CWVM - Is it just a measurement error ,, ??

Thread Starter

Mars Park

Joined Feb 20, 2022
37
1656847140016.png

It is a CWVM circuit consisting of two stages. (Common ground)

The voltage of 578V was measured in the first capacitor and 582V in the second capacitor.
1656847403261.png 1656847436795.png


So, if I measure the second capacitor from the ground, I thought that there would be about 1160V.
But,

1656847681405.png

The multimeter shows 437 volts. The internal resistor of multimeter is 10M. (No resistor in probe.)

Is it because the ESR of capacitor is too high ??

And then i used 80k-6. (high voltage probe)
1656847959447.png
The result is below.
1656848123449.png
1656848147680.png

(It show as a value of one thousandth due to the voltage divide.)

Which value is correct ??
 

MrChips

Joined Oct 2, 2009
26,544
The internal resistance of the DMM is 10MΩ.
For 10% error your measuring device must have an internal resistance that is 10 times (or higher) than that being measured.
Your HV probe has an internal resistance of about 75MΩ, i.e. 8 times higher than the DMM.
 

Thread Starter

Mars Park

Joined Feb 20, 2022
37
The internal resistance of the DMM is 10MΩ.
For 10% error your measuring device must have an internal resistance that is 10 times (or higher) than that being measured.
Your HV probe has an internal resistance of about 75MΩ, i.e. 8 times higher than the DMM.
In conclusion, is the value measured by HV probe close to the actual voltage?
 
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