Thanks,

Jay

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- Thread starter JMB
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Thanks,

Jay

Thanks, again.

Regards Karl

Assume that below a critical voltage (max voltage of the resistor) the power dissipation is only limited by case temperature (given in the datasheet). Calculate simply the stored energy in the cap and then the resulting power given the esr value of the cap. This gives you a simple differential equation for power as function of time.

behind how a 25 watt resistor might handle much more power for a brief pulse.

It's not power, it's energy. Power is energy per second and that is limited. As long as no other factors limit, the energy at a certain fraction of 1s can be as high as you like, provided the integrated energy (integration over 1s) stays the same.

This is just the mathematical side of things, I'm in no way involved with your special topic, so please get also additional advice from people working in this field.

All the best, Hannes

JMB said:

Thanks, again.

Hi,

Energy stored in a capacitor = 0.5CV*V = 2.5J

2.5J = 2.5 watts for 1 second.

Peak power = V*V/R = 2.5kW, peak current = V/R = 50mA.

Time constant = RC = 2ms = time to drop to 37% of full V,

Capacitor mostly discharged, 99% after 5 times RC.

I do not think you can arrive at 10W continuous rating by formala.

/sreten.

Good question but this is very dependent of the ressistor type but in general wirewounded resistors can handle more than film dito.JMB said:I am looking to see if there is a formula (not just a rule of thumb) that might predict a resistor's capacity to handle a rapid high voltage pulse.

If you are up to high voltage you must use high voltage resistors. Caddock has lot's of these.

btw Mr moderator, can you tell me why certain punctuation such as slashes and apostrophes are being interpreted as control codes while i type this reply?

Have you read the FAQ?

http://www.diyaudio.com/forums/misc.php?s=&action=bbcode

http://www.diyaudio.com/forums/misc.php?s=&action=bbcode

Thanks,

Jay

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