GS-1100-Pro only outputting half voltage

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Quaid
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GS-1100-Pro only outputting half voltage

Post by Quaid » 29 Jun 2020, 07:31

A few months back I picked up a Gs-neutron and a Gs-1100-pro as a power supply (for 500V). It worked fine for the neutron as I got expected background neutron counts, but haven't used it since. I recently put together a scintillation detector based off the XP5312-SN PMT. Instead of building my own power supply and pre-amp, I thought I'd use the Gs-1100-pro to power the PMT at 1000V. I tested the output out of curiosity with a new probe (1000:1) and found the voltage was only a little above half of what was being displayed on the screen (600V for 1000V). I tried to adjust the calibration pot on the back of the meter but it only adjusts +/-100V, and wasn't enough to bring it in range.
Gs1100Vout.jpg
Is this a flaw with my measurement or with the Gs-1100-pro? In a previous post about the gs-1100a, I think one of the zener diodes had failed. My measurement seems the same - measuring with 1Gohm probe after letting it warm up for a few minutes. If the pro is the same, could this be the problem here?

Anyway, the detector has been giving pulses at the measured Gs-1100 output (~600V) so I haven't pushed it any higher worrying my measurement is at fault.

If anyone has some guidance, it would be much appreciated :)

Cheers,
Quaid
Quaid || Canada || "Get to the reactor!"

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Steven Sesselmann
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Re: GS-1100-Pro only outputting half voltage

Post by Steven Sesselmann » 29 Jun 2020, 08:29

Quaid,

What you are seeing is perfectly normal. The voltage display on the GS measures the voltage on the high end of the load resistor, therefore it technically displays the correct voltage for infinite impedance, but when you connect a standard multimeter with ~2M impedance, the voltage will read much lower.

With a geiger tube or proportional tube where impedance is "almost" infinite the voltage display on the GS should be within a couple of volts, but if you hook up a PMT with low value resistors in the divider, then the display will be out.

You can easily calculate it...

(voltage on display) = (voltage on BNC) * (resistance in detector)/(resistance in detector + 1 M Ohm)

1M Ohm is the load resistor.

In the case of a typical multimeter the impedance might be around 2M so your voltage will read low by 2/3.

Steven

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