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| Measuring HV using analog voltmeter |
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| 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. |
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