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High voltage supply 5kV DC

KlausST

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Hi,.

Using isolation transformer should make system safe right?
No. An isolation transformer just increases safety against earth/ground potential.

High voltage still remains. If you touch it it may kill you.

Mind: a charged capacitor - even if connected to nothing else - may be dangerous.

Klaus
 
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betwixt

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I built a 'negative ion generator' for a friend a few years ago using a similar circuit. They were advised that negative ions made you healthier and I couldn't dispel the myth.

I never found out what the final voltage was but I think it used around 20 diodes and capacitors and the needle at the output had a blue corona around it and emitted an ozone smell. I think I used a 1M resistor in the input to limit the current and I think the capacitors were 100nF (note nF not uF!). Using large capacitors does not increase the voltage but it does increase the kick if you touch it!

Brian.
 
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I built a 'negative ion generator' for a friend a few years ago using a similar circuit. They were advised that negative ions made you healthier and I couldn't dispel the myth.

I never found out what the final voltage was but I think it used around 20 diodes and capacitors and the needle at the output had a blue corona around it and emitted an ozone smell. I think I used a 1M resistor in the input to limit the current and I think the capacitors were 100nF (note nF not uF!). Using large capacitors does not increase the voltage but it does increase the kick if you touch it!

Brian.
Thanks betwixt for the reply. I developed circuit (full wave cockroft walton) for (voltage multiplication upto 8 times) consisting of 47uF and diodes 1n4007. Initially I used multi-meter to measure the output and checked upto around -950 V DC (AC input 85V rms). It was working fine. However, since multi-meter was rated at 1000V only, I used non-isolated High voltage probe from pintek for further measurement upto2 kV. In the probe datasheet. It was mentioned to ground one end of the probe. I grounded the midpoint of Cocroft-Walton and also ground point of electrode. However at -400 V output one of the diodes got damaged (have to check others) infact physically broke. What could be issue. How do we measure the HV voltage. Is resistor divider arrangement of 20Mohm and 20Mohm reasonable. Thanks
 
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1N4007 should be more than adequate, they are rated at 1KV PIV and each stage of the multiplier handles a fraction of the total output. I suspect the damaged diode was due to other reasons.

Measuring high voltage isn't easy, the problems are that single resistors are generally only rated for low voltage, maybe up to 250V although special higher voltage ones are available. You will almost certainly have to chain several resistors to achieve the overall 'end to end' voltage requirement. The other problem is that as you use higher value resistors, the influence the voltmeter has on the reading becomes greater. You can generally factor the meter into the voltage drop if you know it's input resistance, it will appear to be in parallel with the lower resistor in the divider. Using lower value resistors lessens the effect but at the same time causes more voltage drop due to the higher current drawn from the output.

Brian.
 

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1N4007 should be more than adequate, they are rated at 1KV PIV and each stage of the multiplier handles a fraction of the total output. I suspect the damaged diode was due to other reasons.

Measuring high voltage isn't easy, the problems are that single resistors are generally only rated for low voltage, maybe up to 250V although special higher voltage ones are available. You will almost certainly have to chain several resistors to achieve the overall 'end to end' voltage requirement. The other problem is that as you use higher value resistors, the influence the voltmeter has on the reading becomes greater. You can generally factor the meter into the voltage drop if you know it's input resistance, it will appear to be in parallel with the lower resistor in the divider. Using lower value resistors lessens the effect but at the same time causes more voltage drop due to the higher current drawn from the output.

Brian.
I have seen that in some circuits, people have used 1Mohm resistor in parallel with the input 230 supply and also 2M ohm in series with output point. Should we use resistance in series or parallel with the input 230V ac supply.
 

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I once built a HV probe, for one of those 200mV Ebay digital panel meters, which exhibit a very high input impedance (because they don't have any attenuators). Therefore the shunt impedance of the divider is unaffected.

The probe itself consisted of 10 HV series resistors 20 Meg each. The shunt was 2k, providing a 10,000:1 attenuation. I used high voltage rated resistors like this: https://www.vishay.com/docs/28907/vr25vr37vr68.pdf

The resistor string was enclosed inside a 10 cm long, 6mm diameter PVC pipe which I filled with oil and sealed.

Near the HV end, I put a PVC ring glued to the outside of the pipe, to prevent the fingers slipping too close to the HV.
For safety purposes, the probe had two grounding clips, one of which YOU SHOULD TIE FIRST to a solid earth ground.

Because such a simple probe does not have frequency compensation, it will only provide valid results for DC or 50/60 Hz readings.
 
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Should we use resistance in series or parallel with the input 230V ac supply.
A resistor in parallel is a good safety measure, use a value that will not waste significant power, usually 1M is fine. All it does is ensure no voltage is present across the power input when it is unplugged, it 'leaks' away any current flowing backwards from the multiplier so the risk to anyone touching the power plug pins is reduced.

A resistor in series with the input limits the current that can flow into the multiplier. It offers little advantage because all it does is delay how long it takes for the chain of capacitors to charge up, once they reach full charge they are almost as dangerous as if the resistor wasn't there. A resistor in series with the output is better from a safety aspect because it limits how much can be drawn from the high voltage but please bear in mind that if you have say 5KV and you use normal 0.25W resistors you may need as many as 20 in series to be sure each doesn't exceed it voltage rating. Note that this has nothing to do with power dissipation in Watts, it is because resistors have an inherent risk of internal arcing if too many volts are present across their ends. Even with the minimum number of resistors in series so their voltage rating matches your output, there is a slight risk of one arcing followed by a chain reaction of all the others following suit.

Brian.
 

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Dear All,

Thanks for the support. I am getting more questions in mind as I go further into this development. I have developed a voltage multiplier based on full wave cockroft-walton circuit. I am using general purpose PCB board and in each board I have fitted 8 sections designed to get about 2kV. I have made 2 more such boards which will be connected in series. I have made on general purpose boards as I could not procure PCB due to lockdown.


These are my queries.

1. As mentioned earlier, when using multimeter (fluke 285) I was able to test upto -950V with ac input of around 85V. In this case point A was not grounded. Multimeter measured between B and A and no load. Should point A be mandatorily grounded when using for negative air ionizer purpose?

<blockquote class="imgur-embed-pub" lang="en" data-id="a/17IV5Ll" data-context="false" ><a href="//imgur.com/a/17IV5Ll"></a></blockquote><script async src="//s.imgur.com/min/embed.js" charset="utf-8"></script>
2. When I tried to run same circuit and measure using HV probe Pintek HVP 28HF (non-isolated HV probe). The point A was grounded. Probe measured between B and A(Ground). However, the input section 1 got damaged at DC voltage of around -480V, one of the diodes broke physically. When I bypassed that section and ran the circuit and measured with multimeter (Point A not grounded), it was running fine. What could be the reason for the damage.
<blockquote class="imgur-embed-pub" lang="en" data-id="lzc4Wl4"><a href="https://imgur.com/lzc4Wl4">View post on imgur.com</a></blockquote><script async src="//s.imgur.com/min/embed.js" charset="utf-8"></script>
3. What other method could be used for voltage measurement beyound 1kV using multimeter. When I checked multimeter datasheet, it showed input impedance of 10Mohm <100pf(ac coupled). What does this mean. Can I use a resistor divider of around 30 nos of 20 Mohm resistor (0.25 W general purpose) connected in series (360M and multimeter measured across the last resistor.

4. Should I put an input resistor of 1M ohm in series (R1) with AC supply or in parallel (R2) with AC supply . I have seen some circuits using 1M resistor parallel with input. Similarly, Should any resistor should be put in series with the last output point B after which electrodes are connected. Will a general fuse of 0.1 A be able to protect circuit.
<blockquote class="imgur-embed-pub" lang="en" data-id="9wwI5JZ"><a href="https://imgur.com/9wwI5JZ">View post on imgur.com</a></blockquote><script async src="//s.imgur.com/min/embed.js" charset="utf-8"></script>
5. Any other safety precaution when interconnecting boards to prevent arcing when we test near 4kV to 5KV.

Thanks in advance

- - - Updated - - -

A resistor in parallel is a good safety measure, use a value that will not waste significant power, usually 1M is fine. All it does is ensure no voltage is present across the power input when it is unplugged, it 'leaks' away any current flowing backwards from the multiplier so the risk to anyone touching the power plug pins is reduced.

A resistor in series with the input limits the current that can flow into the multiplier. It offers little advantage because all it does is delay how long it takes for the chain of capacitors to charge up, once they reach full charge they are almost as dangerous as if the resistor wasn't there. A resistor in series with the output is better from a safety aspect because it limits how much can be drawn from the high voltage but please bear in mind that if you have say 5KV and you use normal 0.25W resistors you may need as many as 20 in series to be sure each doesn't exceed it voltage rating. Note that this has nothing to do with power dissipation in Watts, it is because resistors have an inherent risk of internal arcing if too many volts are present across their ends. Even with the minimum number of resistors in series so their voltage rating matches your output, there is a slight risk of one arcing followed by a chain reaction of all the others following suit.

Brian.
Thanks Brian. I asked this question again before I could see your reply. So what is your suggestion if I put 1M ohm in series and parallel at input and 220k (30Nos) in series at output. Will the output series resistors reduce air ionizing capability? Thanks
 

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I built a 'negative ion generator' for a friend a few years ago using a similar circuit. They were advised that negative ions made you healthier and I couldn't dispel the myth.

I never found out what the final voltage was but I think it used around 20 diodes and capacitors and the needle at the output had a blue corona around it and emitted an ozone smell. I think I used a 1M resistor in the input to limit the current and I think the capacitors were 100nF (note nF not uF!). Using large capacitors does not increase the voltage but it does increase the kick if you touch it!

Brian.
Hi Brian,

I tried calculating the number of resistors and value to measure, lets say 6kV voltage using resistor divider method and dissipating 0.1W. It shows that I need around 360Mohm. Considering voltage limit of 200V on each resistor it comes to requiring 30 resistors each of 12Mohm. and at full voltage each resistor will have 200V drop. However, now the multimeter probe is connected acroos 60M resistor and input impedance is 10M. This is affecting effective measurment. I think multimeter method is not suitable.

You mentioned you never measured final voltage. Any other method of measurement and do we really need to measure output voltage for safe operation. Is it mandatory to ground the point A(fig from my previous post).

Thanks


- - - Updated - - -

Can we not connect the inputs to 230V ac outlet (2 pin) and earth the point A?
 

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firstly - note Easy Peasy's comment in post #29. If you consider that in most wiring situations, the neutral AC line is bonded to the Earth line you will see the top diode on the left would be wired directly across the incoming power. This could well be the reason it was damaged. You really should use an isolating transformer, not only for safety to to allow that kind of multiplier to work properly.

I tried calculating the number of resistors and value to measure, lets say 6kV voltage using resistor divider method and dissipating 0.1W.
I'm not sure why you want to do that. The resistors are there to limit the output current to a safe level, not to load the output. They go in series with the high voltage output, not from the high voltage to ground. They need the high voltage rating just in case they do become grounded. It is up to you to decide what you consider a safe maximum output current, I would suggest no more than a few mA, especially if you are using large value capacitors which could sustain the output current for quite a long time. As I stated before, the value of the capacitors does not (within reason) influence the output voltage but smaller values will discharge quicker if you do draw significant current.

If you are using this to ionize air, the current needed is only a few uA.

Brian.
 
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By coincidence a recent Youtube video (posted by Big Clive) features a high-voltage ionizer circuit.
He does onscreen comparison to a Cockcroft-Walton voltage multiplier.

He recommends installing high-ohm resistors in the final output, as well as a carbon fiber filament as a 'spray attachment'.
He mentions the need for some kind of earth connection at the middle positive DC terminal.

LED lamp with tiny 5kV high voltage ion module. (inc schematic)


Big Clive's website resurrected a magazine article about the semi-famous 'Joule thief' boost converter, thereby gaining himself a kind of distinction.

<iframe width="560" height="315" src="https://www.youtube.com/embed/tl0g_z25xqo" frameborder="0" allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen></iframe>
 
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By coincidence a recent Youtube video (posted by Big Clive) features a high-voltage ionizer circuit.
He does onscreen comparison to a Cockcroft-Walton voltage multiplier.

He recommends installing high-ohm resistors in the final output, as well as a carbon fiber filament as a 'spray attachment'.
He mentions the need for some kind of earth connection at the middle positive DC terminal.

LED lamp with tiny 5kV high voltage ion module. (inc schematic)


Big Clive's website resurrected a magazine article about the semi-famous 'Joule thief' boost converter, thereby gaining himself a kind of distinction.

<iframe width="560" height="315" src="https://www.youtube.com/embed/tl0g_z25xqo" frameborder="0" allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen></iframe>


Thanks Brian and BradtheRad for the reply.

I am able to understand.

However, to make this system work for air ionizing, can I link the midpoint to a aluminium metal plate (25cm*25cm*.3cm) kept on an isolating rubber sheet for ground purpose.

If I am able to salvage a isolation transformer (1:1)(not centre-tapped), can I ground the midpoint for the system to work.

Based on all the previous posts, and google research, I know the sytem is overkill. But its just for test purpose. I am trying to procure 0.01uF ,1000V cap once lockdown ends.


Yes Brian, the resistor divider method creates loading. I tried to create a temporary probe system using method mentioned in

https://www.youtube.com/watch?v=RAOw0mHQRvk

Using 30M and using 10M internal resistance of multimeter (able to get 4.3:1) measurement and tested upto 2kV. However, to measure upto 6kV, I had to increase the resistance to 400Mohm using a series of 10Mohm resistors. However, this time it didnt work . It showed approximately zero voltage. Could the current be less than what the multimeter uses for voltage measurement. Any other method to solve this issue?
 

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As a last resort, measure across a few capacitors at a time, then sum the amounts.

I had the same problem with a CW multiplier which I built to generate 1kV. My analog meter is 20,000 ohms per volt. Therefore the 1200 VDC range is a load of 24 M-ohms. It pulled down the output voltage severely. It's because I tried to get by with tiny cap values, 500 pF.

- - - Updated - - -

My simulation shows the last few stages go into opposite polarity if you draw too much current.

As you know this should be avoided with electrolytics. In fact after power-up some capacitors are in opposite polarity for a second. Therefore it may be wise to start with no load attached.
 

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Like many Internet video producers, the guy who made that Youtube should be taken out and shot. Please don't trust anything without doing the math and following safe procedures. Using the flame from a cigarette lighter to fit insulation is not only downright dangerous but it vastly overheats the plastic and can actually make it conductive.

If you use an isolating transformer, yes you can ground the center point.

Brian.
 

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However, to make this system work for air ionizing, can I link the midpoint to a aluminium metal plate (25cm*25cm*.3cm) kept on an isolating rubber sheet for ground purpose.
If the ionizing current is around a few uA in value, then you need not worry about the return path. The leakage across various devices shall take care of that.

However, if you want ionizing current of the order of mA, then you must provide a return path. Consider the following:

As you produce some ionization, the electrodes lose negative charge and this is given to the molecules in the air. After sometime, there will be lots of ions in the air, the electrode has lost lots of electrons and the circuit assembly has become net positive charge.

Here your Al foil comes into play; it collects lots of the negative ions and returns them to the circuit for electrical path completion.

How does ion rockets work?
 

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Easy peasy

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You need a bit of pcb - or metal ( stainless steel ) surrounding the sharp pin points that are delivering the neg electricity - about 2 inches away ( 50mm ) and connected to the pos end of the HV generator - this will give the highest current flow thru the air, neg ions will be made and move from neg to pos ...

the sharp pin points can extend out by an inch or so also to help get the ions into the general air space - you may/will get some ozone too - if you smell it too strongly time to turn down the volts a bit ...
 
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surrounding the sharp pin points that are delivering the neg electricity - about 2 inches away ( 50mm ) and connected to the pos end of the HV generator
Putting the two electrodes so close somehow defeats the purpose: the negative ions produced will just run towards the positive (or ground) electrode. We want the negative ions to mix with the room air, do whatever they were supposed to do and finally diffuse towards the positive electrode.

I would suggest that they should be kept as far away as practical. A conducting sphere will be ideal for this.
 

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convection or any slight breeze will carry the ions away which is why post #37 works ...
 
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Putting the two electrodes so close somehow defeats the purpose: the negative ions produced will just run towards the positive (or ground) electrode. We want the negative ions to mix with the room air, do whatever they were supposed to do and finally diffuse towards the positive electrode.

I would suggest that they should be kept as far away as practical. A conducting sphere will be ideal for this.
Thanks c_mitra for the reply
What about using the aluminium plate (mentioned earlier) connected 1 or 2 feet away from -Ve electrode pins. As you metioned, for few uA current, I neednt do any thing. So can increasing the number of pins (of -ve electrode) and size of plate affect this current. Also how many pins should be used (any reference will be helpful)

- - - Updated - - -

You need a bit of pcb - or metal ( stainless steel ) surrounding the sharp pin points that are delivering the neg electricity - about 2 inches away ( 50mm ) and connected to the pos end of the HV generator - this will give the highest current flow thru the air, neg ions will be made and move from neg to pos ...

the sharp pin points can extend out by an inch or so also to help get the ions into the general air space - you may/will get some ozone too - if you smell it too strongly time to turn down the volts a bit ...
Thanks easy_peasy for the reply.

Is there any size limit for the +Ve (or ground) electrode (like using an aluminium plate mentioned earlier). Will putting the ground electrode 1 or 2 feet away affect the functioning of the system. In final prototype, I may need to use what you suggested (to keep both positive and negative in a box enclosure and use small fan to blow the ions).

At what voltage should i expect ionization, I have seen some articles mentioning that around -4kV DC it starts.

- - - Updated - - -

As a last resort, measure across a few capacitors at a time, then sum the amounts.

I had the same problem with a CW multiplier which I built to generate 1kV. My analog meter is 20,000 ohms per volt. Therefore the 1200 VDC range is a load of 24 M-ohms. It pulled down the output voltage severely. It's because I tried to get by with tiny cap values, 500 pF.

- - - Updated - - -

My simulation shows the last few stages go into opposite polarity if you draw too much current.

As you know this should be avoided with electrolytics. In fact after power-up some capacitors are in opposite polarity for a second. Therefore it may be wise to start with no load attached.

How do we control the loading. Does the number of pins at the -ve electrode affect the loading or series resistor of about 1Meg or 4Meg before the -ve electrode pins.

- - - Updated - - -

Like many Internet video producers, the guy who made that Youtube should be taken out and shot. Please don't trust anything without doing the math and following safe procedures. Using the flame from a cigarette lighter to fit insulation is not only downright dangerous but it vastly overheats the plastic and can actually make it conductive.

If you use an isolating transformer, yes you can ground the center point.

Brian.
Yes Brian. I didnt try with heating method, but used layers of insulating tape on the series resistors. I tried using 390Meg (consisting of series resistor of 39 Nos of 10 meg each) to be able to measure upto 6kV, but i didnt work. Have to check the issue. I am wondering if there is any current limit to which the multimeter can respond for voltage measurement.

Also I am wondering whether I should measure at the final point voltage (say 5kV or 6kV). I am thinking that I measure say upto 3.5kV to see if the system is functioning and then finally increase the to rated input voltage without any measurement. Is there any other indication that I reached desired voltage.
Thanks
 
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