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I need to monitor the current for a negative high voltage supply.

I implemented the following circuit in a simulation program: enter image description here https://www.maximintegrated.com/en/app-notes/index.mvp/id/1867

But this only works when the high voltage is +1kV DC. Any ideas how I could make the circuit work for -1kV DC?

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  • \$\begingroup\$ Since it is galvanically isolated, it won't know if it's 1kV or -1kV. That 9VDC is a DC/DC converter + LDO+ low pass filter. \$\endgroup\$ – Marko Buršič Oct 8 '18 at 13:24
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Given it's optically isolated, and that it has a separate battery supply, it should work as well for -1'000V (or -991V).

You would have to:

  • invert the pin on the shunt to have a positive reading of the current.
  • keep the battery as-is, means the low side would be -1000V - 9V max.
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  • \$\begingroup\$ Hi , by perfoming your changes i get a measurable output voltage but R2 consumes 365 W . Best regards \$\endgroup\$ – Suparman Sy Oct 9 '18 at 9:14
  • \$\begingroup\$ The only change needed is to switch the shunt direction. R2 should not consume that power, since it's powered by a 9V battery, how could it consume 365W? \$\endgroup\$ – Damien Oct 9 '18 at 9:25
  • \$\begingroup\$ Hello again , i suppose it must be a bug in the programm , i use NI multisim . I suppose i have to make a pcb and test it . Best regards \$\endgroup\$ – Suparman Sy Oct 9 '18 at 9:27
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Finally I used a floating voltmeter for the measurement. I use a 9v battery, a 10k sense resistor and a 7-seg display voltmeter with three pins VCC, FGND and the measurement probe, the negative pole of the battery connects to the -1100 V rail, the one side of the resistor and the FGND of the display. The positive pole supplies the display and the probe connects to the other side of the sense resistor. For a current of 280 uA the display reads 2.7 volts, which is ok for me.

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