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I am conducting a project to measure Breakdown Voltage and would like to use a protection resistor, please see below.
When breakdown happens I will have something like
In this case the Power dissipated by the resistor should be P=I^2 * R, which gives
P=I^2 * R=(0.1)^2*1000000= 10 KW.
However, I cannot find resistors with that power rating. I have seen that some HIGH POWER RESISTOR will reach 500 W - 2000W maximum. Others only work for 71 W .
Am I calculating this in the right way?
Any ideas?
If I do parse your question correctly, this is what you are trying to do:
As you have noted, a resistor with a 10KW power rating is kind of rare - you can of course get them custom build - money can buy almost anything.
My opinion: Make it simpler. Make sure to use a so called "impedance protected" transformer. Form a direct connection on the output terminals to the DUT-Cap. No current limiting is required in case of break down, as the transformer input impedance will be high enough to prevent blowing your power supply / trafo.
Please see:
Please consider: Lots of high power application notes - correct? This was turned out by google in thirty seconds. "impedance protected transformer filetype:pdf". Feel free to read these papers or go on the hunt again.
My assumption: Under the assumption, that the 10KW figure is far beyond your actual design requirements: Just get a 1:X impedance protected transformer an test you CAP. If the test current of .1A RMS is required make sure, that you understand the energy levels you are dealing with when forcing such a current on high voltage systems (Assuming >1kV) - sinewy stuff.
