2
\$\begingroup\$

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

PS:

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.

\$\endgroup\$
5
  • \$\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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.