Help with a voltage multiplier...

[Lamont Cranston]

On Wednesday, May 3, 2023 at 9:10:32 PM UTC-5, Phil Allison wrote:

Lamont Cranston wrote:
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So now I\'m using it as a quadrupler, excess parts removed from circuit.
With 150Vac input, I measured Voltage with 10MΩ meter at the output at 184Vdc,
I have series connected Nine, 10MΩ resistors to make a 10 to 1 divider with my DVM.
I now measure 684Vdc. Difference, 10MΩ load vs 100MΩ load.
Using this data, I calculated a 43MΩ output impedance of the voltage quadrupler.
I expect with no load it has an output of about 958V volts with 150Vac input.

** This link might be useful to you:

https://www.extremeelectronics.co.uk/calculators/cw-voltage-calculator/


.... Phil

Thanks Phil.
Mikek

 

[Lamont Cranston]

On a somewhat different note:
Looking through some experimental literature on oil/water separation using using electric fields.
There is AC, sine and square, 50Hz to 50kHz, DC and pulsed DC, so I think the starting experiment is with AC, 60Hz.
Experiments showed little affect of frequency on separation efficiency.
I\'m seeing glass vessels wrapped with a copper foil and an electrode in the center of the glass
vessel in the solution. Applying the HV to this sets up the electric field, but I expect very little current.
The AC out of the neon sign transformer isn\'t to bad to measure as the transformer output impedance is 295K,
although 1000V open circuit measures 971V when loaded with a 10MΩ meter. ( I did build a 90MΩ prob e)
I\'m curious to know what current the Vessel with the electric field is drawing?
So, I\'m thinking of doing the usual, putting a resistor in series with the load and measuring the voltage drop across the resistor.
In this case the current is so low, I\'m thinking 1MΩ resistor in series will only drop 1V at 1ua, and I suspect current to be lower than that.
Does that seem like the way to do the current measurement?
Living vicariously through my son\'s experiments at his work, Mikek

Fun fact: I\'ve seen the neon sign transformer say 30ma, impedance protected.
The transformer we\'re using is 9000V at 30ma. I measured the output impedance at 295kΩ.
9000V / 295kΩ = 30.5ma, that could not have come out better!

 

[Lamont Cranston]

On Tuesday, May 2, 2023 at 12:43:15 PM UTC-5, Fred Bloggs wrote:

On Monday, May 1, 2023 at 9:55:15 PM UTC-4, Lamont Cranston wrote:
In my junque box I have a High Voltage, Voltage multiplier.
I pulled it out to help my son on a project and it doesn\'t work as I expected.
Testing of the parts tells me they are all good. But the voltages don\'t seem right. I\'m using a Bryman 235 DVM with a 10MΩ input impedance. So, I\'m don\'t expect I\'m loading the circuit.
I took a picture and put voltage reads on the picture with some info about the diodes and caps. The diodes are about 1\" long and have a 10v drop at 2ma.
I would like to know the voltage ratings of the diodes and caps, but don\'t
find any info on the net.
Any Idea, why my voltages are way off?
Mikek

https://www.dropbox.com/s/ox8xomn4aiutprf/HV%20Multiplier.jpg?dl=0
Bottom row second capacitor from the left is not holding its charge OR the fourth diode from left is not conducting OR you have a cold solder joint in that area.

After removing all the parts so I would have diodes for a full wave bridge, I checked the caps out off circuit. I did have one bad one,
The capacitance was good, but the RP was 1.9M the good caps were over 190MΩ. D showed it but not dramatic, Q was 44 vs 450.