# Crockcroft-Walton Voltage Multiplier Large Voltage Loss

I have a CW voltage multiplier that uses 30kV, 1nF capacitors and doubled 20kV, 5mA, 100ns diodes (for a total of 40kV). My ac source is a sine wave (viewed on the primary side of the transformer because the secondary is beyond the capacity of my oscilloscope) around 20-24kHz and produces 15kV DC when half-wave rectified (measured after the 150Mohm resistor.)

simulate this circuit – Schematic created using CircuitLab

The output voltage of the CW multiplier is only 1.5kV. I have tested the diodes and capacitors. 100ns is compatible with most frequencies below 1MHz, I believe.

I am thinking that it might be perhaps that the 12 $$\\Omega\$$ resistor is not allowing enough current to charge the capacitors, but maybe it is some thing else such as some reactance I am not taking into account?

• Arcing (or almost invisible leaking) through the air comes to mind. I don't know the size or arrangement of your parts. Can you consider immersing this in pure mineral oil and also take a moment to make sure that the distances between nodes are appropriate? Have a look here for an example of what I'm thinking about. – jonk Apr 13 '19 at 20:56
• How are you measuring voltage? – user_1818839 Apr 13 '19 at 21:00
• Consider R2 and the input impedance of whatever you are using to measure voltage. – JRE Apr 13 '19 at 21:08
• @jonk the multiplier is immersed in mineral oil and the 15kv half wave rectification of the input will charge the capacitor to 15 kilovolts. – user198606 Apr 13 '19 at 21:53
• @Brian Drummond two ways. the first with a 100 volt voltmeter with the voltage split between the internal impotence of the voltmeter and an add-on resistor in the Mohms. The second, with a regular multimeter with a hundred times probe. The meters work just fine when measuring 15 kV when half wave rectified. – user198606 Apr 13 '19 at 21:58