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I am looking to set up a voltage-measuring system for the output of a recently purchased HV power source. The source supplies around 0 - -60 kV DC and 10 mA to my load. Unfortunately, the source does not have a voltage measuring system, so I will be tasked in its creation. This will be of particular difficulty because I do not have enough money to purchase sufficient HV probes that could simply be attached to a multi meter. Thus, I will need a way to lower the incoming voltage from -60 kV to something manageable for the multi meter.

Currently, this is the schematic that I am considering to use in order to lower the voltage:

schematic

simulate this circuit – Schematic created using CircuitLab

The multi meter that I currently own can handle up to 660 VDC. Using my apparatus above coupled with the multi meter, would the multi meter be able handle the voltage that is delivered to it and give out an accurate reading of the line voltage (assuming that a series of calculations are performed)?

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  • \$\begingroup\$ Maybe look into a capacitive divider. I know they've been mentioned here when discussing how to measure high DC voltages. \$\endgroup\$
    – JRE
    Commented Sep 12, 2018 at 16:14
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    \$\begingroup\$ This is extremely dangerous. You have to make sure that all of your connections will not leak current at these high voltages. Don't plan to use a PCB. Please, please find a way to buy a good bit of commercial test equipment. \$\endgroup\$
    – Elliot Alderson
    Commented Sep 12, 2018 at 17:37
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    \$\begingroup\$ Any high voltage like this will also create tiny quantities of static charge (as part of a gradient from one end to the other of a galvanic path) that immediately set themselves up in order to cause the current to flow. Those charges can be independently measured. That's another approach to consider. How often are the measurements? What kind of resolution do you require? Is there a separate accuracy requirement? \$\endgroup\$
    – jonk
    Commented Sep 12, 2018 at 18:11
  • \$\begingroup\$ What is (1) the output impedance and (2) output capacitance of this -60 kV 10 mA DC PS? A schematic would help. \$\endgroup\$
    – Mike Waters
    Commented Sep 12, 2018 at 19:26
  • \$\begingroup\$ Even wire is not easily arranged at high voltage; instead of 'a multimeter', consider using an electrostatic meter (a moving-vane capacitor, or even a piezoelectric crystal or bimorph) that draws no current. \$\endgroup\$
    – Whit3rd
    Commented Sep 13, 2018 at 6:16

3 Answers 3

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  1. You are unlikely to find a single resistor that can handle 60kV. You will probably have to use many resistors in series. Make sure to consider the tolerance when calculating the voltage across each resistor.

  2. The 100M resistor would be dissipating 36 W.

  3. Be careful of the loading on the supply. This contraption draws 0.6mA.

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From my knowledge, the accuracy of your reading will depend on the sensitivity of your multi meter. According to your divider,about 600uA flows through the resistors. So a small current like 100uA drawn into the multi meter will change the readings.

Almost every digital multi meter uses an electronic amplifier (op amp) in the input stage making the impedance fixed. So a good digital multi meter will do the job for you. But most of the analog multi meters will draw a varying current depending on the input voltage. So that will be a no go.

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  • \$\begingroup\$ As TEK mentioned, using multiple resistors in series rather than having a one big resistor will help the tolerance problem too. Negative errors may be cancelled out by positive errors. \$\endgroup\$
    – Arosha Dissanayake
    Commented Sep 12, 2018 at 16:21
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    \$\begingroup\$ @AroshaDissanayake Increasing accuracy when combining resistors doesn't usually work in practice. You are assuming that when you buy a batch of resistors that their mean value is the nominal value and their deviation has a normal distribution. It is extremely unlikely that you would get that in the real world. \$\endgroup\$
    – Elliot Alderson
    Commented Sep 12, 2018 at 17:35
  • \$\begingroup\$ @ElliotAlderson yeah, you may be right. But more importantly using a series of resistors divides the heat dissipation among them right? In this case, we have to solve that too. \$\endgroup\$
    – Arosha Dissanayake
    Commented Sep 12, 2018 at 17:39
  • \$\begingroup\$ Let us continue this discussion in chat. \$\endgroup\$
    – Arosha Dissanayake
    Commented Sep 12, 2018 at 18:55
  • \$\begingroup\$ @ElliotAlderson for a one-off project you can choose the best subset from a batch to even things out. And if it happens that they're all high, well, you can buy another batch. It's not just blind reliance on statistics. \$\endgroup\$
    – hobbs
    Commented Sep 13, 2018 at 1:16
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A small change would be to use the multimeter in current measuring range, instead of voltage measuring.

This way you can reduce the measurement current to microamperes, and it would also be much less sensitive to the impedance of the multimeter. And the voltage on the wires going to multimeter will be less than a volt.

It is still a good idea to leave R2 in place to limit the voltage in case the meter is not connected. For µA level measurement current, your value of 1 Mohm is suitable to limit the voltage to quite safe ~60 volts.

And like other answers already mentioned, you will have to construct R1 from multiple resistors as a physically long chain, to avoid arcing. For example, 100x of 10Mohm resistors in series would give 1 µA = 1 kV response for the meter.

The advantage is that in current measurement mode, the multimeter's impedance is very low (at most a few kilo-ohms) compared to R2. This way all measurement current goes to good use in the meter, instead of being wasted as heat. Also the exact impedance matters less, because in any case it is much smaller than R2, whereas in voltage mode the impedance is several megaohms and is thus similar to R2 value.

By reducing the measurement current to microamperes, the power lost to heat is reduced to e.g. 60 µA * 60 kV = 3.6 W, which is only 0.036 W per resistor.

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  • \$\begingroup\$ @SuperNerdsTeam For measuring current. That way all the measurement current goes to good use, instead of being wasted as heat in voltage divider. \$\endgroup\$
    – jpa
    Commented Sep 13, 2018 at 5:25

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