Measuring HV using analog voltmeter

[MRMILSTAR]

I am building a quarter shrinker. I plan on using an analog voltmeter to measure the charge on the capacitor. The meter I plan on using is an old Weston DC volt meter with a maximum reading of 2500 volts on the scale. It has a measured resistance of 85 ohms and a full-scale current draw of 1 ma. I want to set it up to measure a maximum of 25 KV. To do this I am using a multiplier resistor of 25M ohms. The resistor is a Caddock HV resistor rated for 30 KV and 15 watts. Obviously, the true voltage will then be 10 times the meter reading. The highest voltage that I actually plan on using is about 20 KV.

In some application notes I see that a voltage divider using 3 resistors is used comprised of the meter internal resistance, another resistor in parallel with the meter resistance, and the multiplier resistor. Is this really necessary? I want to use a voltage divider consisting of only 2 resistors: the meter resistance and the multiplier resistor.

I am also wondering if I should ground the meter terminal that does not have the multiplier resistor connected to it to earth to avoid the possibility of the meter floating up to an excessively high voltage.

The entire quarter shrinker will be completely isolated from the mains except during charging. After the final charge voltage has been reached, the charging connections to the capacitor will be physically disconnected before firing.

Thoughts?

[OM222O]

how did you measure the resistance of the meter? 85ohm is incredibly low! even some ammeters have higher resistance than that!

if you use high value resistors (like the 25MEG you mentioned) you can no longer ignore the resistance of the meter as it will be put in parallel with your voltage divider resistors, so you have do the calculation based on that. easiest fix would be adding a 1Gohm or even 10Gohm resistor in series with the meter to increase it's input resistance beyond the divider circuit, so you can ignore it again.

[TimFox]

A simple voltmeter is a (large) resistor in series with a (low) current meter, as in your proposed circuit.
However, your ammeter requires a relatively high current for a high voltage sensor.
25 megohms and 1 mA will give a full-scale voltage of 25 kV, but the power dissipated in the resistor will be 25 watts, exceeding the 15 watt rating of the resistor.  At your expected maximum of 20 kV, you will still dissipate 16 watts.  You would be better with, say, 100 microamps for the meter and 250 megohms.

[German_EE]

"25 megohms and 1 mA will give a full-scale voltage of 25 kV, but the power dissipated in the resistor will be 25 watts, exceeding the 15 watt rating of the resistor."

It would be better to use 5 x 5M 5W ohm resistors in series, this will divide the 25W along the resistor chain and be a safer HV installation.

[MRMILSTAR]

The largest voltage I plan on measuring is 20 KV which would dissipate 16 watts which slightly exceeds the 15 watt power rating of the resistor. However, the vast majority of the charges will from about 10 KV to 15 KV which is well within the power spec. In addition, the charging time will only be a few minutes at most so the duty cycle of the resistor will be quite low affording the ability to slightly exceed its power rating.

[MRMILSTAR]

how did you measure the resistance of the meter? 85ohm is incredibly low! even some ammeters have higher resistance than that!

if you use high value resistors (like the 25MEG you mentioned) you can no longer ignore the resistance of the meter as it will be put in parallel with your voltage divider resistors, so you have do the calculation based on that. easiest fix would be adding a 1Gohm or even 10Gohm resistor in series with the meter to increase it's input resistance beyond the divider circuit, so you can ignore it again.

I measured the resistance with my Fluke multimeter as 85 ohms. The Weston meter has a scale which reads a maximum of 2500 volts. At that scale it is designed to be used with a factory-specified external multiplier resistor. I can't relate the part number of the specified resistor to any spec sheet that I can find because Weston is not around anymore. My calculations indicate that it is probably 2.5M ohms. With this external multiplier resistor the faceplate of the meter indicates a sensitivity of 1000 ohms/volt which is in-line with the 85 ohm meter resistance that I measured.

I am not ignoring the meter resistance, I am using it as one of the 2 resistors in a 2-resistor voltage divider. My point was that I do not think that I need a 3-resistor voltage divider (meter resistance, parallel resistor across meter, and series resistor. I think I only need a 2-resistor voltage divider (meter resistance and series multiplier resistor).

[MRMILSTAR]

A simple voltmeter is a (large) resistor in series with a (low) current meter, as in your proposed circuit.
However, your ammeter requires a relatively high current for a high voltage sensor.
25 megohms and 1 mA will give a full-scale voltage of 25 kV, but the power dissipated in the resistor will be 25 watts, exceeding the 15 watt rating of the resistor.  At your expected maximum of 20 kV, you will still dissipate 16 watts.  You would be better with, say, 100 microamps for the meter and 250 megohms.

I wanted a meter that had 25 with some factor of 10 as the full scale reading (25, 250, 2500, 25000) for easy reading of the true voltage. It also simplifies the task of finding a suitable HV multiplier resistor. 25M ohm, 30 KV, 15 watt Caddocks are readily available. I actually need 2 of these meters. One is for charging the capacitor and the other is for measuring the residual discharge voltage into a resistor bank. I found 2 identical Weston DC volt meters with a full scale reading of 2500 volts. Finding 2 of these meters that are identical is hard enough without also wanting a desired full-scale current draw. Accordingly, I have to take whatever current requirement that the meter needs and use an appropriate power resistor however inefficient that may be. Efficiency is not really a problem here. For the charging transformer I am using a NST rated for 1 KVA (60 ma at 15 KV AC) so the multiplier resistor power dissipation is only about 2% of the available power. The charging period is only a few minutes at most.

[TimFox]

Yes, your circuit will work.  There is no need for a “voltage divider” when connecting a series resistor and ammeter to measure voltage.
I always derate resistor power, due to early experience with fully or over powered resistors.  Your Caddock part is a good example of high-voltage resistor specification:  the resistor family is rated for 30 kV, but the 25 meg value will be limited by power.  Manufacturers specify a resistance value for a given family, below which the power limit governs.  To be conservative (always a good idea with high voltage), I would use two resistors in series, or GermanEE’s suggestion.