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Negative air ion generator help

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slow_rider

New Member
Hello all,

I'm building a negative ion generator for a friend. I'm using a full wave Cockroft-Walton voltage multiplier (http://upload.wikimedia.org/wikipedia/en/thumb/4/47/Full_wave_Cockcroft_Walton_Voltage_multiplier.png/180px-Full_wave_Cockcroft_Walton_Voltage_multiplier.png 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.

Any help will be really appreciated!
Thanks
 

Nigel Goodwin

Super Moderator
Most Helpful Member
Hello all,

I'm building a negative ion generator for a friend. I'm using a full wave Cockroft-Walton voltage multiplier (http://upload.wikimedia.org/wikipedia/en/thumb/4/47/Full_wave_Cockcroft_Walton_Voltage_multiplier.png/180px-Full_wave_Cockcroft_Walton_Voltage_multiplier.png 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.
You have no load.

I suggest you try googling for circuit digrams and see how it's usually done.
 

slow_rider

New Member
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.

Thanks for the help!
 
Last edited:

Nigel Goodwin

Super Moderator
Most Helpful Member
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.
 
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