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How to calculate stored wattage of a Supercapacitor

How would I figure out how to calculate the stored power (in watts) of a 500 Farad capacitor at 2.8 volts? I know the basic volts times stored amps calculation for a battery but how would I figure this out for a capacitor?

Thanks in advance -Ray, KD2JID
 

Diver300

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Energy stored in a capacitor is measured in Joules (Watt-seconds), not Watts.

The energy stored in a capacitor is 0.5*C*V*V, so it would be 0.5*500*2.8*2.8 = 1568 J

That does depend on the capacitance not changing with voltage, and I don't know if supercapacitors do behave like ideal capacitors. Also, the voltage falls during discharge, so 1/4 of the energy would be delivered at less than half the starting voltage.
 

ronsimpson

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A battery has constant voltage during the discharge cycle. (almost) Nearly constant voltage for most of the charge cycle.
A capacitor voltage changes with the amount of charge.
Example; if you have digital logic that needs 4.5 to 5.5 volts to run, then the ICs will stop functioning just below 4.5V while the capacitor has most of its charge.
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AnalogKid

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As above, the total energy in, well, anything is measured in Joules. But I find it easier to discuss in watt-seconds. So the cap will deliver 1 watt for 1,568 seconds, 26.13 watts for 1 minute, or 1,568 watts for 1 second. Note that this is for the complete discharge of a theoretically perfect battery.

ak
 

alec_t

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They hold energy, not power. Power is the rate at which that energy is used.
 

Les Jones

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Neither holds an amount of POWER. They hold an amount of ENERGY. Power is the amount of energy per unit of time. So if a device (Battery. capacitor or inductor.) held 100 joules then if the energy was consumed in one second the power would be 100 watts. If it was consumed in 100 seconds the power would be 1 watt.

Les.
 

rjenkinsgb

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I'm just wondering if their is a similar conversion calculation to watts so I can understand how much power each one holds.
Batteries are commonly measured in Watt-Hours; or you can convert to that by multiplying the nominal battery voltage by its amp-hour capacity.

One watt-hour is 3600 watt-seconds which is 3600 joules.

At 1568 Joules, the cap holds a bit less than half a watt-hour.

It's not that much usable energy though, as the voltage drops linearly with a steady discharge current, vs. a battery which gives useable voltage down to full discharge.

If you are trying to use a supercap as a battery replacement, you need to allow the difference in energy stored between the minimum practical output voltage and the full charge voltage.
 

shortbus=

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No, know the differences between batteries and capacitors. I'm just wondering if their is a similar conversion calculation to watts so I can understand how much power each one holds.
You would be surprised at those who don't.
 

KeepItSimpleStupid

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I know the basic volts times stored amps calculation for a battery but how would I figure this out for a capacitor?

The volts * amps calculation for a battery really doesn't mean anything.

lets take a car battery for example:
There are couple of things:
Warranty
Cold cranking amps
It is a non-deep discharge battery type.

CCA is another way of telling you the batteries internal resistance. That limits the absolute maximum current.
In this application we don;t care abut the "gas guage" or Watt-hours the battery can provide.

Once you deplete the voltage below some value, the batteries lifetime has been seriously impacted.

You would use a deep-discharge battery for a trolling motor.

The only thing that makes any sense is the design curves.

The single numbers are good for comparisons and not design.

A Warranty costs you money.

Someone went to a water heater manufacturing plant. Some heaters had a 3 year and some a 10 year warranty.
They were maufactured on the same production line. The only difference was the cost.

There is a lot of battery information at www.batteryuniversity.com

In school I was told that a 1F parallel plate capacitor would be the size of a football field.

They exist now and early ones had a very low voltage between 2 and 3 Volts. They are difficult to charge quickly.
 

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