few Joules in a few nanoseconds."

This is going to pose some major obstacles.

In order to reduce the discharge time the capacitor should be as small as possible. To increase de level of energy storage I'd set the maximum voltage at 1000V, which is already hazardous.

C = 2E/V² or 2*5/(1000*1000) = 10μF

When loaded by resistor R, a capacitor could be considered fully discharged after 5*R*C secs. R = t/(5*C) = 5.10𐄐⁹/(5*10.10𐄐⁶) = 0.1mΩ.

This is about the resistance of 10cm of 15AWG copper wire and far below the ESR of a physical capacitor.

The most important issues that you will face regarding the capacitor with trying to discharge so quickly are probably the equivalent series resistance and the inductance of the leads. Both of these can be mitigated by using more capacitors in parallel instead of a single large capacitor. Both resistance and inductance in parallel reduce in this configuration.

There are other non-capacitor related issues such as the switch and other inductances in the system as well as how to control the current. In any case, take care with your experiments.