Electronic – limit to capacitor peak power density

capacitorpower

If so, what is it?

P=V^2/R, so increase voltage and keep R low. R is the resistance of the load, which is equal to the internal resistance. (With superconducting internal and external load R=0 but we have a current density limit of superconductors and inductance. Power could then be transferred to an antenna or motor. I guess then we'd still have impedance matched load.)

Increasing dielectric thickness d allows for higher voltage, but we also need to increase insulation at the fringe to avoid corona discharge, which increases volume/mass by d^3, while power increases with d^2, so power density actually decreases.

Could a capacitor made of two charged hollow spheres in the vacuum of space be scaled to ever increasing power/mass (on even volume) density? Large spheres/distance would limit "corona" discharge.

Do we get peak power density by fully exploiting the highest dielectric strength material (diamond?), regardless of capacitor size?

I'm asking, because

https://de.wikipedia.org/wiki/Energiespeicher#Speichern_elektrischer_Energie

said the maximum power of a normal capacitor is 10 kW (0.01 MW in column "max. Leistung in MW"="max. power in MW"), which seemed wrong.

https://en.wikipedia.org/wiki/Power-to-weight_ratio#Electrostatic,_electrolytic_and_electrochemical_capacitors

mentions a 3300 V, 100 kA capacitor, which works out to 330 MW.

Best Answer

Your post is unclear so I just offer the following observations.

Wikipedia's Power to weight ratio mentions a 3300 V, 100 kA capacitor, which works out to 330 MW.

That is in the section Electrostatic, electrolytic and electrochemical capacitors and the details are:

Type                      General Atomics 3330CMX2205 High Voltage Capacitor
Capacity                  20.5 mF
Voltage                   3300 V
Energy to weight ration   2.3 kJ/kg
Power to weight ratio     6.8 MW/kg @ 100 kA

330 MW is not "the power of the capacitor". It is the peak power the capacitor can deliver.

We can work out roughly how long this will last as follows:

$$ Q = CV $$ $$ V = \frac {Q}{C} $$ $$ \frac {dV}{dt} = \frac {dQ}{dt}\frac {1}{C} = \frac {I}{C} = \frac {100k}{20.5m} = 4.8 \ \text {MV/s} $$

Your capacitor, if it discharged linearly would be flat in \$ \frac {3300}{4.8M} = 0.7 \ \text {ms} \$.

However, I suspect that 100 kA can only be delivered into a short circuit and in that case the output voltage is zero so the power delivered to the load is zero and all the energy is dissipated in the internal resistance of the capacitor. This is probably the main flaw with your thinking. If at 3300 V the device can supply 100 kA max then it implies the internal resistance is \$ \frac {V}{I} = \frac {3300}{100k} = 33 \ \text m \Omega \$.

Now, using your maximum power transfer idea we would use a 33 mΩ load, current would be reduced to 50 kA, voltage reduced to half of 3300 and the maximum instantaneous power to the load would be \$ 50k \times 3.3k /2 = 82.5 \ \text {MW} \$.

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