# Trouble Measuring High Voltage with Voltage Divider

Discussion in 'General Electronics Chat' started by HoldenRyeField, Dec 18, 2010.

1. ### HoldenRyeField Thread Starter New Member

Dec 18, 2010
1
0
Hi,

I am trying to measure the output of 10kV DCDC converter (EMCO F101).

The output runs through a voltage divider:

V+(out) DCDC
|
|
120MOhm
|
|
10kOhm
|
|
V-(out) DCDC

I measure voltage across 10kOhm resistor with digital multimeter (MS8264, 10MOhm impedance). In a test case, I apply 2V input to DCDC and expect 2/15*10kV=1333V output. Thus I expect voltage across 10kOhm to be 10E3/(120E6+10E3)*1333=111mV. However, the dmm measured 186mV.

What is the reason for the discrepancy? Is there a better way to measure the output?

Apr 5, 2008
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3. ### timrobbins Senior Member

Aug 29, 2009
318
18
The output of the dc/dc is very load dependant, and the data sheet indicates the output voltage is significantly higher for lighter loads - you have a very light load compared to rated 10Mohm at 10kV.

Of course you could always enquite from EMCO what to expect - or if they have characteristic curves for input voltage and output voltage and loading.

And as bertus indicates - unless you are experienced in setting up hv measuring apparatus, then at least use a commercial probe and follow all the probe usage recommendations.

Ciao, Tim

4. ### Jaguarjoe Active Member

Apr 7, 2010
770
91
The output voltage is also dependent on the input voltage. Read the 4 footnotes under the table in your link.

5. ### gootee Senior Member

Apr 24, 2007
447
51
I agree that you should get a proper HV probe. Your or someone else's life might depend on it. But for the curent setup, did you at least try to measure the actual resistances in the divider, first, and use those values for your calculations? Power off and then short everything to ground first. Then remember to short your ohmeter probes together before measuring and subtract the indicated resistance value to correct the resistance measurements.