# How to make 15kDCV low current power supply?

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#### newfreewill

Joined Mar 8, 2023
9
Looking for some help, please. I don't know much about electronics - studying hard at the moment.

I need to make as cheaply and simply as possible a 15kV (or thereabouts) DC power supply with low output current - ideally around 20-25 microA.
(The application is boring; if you know the answer to the power supply, you'll probably guess its application. Please let's not have a thread about the application.)

A 3.7V 18650 battery should have enough juice, but I don't care too much about the input.

This thing is frustratingly close:
https://www.ebay.com/itm/382131734381

It outputs 15kV AC
The specs say output current is 0.4A, but of course that's not right, it must be 0.4mA.

I guess I can reduce the output current by replacing the resistor?

Is there a cheap and simple way to make a rectifier for the output?
Or a similar transformer with lower voltage output that I can add a voltage multiplier to?
Or any other kit or cheap prebuilt unit that gives around 15kDCV at around 20-25 microA or could be modified to be close?

Thanks, guys

#### LowQCab

Joined Nov 6, 2012
3,062

Ok,
then in that context, at least I won't.
You've stated that You don't know what you're doing,
but You want us to tell You how to build something that may kill You, or someone else.
Brilliant ........
.
.
.

#### WBahn

Joined Mar 31, 2012
28,469
In general, if you have a voltage supply that can deliver AT LEAST the current that you need, then the load will limit the current to what it actually needs.

Your car battery is capable of delivering hundreds, even over a thousand, amps. But when you open your car door and the interior light comes on, it's only drawing a fraction of an amp.

Since we don't know what the application is, we can't guess whether it will limit the current or not.

Rectifying the output is pretty straight forward, but you have to be sure to use components rated for the voltages involved.

I would agree that the spec is most likely supposed to be 0.4 mA, based on power-in/power-out considerations.

#### newfreewill

Joined Mar 8, 2023
9

Ok,
then in that context, at least I won't.
You've stated that You don't know what you're doing,
but You want us to tell You how to build something that may kill You, or someone else.
Brilliant ........
.
.
.
I know about the application; I don't know as much as I need to about the circuit, and I'm not afraid to say so.
Kill people with a 3.7V battery drawing 20-25microA. Right...

#### Ya’akov

Joined Jan 27, 2019
7,418
I know about the application; I don't know as much as I need to about the circuit, and I'm not afraid to say so.
Kill people with a 3.7V battery drawing 20-25microA. Right...
Welcome to AAC. Thanks for joining us.

As you said, you don’t know much about electronics and you are asking about potentially lethal voltages. Recently, several people have been killed ignorantly using high voltages for craft projects so we are on special alert for such things. We won’t help you hurt yourself or others.

The problem is compounded by your self-professed and demonstrated ignorance. By attempting to isolate parts of the problem you are trying to solve you are not providing the information we need to actually help you solve it.

For example, ff you are going to get 15kV@25㎂ out of a 3.7V source you are going to need to get about 10A out of it. You can’t get anything for free, and you get even get anything for cost—there are always losses—but as a first order estimate W=VA will get you idea of that is going on. W is for Watts, which is a measurement of power. You can see that power is a product of the voltage, V (in Volts) and current A (in Amps).

You can trade volts for amps and vice versa, but you can’t make more power than you start with. Since you want to get about 4000 times more voltage than you are putting in, you will have to provide about 4000 times more current than you want out. So, the battery is going to be getting very warm as it spews a lot of current to make up for the low voltage.

Imagine the damage you could do with 3.7V@10A using it to create heat, or turning it into mechanical energy with a motor or solenoid. That’s the power you are playing with, and the “25㎂” isn’t a mitigation.

Since we have to guess at the application, I am inclined to guess this is for an application that is intended to harm another person, since that is the safest bet when trying to decide if your post violates AAC rules. That being the case, unless you decide to make your next post an explanation of the application, this thread will be closed.

#### newfreewill

Joined Mar 8, 2023
9
In general, if you have a voltage supply that can deliver AT LEAST the current that you need, then the load will limit the current to what it actually needs.

Your car battery is capable of delivering hundreds, even over a thousand, amps. But when you open your car door and the interior light comes on, it's only drawing a fraction of an amp.
Right.

Since we don't know what the application is, we can't guess whether it will limit the current or not.
https://reliantfinishingsystems.com/powder-coating-gun-settings/

Rectifying the output is pretty straight forward, but you have to be sure to use components rated for the voltages involved.

I would agree that the spec is most likely supposed to be 0.4 mA, based on power-in/power-out considerations.
MOD EDIT: Reworked QUOTE tags to separate the poster's responses from the quoted material.

Last edited by a moderator:

Joined Jan 15, 2015
7,192
I have used old Neon Sign Transformers which outputted 9 KV and 15 KV at 30 mA of current. All that is needed are some HV Diodes to get you to DC. Not knowing the project I have no idea if you want HV filter caps in the mix. As you know voltages like this with currents like this can make you very dead very fast.

You may also want to look into HV Multipliers as used on older CRT TV sets.

If you go with an inverter type HV Ignition Coil Generator I suggest a Google of HV Diodes.

Also consider this:

If you are going to get 15kV@25㎂ out of a 3.7V source you are going to need to get about 10A out of it. You can’t get anything for free, and you get even get anything for cost—there are always losses.
There is no free ride.

Ron

#### newfreewill

Joined Mar 8, 2023
9
Thanks, Ron,

No, that's too much for me, thanks. As I said, about 1/1000 of that current.

No idea where I might find a CRT TV

I know I'm going to need HV diodes.

Didn't understand or care for Ya'akov's comments. That's the unfortunate thing about forums - unhelpful, condescending, rude types. He even threatened me at one stage. Don't know what a free ride has to do with anything. The eBay circuit outputs 15kV@0.4mA and draws less than 2A from the battery.

#### WBahn

Joined Mar 31, 2012
28,469
Looks like you are trying to make a supply for doing powder paint coating?

If so, why are you trying to make one instead of purchasing a commercially available one?

If the answer is because you want to save money, be aware that there is a real likelihood that by the time you discover and work through all of the real-world gotchas that are almost inevitably involved, you will probably have been money ahead to just bite the bullet and get a ready-made unit (perhaps used?).

It's not going to be as simple as getting a cheap HV supply that is fixed-voltage. You probably need to be able to vary the voltage in order to control the quality of the result, right?

How long do you expect to use the system on that single 18650 battery?

These batteries typically have capacity ratings of about 3000 mAh (2200 mAh to 3500 mAh) and are suitable for continuous currents of between 10 A and 30 A, depending on the specifics of the cell chosen.

To get 25 µA at 15 kV, you would need to draw about 100 mA at 3.7 V, so that's well within the capabilities of the battery. That's with 100% conversion efficiency. So figure perhaps 80% efficiency and you are about about 135 mA, meaning that a 3000 mAh cell might be expected to last about 24 hours.

So that all seems reasonable.

#### Papabravo

Joined Feb 24, 2006
20,122
Thanks, Ron,

No, that's too much for me, thanks. As I said, about 1/1000 of that current.

No idea where I might find a CRT TV

I know I'm going to need HV diodes.

Didn't understand or care for Ya'akov's comments. That's the unfortunate thing about forums - unhelpful, condescending, rude types. He even threatened me at one stage. Don't know what a free ride has to do with anything. The eBay circuit outputs 15kV@0.4mA and draws less than 2A from the battery.
He's a moderator and they don't care for people who ignore their requests like you just did.

#### WBahn

Joined Mar 31, 2012
28,469
Thanks, Ron,

No, that's too much for me, thanks. As I said, about 1/1000 of that current.

No idea where I might find a CRT TV

I know I'm going to need HV diodes.

Didn't understand or care for Ya'akov's comments. That's the unfortunate thing about forums - unhelpful, condescending, rude types. He even threatened me at one stage. Don't know what a free ride has to do with anything. The eBay circuit outputs 15kV@0.4mA and draws less than 2A from the battery.
We get people from time to time trying to make high voltage devices for the purposes of shocking people, either as a prank or as a weapon. That, or anything like it, violates the User Agreement. If the mods have reason to suspect that this might be the case (and refusing to disclose the intended use is a common red flag), then we need to be satisfied that this is not the case or we need to close the thread.

As for the free ride, that stems from your earlier comment about only drawing 20 to 25 µA from a 3.7 V battery. A lot of people unfamiliar with electronics (which you self-identified as) do not realize that you can't draw a certain current from a low-voltage source and somehow increase the voltage while delivering that same current. Even if they understand the concept of the conservation of energy, they may not know enough to know how that relates to voltages and currents.

#### WBahn

Joined Mar 31, 2012
28,469
He's a moderator and they don't care for people who ignore their requests like you just did.
In fairness, his next post linked to what is, presumably, the application. This is easy to get lost in the shuffle because of poor use of quote tags.

I'll go in and patch that up.

#### Danko

Joined Nov 22, 2017
1,602
It outputs 15kV AC
The specs say output current is 0.4A, but of course that's not right, it must be 0.4mA.
I guess I can reduce the output current by replacing the resistor?
Is there a cheap and simple way to make a rectifier for the output?
DC 12V to 20000V High-Voltage Electrostatic Generator 20 kV Step-up Power Module
US $9.13 https://www.ebay.com/itm/1428895478...V5FKO-N7zt7yYcbvveUja5TPcNoTTjeoaAhItEALw_wcB #### WBahn Joined Mar 31, 2012 28,469 I know about the application; I don't know as much as I need to about the circuit, and I'm not afraid to say so. Kill people with a 3.7V battery drawing 20-25microA. Right... These batteries can deliver tens of amperes. If the circuit pumps that up from 3.7 V to 15 kV, then it could deliver in excess of 10 mA if it is being limited by the battery. This is well into the vicinity of dangerous shock territory and not far from the range of lethality. Depending on the details of how the circuit behaves when connected to a relatively low-impedance path (like a person), it might well put out considerably more current than this. So the shock hazards associated with what you are trying to do can't be ignored. If you aren't willing to consider them and do what is necessary to ensure that, under no conditions, can it deliver harmful currents to a human, then you really should walk away from the project. The good news is that taking the steps needed to ensure this probably aren't too onerous, it's just that they can't be ignored, whether it be due to carelessness, ignorance, or neglect. #### LowQCab Joined Nov 6, 2012 3,062 Some projects are just not good DIY projects. This is definitely one of them. Go to Eastwood Products and buy a very nice, reliable, Powder-Coating-Setup. It will work for years in a commercial-production-environment, and it has a warranty, and it won't KILL YOU. . . . Thread Starter #### newfreewill Joined Mar 8, 2023 9 Welcome to AAC. Thanks for joining us. As you said, you don’t know much about electronics and you are asking about potentially lethal voltages. Recently, several people have been killed ignorantly using high voltages for craft projects so we are on special alert for such things. We won’t help you hurt yourself or others. The problem is compounded by your self-professed and demonstrated ignorance. By attempting to isolate parts of the problem you are trying to solve you are not providing the information we need to actually help you solve it. For example, ff you are going to get 15kV@25㎂ out of a 3.7V source you are going to need to get about 10A out of it. You can’t get anything for free, and you get even get anything for cost—there are always losses—but as a first order estimate W=VA will get you idea of that is going on. W is for Watts, which is a measurement of power. You can see that power is a product of the voltage, V (in Volts) and current A (in Amps). You can trade volts for amps and vice versa, but you can’t make more power than you start with. Since you want to get about 4000 times more voltage than you are putting in, you will have to provide about 4000 times more current than you want out. So, the battery is going to be getting very warm as it spews a lot of current to make up for the low voltage. Imagine the damage you could do with 3.7V@10A using it to create heat, or turning it into mechanical energy with a motor or solenoid. That’s the power you are playing with, and the “25㎂” isn’t a mitigation. Since we have to guess at the application, I am inclined to guess this is for an application that is intended to harm another person, since that is the safest bet when trying to decide if your post violates AAC rules. That being the case, unless you decide to make your next post an explanation of the application, this thread will be closed. Thread Starter #### newfreewill Joined Mar 8, 2023 9 If you read my original post - all of it, not just bits to pounce on, you will see that I suggested a circuit that isn't lethal and asked how to limit its output to 1/20th of the current and change it to DC. Obviously, I'm not trying to hurt anyone. I posted the application. Does that help? Will you now try to help me and answer my questions? Or you just want to keep up the constant barrage of personal comments - 'You', 'you', 'you', sixteen in this post - and how ignorant I am. Its not about me, but a circuit. #### Danko Joined Nov 22, 2017 1,602 This is the same thing I showed, but higher voltage. This one? US$3.23?
It is high frequency HV module.

Click on "-Select-" and choose "20000V"
This one with price US \$9.13 is DC 20kV module.

Description:
Input voltage: DC 12V
Rated power: 5W
Mounting hole diameter: 5mm
Hole center distance: 112mm
Size: 60*70*30mm
Note: Small red and black line is input, red is 12V power supply positive input, black is negative input,
thick red line is high voltage positive output, thick black line is high voltage negative output.

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#### newfreewill

Joined Mar 8, 2023
9
Looks like you are trying to make a supply for doing powder paint coating?

If so, why are you trying to make one instead of purchasing a commercially available one?

If the answer is because you want to save money, be aware that there is a real likelihood that by the time you discover and work through all of the real-world gotchas that are almost inevitably involved, you will probably have been money ahead to just bite the bullet and get a ready-made unit (perhaps used?).

It's not going to be as simple as getting a cheap HV supply that is fixed-voltage. You probably need to be able to vary the voltage in order to control the quality of the result, right?

How long do you expect to use the system on that single 18650 battery?

These batteries typically have capacity ratings of about 3000 mAh (2200 mAh to 3500 mAh) and are suitable for continuous currents of between 10 A and 30 A, depending on the specifics of the cell chosen.

To get 25 µA at 15 kV, you would need to draw about 100 mA at 3.7 V, so that's well within the capabilities of the battery. That's with 100% conversion efficiency. So figure perhaps 80% efficiency and you are about about 135 mA, meaning that a 3000 mAh cell might be expected to last about 24 hours.

So that all seems reasonable.

I said I didn't want to make this a thread about the application because I really don't want to explain about the powder coating - it wouldn't help either of us. (Unless there's any electrical aspect that you need for the power supply, or you're interested in powder coating.) The setup I need isn't commercially available, I just need the power supply. The whole system is tens of thousands of dollars, and I'm sure isn't sold 2nd hand.

No, I don't need to vary the voltage. I won't need to vary the current either, but it would be handy to experiment with in the first case. Most powder coating guns are not adjustable - you might get to choose one of two voltages. The article I linked explains the settings for small bits of metal with hard-to-reach surfaces, which is what I'm working with.

The kit I linked in the OP was just meant to be an example. I think I understand how it works, so maybe it's a good starting point. I don't really care what the input is. Mains would be more convenient, but I also like the idea of working with as little power as possible. I'm glad the battery is doable, thanks. I won't spend much time powder coating - less than an hour/week, and I can have a charged battery at hand.

Now that you know the application - how about my starting suggestions for modifying the kit? Could I limit the output current by increasing the resistance into the transformer (where there's a resistor already?) If so, maybe I could put a variable resistor there? What would be a good starting value?

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