Sunday, September 9, 2012

One Gigaohm high voltage probe

Measuring Geiger's power supplies is a difficult task, because usually the voltmeter is a huge load to the supply, therefore the measured voltage is unexpectedly low. For a good and realistic measurement, the used voltmeter must have a very high input impedance.

You can find there types of voltmeter around:

Chinese digital voltmeter: The cheaper ones, from 3 to 10 euros. They have only 1 MΩ of input impedance. At the left you can see how to measure the impedance of one of these voltmeters with the aid of another one.

The one at the left is in the 2V scale, and the one at the right is acting as a ohmmeter with 2 MΩ scale. You can see the impedance of the voltmeter (at left) is only 1004 kΩ.

Decent digital voltmeter: Most of them have an input impedance around 10 MΩ.

Analog voltmeter: The good ones usually have a sensitivity of 20.000 Ω/V, so about 20 MΩ at 1000 volts scale. Even 20 MΩ is a huge load for most Geiger tube's power supplies so for a precise measurement, you need input impedances of 100 MΩ or even greater.

1 GΩ resistor

Yes, they exists, a one gigaohm resistor, or 1000 MΩ. In fact, there are even higher values.  For example the CDV-715 have a 220 GΩ resistor in the ionization chamber circuitry. Of course there are not very common to find, but some manufacturers make them.

The one at the picture is a 1GΩ 1/2 watt resistor, and it is perfect to transform a cheap 1MΩ voltmeter in a really nice high voltage, high impedance voltmeter. How? Making a voltage divisor with the resistor :

Placing a 1 GΩ resistor in front of 1 MΩ voltmeter, you will get almost a 1/1000th voltage divider, so you can measure volts in the millivolts scale, but with an input impedance of 1001 MΩ !!!. In fact, it is not a 1/1000 divisor: You will read 999 mV with 1000 volts input... just close enough ;-)



A piece of plastic from a ball pen, a metallic tip, a socket and the resistor. That's all you need. Just solder the parts and glue the plastics. That's all!

You don't need a high voltage to test it. Just use your 12 volts power supply. It must read 12.0 +/- 0.1 at the 200 mV scale and 12 +/- 1 at the 2000 mV scale.

Does it worth it?

Examine the following table and extract your own conclusions. It was obtained measuring an unregulated power supply for a SBM-10 geiger tube:

Voltmeter Impedance Value
Cen-Tech 92020 + 1GΩ resistor 1 GΩ 400 V
ICE 680E 20 MΩ 375 V
Fluke 79 series II 11 MΩ 336 V
Uni-T UT81B 5 MΩ 307 V
Cen-Tech 92020 1 MΩ 214 V

Just a final tip: Do not try to measure over 2000 volts with this probe!!!

2kV is high enough to measure all geiger tubes and photomultipliers you can find.


  1. Where do you find such a 1G resistor and how much does it cost?

    1. They can be found on eBay from time to time. Cost depends a lot from one seller to another one. I have several resistors with different values from 1Gohm and up, and the most expensive was around $10.

  2. Excellent article. I have hunted around the net looking to solve accurately measuring the HV side of my Geiger circuits. My sparkfun DMM has 1 M ohm internal impedance... this article has enlightened me to how to measure more accurately. Thanks heaps! now to find a 1 G ohm res....

  3. Isn`t it a 500 to 1 divider because your meter is a 1M resistor in parallel with another 1M resistor!

    1. No. It's a 1000:1 divider. The volmeter in the schematics is only an ideal volmeter showing the voltage across the 1MΩ resistor (If you have ever use the old Electronic Workbench simulator you will understand it instantly). As you can see at the first photo, the real input impedance is 1004 kΩ

    2. He's using the nominal 1M ohm input of the DMM as the lower leg of the divider. It might be clearer if the diagram had a dotted box surrounding both the digital readout and the 1M resistor. Depending on the meter, you might not see the same ratio. For example, the Fluke 79 mentioned in the table above has a listed 20 Mohm input impedance. In that case you'd expect to see 1000 / 20 instead of 1000 / 1.

  4. The geiger tube works with DC, right? But what if you are measuring high voltage AC instead of DC? Would you have to connect a 1 gigaohm resistor to ground as well as to the probe?