C1 and C2 are microwave capacitors that each have a 10M bleed resistor in parallel with them. I'm charging them with a cheap HV model from amazon. I've been measuring the voltage with a 7:1 divider as shown. This is reading about 3.5kV (500V across the 2M resistor). The issue you I am having is figuring out the true voltage without the DMM/voltage divider. Since the model is so low current a significant portion of the total current is going through the divider, so when the divider is removed the actual voltage will be higher. Any ideas how I can go about getting a more accurate read other then just making a bigger voltage divider?

 

The best thing is to use a higher ratio voltage divider. Be very careful.
https://www.rossengineeringcorp.com...viders/hv-voltage-dividers-detailed-info.html

Ratio and Loading
In the standard 1000/1 or 10,000/1 Ross voltage dividers the low voltage capacitors and resistors are selected to provide correct ratios when shunted by 1 megohm, 20-50pF oscilloscope (or other load), plus 2, 3, 6, 15, or 20 feet (please specify the shortest practicable length) of low capacitance RG59 or special Ross Engineering Corporation coaxial cable unless otherwise required. Standard cable lengths are 3 or 6 feet for 15kV or 30kV dividers, 15 feet for dividers above 30kV, For the compact types, with metal oxide resistors, DC ratio is generally stable within better than 0.1% to 1% in an operating range of over 80 degrees C ambient temperature change for most models, and even better than 3-5 PPM/ºC (0.0003%) for most high accuracy types. In 0.1% dividers with less than 4 megohms per kV, units can drift up to 0.1% - 0.3% as they warm up when used continuously at maximum rating, less at below maximum rating. However, repeatability remains at 0.1% after stabilizing. The actual voltage divider ratio is determined by the following formula:

 

Just figure the 10 Meg input to the meter and the 2 Meg lower resistor in the divider as one resistor with resistance to the parallel combination of the two -then you re-calculate the divide ratio.

 

You probably cannot find a 10M 5kv resistor.
You could use 10 x 1M resistors where each resistor will see only 350V. Or 20 X 470k resistors. Check the data sheet for any resistor you use to see what the max voltage is.

 

Hi
the only way to get more accurate readings is to increase divider resistance. As we can see, the resistance must be much higher than 5 megohms ( two resistors 10 megohm each ) . 50 megohm or 100 megohm is acceptable, imho. So, simply take 5 or 10 resistors, 10 megohm each , and connect them in series

 

Why do you have the 10 meg resistors across the capacitors? Those draw more current than your 12 megohm voltage divider. How accurate does the voltage reading need to be? And what is the maximum voltage range on the voltmeter that you are using?? And how accurate is your meter specified to be??

 

I guess, those are conventional micrawave capacitors with built in discharge resistors, for safety

 

The ones that I have seen with resistors, the resistor is across the the terminals on the outside so that it does not affect the capacitor. OR it may just be a leaky capacitor.
But no matter which, the two ten meg resistors have more effect than the 12+ten meg divider. If you have the specifications for your meter then you can calculate the resistance for a 10: division, Then by simply adding series resistors you can produce the10: division, which will load the source a bit less, and make the math simpler.
What are the voltage and current rating of the " cheap HV model from amazon. " Of course, it may not have any rating information provided, which is very typical of amazon.