Making a HV prove

[Alberto]

Hello.

I want to make a HV probe using a voltage divider and a needle voltmeter (my cheap DMMs get grazy when I try with my old HV prove)

If I get the point, the higher the value of the resistors, the better right? For example, lets make one up tu 150KV. Using HV Vishay resistor, each one can handle 10Kv so I can make a string of 20 resistors plus one more where I would measure the voltage drop. I know I have to keep in mind the value of the internal resistor of the voltmeter. What would be the ideal value of those 20 resistors? 50Mohm each?

And what about if I want to make one of 70Kv? Resistor of half the resistance of the first ones?

And one last question, to calibrate it, will it´s behaviour be linear? I mean, if I connect to a well know source of for example 500 volts and  the value is 1 volt measured for every 1.000 volts, when I connect it to 50Kv, the ratio will be the same?

Thank you

[TizianoBll]

Hi,
when using a digital dmm I suggest wrapping it with some aluminum foil connected to common (the negative terminal), leaving out only the display. Doing that stopped my dmm from resetting.

Analog dmm won't reset but, under strong electric fields, the needle will polarize and stick to the scale (so it should face away from the HV side).

Low Value resistor -> it will load the circuit and heat up (causing some error due to temperature coefficient)
High Value resistor-> It won't heat up, but leakage currents, humidity and corona will cause higher errors -> worse stability, usually high value resistors have worse specifications.

Usually a good value is 50-200uA flowing at nominal voltage. For example: 150kV nominal, 75uA current -> 2Gohm @ 11.3W.

For calibration you apply 1000V, measured directly by a precise DMM, and compare it with the divider. If the divider is good enough the ratio will be mostly the same at all voltages.


Will it be linear? It depends:

I have a 270Mohm 100kV divider. It consists of two URSS resistors in series. At nominal voltage it dissipates 40W, the resistor heats up and the linearity error is 2.5% (4kV ratio vs 100kV ratio). It is not good due to the high temperature coeff. of these resistors (0.2% per °C). But still usable for monitoring (and the error is easy to calibrate out). So watchout for tempco.

If you use higher value resistors they won't heat up but, any leakage or corona (not linear, depends on weather, difficult to calibrate out) will cause some error (measure higher).

[Alberto]

...

Thank you a lot. That just is the value I needed.

I didn't express myself properly. I'm going to make a complete voltmeter. For example, for a 50Kv voltmeter I would use a string of 10 resistors of 47Mohm plus one resistor of 9400 ohm what would give me a max current of 50.000/470.008.400 = 106uA a ratio of 1:50.000. I'm looking for that ratio because I have some needle 1v voltmeter.

Another question is that those voltmeters have a internal resistance of 1.140 ohms. That resistor is in series with the voltmeter coil so I'm not sure if I have to take it in account for the calculations.

[TizianoBll]

Of course you need to consider the impedance of the voltmeter.

I'm afraid that you can't use that meter, infact:
1V/1.14kOhm = 877uA, way more that your 106uA. You should use a 100uA ammeter, it would be perfect.

If you use a uAmmeter you can add two protection diodes (so if, in case of an overload, or if the ammeter goes open circuit they'll clamp the voltage).

[Alberto]

Yesss I forgot about that. Thank you, I'll use a 100 uA ammeter