Insulation Resistance Testing - Issue with Hi-Pot Tester

AI Thread Summary
A HiPot tester with a minimum voltage of 100VDC is being used to test for leakage between a string-pot and its housing, but the required testing voltage is 45VDC. To achieve this, the user considers using two power resistors to divide the voltage down to 50V, ensuring the resistors can handle potential leakage without burning out. The discussion confirms that calculating the wattage rating for the resistors involves using the formula P=V^2/R, with considerations for worst-case scenarios, such as one resistor being shorted. The example provided indicates that a 1000-ohm resistor would need to be rated for over 10 watts under certain conditions. Proper resistor selection is crucial for safe and effective insulation resistance testing.
elmos
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So, I have a HiPot tester with a minimum capable voltage of 100VDC.

I'm testing for any leakage between a string-pot and the housing case for the unit it goes into, with the requirement being 45VDC applied to it. With the HiPot tester incapable of supplying a voltage that low, I am thinking of using two power resistors to divide the voltage down to 50V, and then supplying that voltage to the EUT. I'm using power resistors to prevent the resistors from burning up in the event that I have leakage.

My question is: "what rated wattage should I choose for the resistors?" Is it a simple P=V^2/I problem? I'm guessing that leakage could result from fairly high resistances (kOhm range), but then it's entirely possible to have it down to 1 ohm or less if the wiring is bad.

I'm not a EE by trade (Aerospace), and I've been thrown into putting together this test for a qualification effort, so any advice would be much appreciated.
 
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If you put resistors across your tester, the tester will read the value of the resistors.

Would it matter if you tested at 100 volts? If it was OK at 100 volts, it would be OK at 50, surely?

To answer your question, you work out the worst case and design your resistors accordingly.
Is it a simple P=V^2/I problem?
Yes, it is.

Assume the lower resistor is going to get shorted out. The top one then has 100 volts across it. If it was 1000 ohms, it would have to be rated for better than 10 watts. E^2/ R = 10 watts.
Normally, it would have 50 volts across it so it would be dissipating 2.5 watts.
 
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