Cockcroft-Walton Multiplier with square wave 0-338V. Need negative voltage? - Page 1
Forum section: electronics / projects

author: kathan
author_rank: Newbie
timestamp: April 11, 2017, 12:54:30 pm
content: As title suggest i need to know if the C-W multiplier will work with 0 to 338V Square wave?My Main task is to generate ~4-5Kv for electro-static Air filter i am working on. So far i came up with Rectifying Mains AC(220-240) voltage to DC via bridge rectifier and filtering capacitor. Then using H-Bridge converting 338Vdc to square wave with 10KHz Frequency and feeding it to C-W multiplier.the reason for choosing higher frequency was to reduce the size of capacitor used in multiplier. Also i need DC voltage output, hence C-W multiplier. Any other suggestion's are welcomeI am kinda stuck and any help would be awesome.Thank youMy Refrence was thishttp://blazelabs.com/e-exp15.aspandthis

author: Zero999
author_rank: Super Contributor
timestamp: April 11, 2017, 01:51:08 pm
content: I haven't watched the video.The polarity can be inverted by reversing all of the diodes and capacitors, if polarised.

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: April 11, 2017, 02:16:22 pm
content: When using higher frequencies, remember that stray capacitance can become an issue, so you will start getting losses at some point.You probably want to keep away from audible frequencies.

author: kathan
author_rank: Newbie
timestamp: April 11, 2017, 03:00:44 pm
content: Quote from: Hero999 on April 11, 2017, 01:51:08 pmQuote from: mikeselectricstuff on April 11, 2017, 02:16:22 pm@Hero999 i do understand that, the issue is can i use CW multiplier without negative supply. thank you for your reply.@mikeselectricstuff I took losses into consideration but the main issue is can i use "0 +338" instead of "-338 +338". And how can i get '+-' voltage from '+ and ground' output from rectifier and i am using mains(240V) ac directly as an input to bridge rectifier. Btw i have been following you on YouTube for a long time, you are awesome.!

author: NiHaoMike
author_rank: Super Contributor
timestamp: April 11, 2017, 03:05:18 pm
content: The current used by an electrostatic air filter is so tiny that the capacitors are still pretty small even at mains frequencies.http://bigclive.com/ioniser.htm

author: kathan
author_rank: Newbie
timestamp: April 11, 2017, 03:16:17 pm
content: That's an awesome link thank youBut is the voltage output of ionizer enough to be used for electrostatic filtering of air?

author: Zero999
author_rank: Super Contributor
timestamp: April 11, 2017, 03:21:47 pm
content: Quote from: kathan on April 11, 2017, 03:00:44 pmI don't understand the question. There is no need for either a positive or negative supply. The input is AC (whether taken from a mains transformer or oscillator) and the output is DC, with the polarity determined by the diodes.

author: kathan
author_rank: Newbie
timestamp: April 11, 2017, 03:41:14 pm
content: Quote from: Hero999 on April 11, 2017, 03:21:47 pmBasically i wanted to convert mains ac voltage to dc then using H-bridge set is frequency to around 10Khz to decrease size of capacitors as calculated from this linkhttp://blazelabs.com/cw-brm-java.aspbut CW multiplier needs positive and negative cycles to work, so i was asking if it is possible to use positive and ground level for each half cycle instead of positive and negative level for square wave.

author: Zero999
author_rank: Super Contributor
timestamp: April 11, 2017, 04:05:19 pm
content: Quote from: kathan on April 11, 2017, 03:41:14 pmIncreasing the frequency will also increase the losses.To answer the question. You don't need a bipolar squarewave. The device producing the squarewave just needs to be able to sink and source enough current.A squarewave going from 0V to 340V is equivalent to a bipolar square wave going from -170V to +170V, plus 170V of net DC. The DC in the signal will be blocked by the first capacitor but if you connect the 0V connection of your voltage multiplier to +V, then it will give you 340V on top of whatever the expected calculated value will be (assuming it's configured to give a positive output). Alternative, you can achieve greater multiplication ratios by using a full bridge driver to give a square wave going from - 340V to +340V.

author: kathan
author_rank: Newbie
timestamp: April 11, 2017, 04:23:56 pm
content: Quote from: Hero999 on April 11, 2017, 04:05:19 pmWow Thank You that did cleared a lot of stuff so if i connect square wave of 0 to 340v to +Vin and +340Vdc to 0V terminal's of multiplier it should work, Right?. And i am using IR2104 mosfet Driver along with IRF840 to Generate square wave. I also liked the Ionizer approach suggested by @NiHaoMike. Either solution would work i guess?

author: Zero999
author_rank: Super Contributor
timestamp: April 11, 2017, 05:30:41 pm
content: Quote from: kathan on April 11, 2017, 04:23:56 pmYes, that will work. To cut down on the number of stages, how about adding a step-up transformer between the output of the driver and the voltage multiplier? Transformer tend to have higher efficiencies than diode & capacitor multipliers.