measuring max current of a 20kv supply

[hazuki]

Hi all. I picked up a high voltage supply last week (http://www.amazing1.com/products/negative-ion-generator-12vdc-input-20kv-output.html). It takes 12V in and puts out 20kV. The website claims that the output current is 50 uamps. I really don't trust that measurement, so I'd like to do a measurement myself of the max current. I have an oscilliscope and a multimeter, neither of which can do that by themselves. Is there a cheap way I can make a measurement? My first instinct is to use a shunt resistor and voltage dividers in order to get a signal to the scope. However, doing that loads the supply so I won't get the max voltage.

Any ideas?

Thanks!

[Psi]

20x 20Meg resistors = 400M = 50uA flowing @ 20kv

So 20x  20Meg resistors in series then scope the FIRST resistor to ground using x10 probe.
(Make sure you know what ground is first!)

Then
I = Volts across resistor / 6,666,666 ohms
(6.6Meg is the 20M resistor in parallel with the 10M from the scope input.)

Say you get ~333V on the scope for the voltage across the resistor
333V / 6.6M = 50uA

However, maybe wait for someone else to confirm before trying that.
I'm always a bit worried when combining test equipment with high voltage. There's potential for damage.

[c4757p]

Make sure those 20M resistors can hold off 1kV each! Aren't standard 1/4W axials typically rated for something like 250V? Even if they don't break over, the resistance could drift significantly under a high voltage stress.

[Psi]

Personally i reckon a regular 0.5W TH resistor would hold off 1kV for long enough for the test.
but yeah, good point

You could use 80x 5M resistors and redo the calculation but that maybe getting a bit silly

[IanB]

Do you have a meter that can measure microamps? Many DMMs or analog meters can do that. Just place the current range of such a meter across the output and measure the current. If your multimeter does not have a microamp range, it is not hard to obtain a meter that does. Even a simple 100 uA panel meter will do. Or Dave's uCurrent 

[c4757p]

That's just the short circuit current, though. I think he wants to measure the maximum current it'll give near full output voltage, a tricky thing to test at 20kV.

Out of curiosity, why?

[Psi]

I wouldn't be putting 20kv into my meter on uA.

Even if you did, its going to load the supply so the reading isn't going to be the same as for 20kV

[IanB]

You're trying to measure maximum current. It's the same as a current limited power supply. You measure the maximum current by shorting the output and seeing how much current flows.

Voltage is not a concern. My ordinary DMM on the µA range has a resistance of 100 ohms. So at 50 µA the maximum voltage will be 5 mV. Nothing to worry about.

[IanB]

That's just the short circuit current, though. I think he wants to measure the maximum current it'll give near full output voltage, a tricky thing to test at 20kV.

Out of curiosity, why?

It's for safety. Ion generators used to be all the rage, they theoretically clean the air of microscopic particles and reduce the effect of allergies. Unfortunately at the cost of depositing all those particles near the ion generator and making horrible dirt stains 

However, they must not kill you when you touch them, so they are limited to safe, low currents.

[Psi]

It depends if he wants to know the short circuit current or the current at 20kV

Loading the supply with say 10uA, may drop the output to 19kV for example.
So in that case the supply cannot provide 50uA at 20kv

[c4757p]

Oh, I assumed the measurement was to test the output current to verify that it was suitable for some application. Yeah, if it's just a safety check, just short it out with the multimeter set to measure uA. I'd connect the meter first and turn the meter on, then turn the PSU on, just for good luck 

[IanB]

It depends if he wants to know the short circuit current or the current at 20kV

Loading the supply with say 10uA, may drop the output to 19kV for example.
So in that case the supply cannot provide 50uA at 20kv

Sure, loading the output will drop the voltage. But it doesn't really matter. High voltage power supplies are configured as constant current sources, so it hardly matters how much you load the output. The output current will be almost the same. It's only the voltage that varies.

[hazuki]

Yes, it's for safety. I like the idea of using the microamps range on a DMM, first connecting to the supply and then turning it on. But can a DMM withstand 20kV on the microamps terminal? I know where i can borrow a fluke 87-V, I'd just rather not destroy it.

This voltage generator uses a voltage multiplier which effectively stores a charge on capacitors until the discharge. If I short the supply with the DMM before turning it on, I won't get the peak current since the caps aren't charged to maximum. Is my reasoning correct?

Many thanks!

[c4757p]

It limits the current by dropping the voltage. Ohm's law - if it's putting out 50uA into a 100 ohm sense resistor, then it's putting out 50uA * 100R = 5mV. Yes, 0.005V. The meter doesn't have to withstand the voltage.

[IanB]

As I mentioned above, there won't be 20 kV on the microamps terminal. The maximum voltage is (current) x (meter resistance). 20 kV is impossible. Suppose the meter resistance was 100 ohms like mine, and the voltage was 20 kV. The resulting current at 20 kV would be 200 A and the power would be 4 MW. Not gonna happen.

I have owned an ion generator and if properly designed you can touch the output with your fingers and nothing will happen. Measure away. You have more chance of damaging your meter by dropping it than by measuring that power supply on a current range.

As to the peak current, Ohm's law says you will get more current with a lower circuit resistance. So if you short the output you will get the maximum current possible. Any higher resistance on the output can only produce a smaller current.

[hazuki]

So I used a multimeter in voltage range to measure the short circuit current. Shorting the output with a 10k yielded 10.4mV across, for a total of 1.04uA. I also tried with a 100k resistor which gave me 112mV, yeilding 1.12uA.

So I think that 50uA rating is for peak current rather than continuous. Interesting. I guess the peak current would be limited by the output impedance of the supply, right? I don't think there is an easy way to measure that!

[Psi]

Maybe check with the scope, its probably not pure DC. Could be unfiltered DC or AC
The DMM is assuming pure DC or 50/60hz on AC range and its probably not trueRMS either.

Just be very careful that you only probe it while the load resistor is in place.
And check your scope earth clip connection isn't going to short anything out.

[hazuki]

I checked with my scope (voltage across a 100k resistor) and found a waveform roughly sinusoidal at 4.5MHz with the top peak around .100V and the negative peak around -.700V for a total swing of .8V which is equivalent to 8uA.

I'd like to check the peak current, which I think would be the current when the high voltage first discharges through the air. Would it be safe on my scope to measure this? I was thinking of tying leg 1 of the resistor to the HV supply return and channel 1 of my scope, tying leg 2 of the resistor to channel 2 of my scope, and then letting the HV supply output arc directy to leg 2. The net voltage across the scope should be on the order of volts or millivolts because of that resistor. Is there any danger of destroying my scope? Note that the HV supply is powered by a battery and has no reference to earth.

Thanks

[IanB]

Have you measured the current with the µA range of your DMM yet? It will probably give a similar number to your measurements across a load resistor, but it would still be useful data point for consistency.