My Cockroft-Walton Multiplier is not multiplying the voltage as projected

intrepidelder

Currently I have been messing around with relatively low voltage diodes and caps and arranging them in the specified order that has been found on the internet. My complication is this: My input is simply 120 V ac from a wall outlet. when connecting one stage ( one stage consisting of 2 Diodes and 2 Caps) the ouput voltage is 260 v dc. This is great! and exactly the result that I had anticipated. Following this I attatched one more stage, but my output voltage was only 312v dc. This isf ar below what is anticipated in the calculations. I should be receiving 480 vdc at the very least! I have exhausted all my resources and turn to anyone one who can aid me in my project.

I must clarifiy that the Caps I have are rated for 500 v and the diods are rated for 400 v, so the componet's limits are not being exceeded. I have arranged the diodes correctly so that they are facing the correct direction. For a refrence of the layout that i am using you can use this link as a reference: http://blazelabs.com/e-exp15.asp

I would greatly appreciate some help

-IntrepidElder

uart

What type of diodes are you using?

What is capacitance value (and voltage rating) of your caps?

What, if any, loading do you have on the output? (including the measurement circuit).

Do yourself a favor when playing around with a circuit like this and stick a relatively large resistor in series with the output (at least 300k while you're experimenting). And BE CAREFUL.

DragonPetter

What kind of capacitors do you use? I think certain large types lose their ability to store charge(their capacitance goes down) if they are charged around half their voltage rating. I forget where I read this, but I thought it was for electrolytics. So if you are drawing a load from it, your effective source resistance will be much higher. Is the circuit unloaded?

intrepidelder

The 4 caps are 0.001µF 500V 20% HI-Q Ceramic Disc Capacitor and the Diodes are 1N4004 Micro 1-amp Rectifier Diodes. I did put so resistors in series, but all that did was drop the voltage by two. I have no load yet, and am using a Voltmeter to ckeck the voltage that is all I have hooked up to it. so yes the current is unloaded.


-Intrepid Elder