I have the circuit that you see below.

My load is a photomultiplier tube equivalent to a 3.63Mohm resistor.

My power supply can deliver max 4W (2kV and 2mA). I use it on 1100V and the load draws ~350uA. I want to monitor the current with the shunt resistor technique. Only high side current sense would work, because the load is tied to the circuit ground. I possess 0.1% tolerance resistors up to 500 ohm and 1% tolerance up to 1Mohm. I can spare up to 10 volts to drop across the shunt resistor.

I can't find a circuit with current amplifier able to withstand the 1100 volts relative to the ground on its inputs.

Any ideas?

schematic diagram


I've seen the Maxim application note with the optocoupler, but it won't work on my simulation package (multisim). I'm currently awaiting the ICs to test it for real.

  • \$\begingroup\$ You have two problems: the first, obvious one is that you need a sensing apparatus for the measurement of the voltage across \$R_\mathrm{shunt}\$ with high isolation, the same as the resistor itself. The second, more hidden one is that your load is a photomultiplier tube, which can sink currents peaks with rise times as low as 100ps: this implies that your shunt and the associated sensing apparatus should work well for low signals in the GHz range. Basically, you need a high frequency high voltage shunt and a high frequency high sensitivity voltmeter/oscilloscope: this is a hard problem. \$\endgroup\$ Sep 12 '18 at 7:25
  • \$\begingroup\$ Hello , thank you for replying. My goal is to monitor the current with a A/D converter or a lcd / led display. thank you \$\endgroup\$
    – Chris Sy
    Sep 12 '18 at 7:55
  • 2
    \$\begingroup\$ Have you considered using a floating power source to power the local sensing? The data could then be passed over a suitable optocoupler to a low voltage controller. \$\endgroup\$ Sep 12 '18 at 8:53
  • \$\begingroup\$ What do you expect to measure? The output of the PMT is a current - the current thru the device - and usually you drop this current across a 50 \$\Omega\$ resistor to ground (i.e. the low side) and that voltage is the your output. \$\endgroup\$
    – D Duck
    Sep 12 '18 at 19:19
  • \$\begingroup\$ D Duck hello , i want to monitor the input current from the HV supply that powers up the pmt . The pmt's output is handled by a different part of my circuit . Best regards \$\endgroup\$
    – Chris Sy
    Sep 13 '18 at 6:42

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