I'm building a negative ion generator for a friend. I'm using a full wave Cockroft-Walton voltage multiplier (**broken link removed** with the diodes reversed since I want negative voltage output). I've built 10 stages on an a thick plastic panel. At this point I have two questions:
1. Because of the number of capacitors in the circuit the inrish current will be big, how can I limit it (and how to calculate)?
2. I want to keep the Vout high but limit the current in case someone will touch one of the needles of the ion generator. I have a bunch of 1MOhm 5W rated resistors, I'm thinking to limit the output to 1mA but for that I need to fill a hole in this formula:
Eout = 2n * Epk - Vdrop
n = Stages
Epk = AC input peak volatge
==>
Eout = 2*10 * 311 - Vdrop
Eout = 6220 - Vdrop
Vdrop = [ Iload / (6*f*c) ] * (n^3 +2n)
Iload = Load in Amps
f = freq. Hz
c = Capacitance in Farad
n = stages
(after calculations) ==>
Vdrop = (Iload / 0.000003) * 1020
So the big question is - what is my load? If I'll know that I can calculate the resistor value according to ohm's law.
I'm building a negative ion generator for a friend. I'm using a full wave Cockroft-Walton voltage multiplier (**broken link removed** with the diodes reversed since I want negative voltage output). I've built 10 stages on an a thick plastic panel. At this point I have two questions:
1. Because of the number of capacitors in the circuit the inrish current will be big, how can I limit it (and how to calculate)?
The inrush current will be small, because the capacitors are small.
2. I want to keep the Vout high but limit the current in case someone will touch one of the needles of the ion generator. I have a bunch of 1MOhm 5W rated resistors, I'm thinking to limit the output to 1mA but for that I need to fill a hole in this formula:
Eout = 2n * Epk - Vdrop
n = Stages
Epk = AC input peak volatge
==>
Eout = 2*10 * 311 - Vdrop
Eout = 6220 - Vdrop
Vdrop = [ Iload / (6*f*c) ] * (n^3 +2n)
Iload = Load in Amps
f = freq. Hz
c = Capacitance in Farad
n = stages
(after calculations) ==>
Vdrop = (Iload / 0.000003) * 1020
So the big question is - what is my load? If I'll know that I can calculate the resistor value according to ohm's law.
I've searched online for a number of times and found the general way these devices are built, but each one is a little different so there are only general guidelines.
Normally, at no load I can say Vdrop = 0 so Eout is 6220V idealy. I can use that figure to calculate the resistor and the fact it will drop when being touched will make sure I "overprotect" the circuit.
Well I won't comment on the usefulness or not of negative ion generators, but I was under the impression that the voltage needs to be within certain limits? - or it sprays metal ions in the air for you to breath in.