Measuring High voltage

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Dren
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Posted by Dren: Fri. Apr 3, 2009 - 12:32 PM
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Hello,

how would you go about measuring the voltage on a VanDerGaff generator ?

Let's say the voltage is 50KV, so then dividing it by x20K using a pot divider would give you 2v5 which can go into an ADC. However, if this was set up then it would probably cause the high voltage to arc down to ground.

I was thinking of something like a gold leaf electroscope (?) something I vaguely remember from school. Basically a thin sheet of gold that is repelled from a ground plane, the angle it lifts is a dependant on the voltage. Of course there's then the problem of measuring this deflection (I'm thinking of an LED and photodiode reflecting off the gold leaf).

Any ideas greatly appreciated

Dren

<º))))><

I am only one lab accident away from becoming a super villain.

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Learning and Information, General Electronics
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Jokerman
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Posted by Jokerman: Fri. Apr 3, 2009 - 01:14 PM
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Due to its very high impedance it might be better to use a capacitive voltage divider.

You might be able to wind your own capacitors using foil and plastic sheet. I guess the absolute value of the capacitor is not important as long as you wound the two caps with a known ratio of their surface areas.

Regards, -=mike=-

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brandon
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Posted by brandon: Fri. Apr 3, 2009 - 01:32 PM
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Don't connect a Van de Graaff generator to a gold leaf electroscope. I did that at school and had to spend 3 hours repairing the gold leaf that had flown off and stuck itself to the glass!!

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ka7ehk
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Posted by ka7ehk: Fri. Apr 3, 2009 - 04:06 PM
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There is no good reason for a resistive voltage divider to arc! To ground or anywhere.

But, what you DO have to watch out for is the end-to-end voltage rating of the resistor(s) you use for the divider. Most 1/4 watt axial resistors are rated at a few hundred volts, maximum. So, what do you do? You connect a whole bunch of equal value resistors in series. If the VanDerGraf generator puts out 50KV and your resistor can handle 500V, then you need 100 in series. That would also give you 1:100 attenuation if you look at the voltage across the last resistor. That voltage can be attenuated further to be suitable for your ADC input.

Half watt and 1 watt resistors should have a higher voltage rating, so you should not need so many. For something like this, you will need to use resistors with KNOWN characteristics, so your corner electronics shop would NOT be a good source. You could safely assume 250V per resistor, and that would only need 200. You could easily construct this in a zig-zag pattern to reduce the total length and there would not be a voltage problem because the voltage difference between any two adjacent terminals is only 500V. Not a small project!

Jim

Jim Wagner Oregon Research Electronics, Consulting Div. Tangent, OR, USA http://www.orelectronics.net

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brunomusw
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Posted by brunomusw: Fri. Apr 3, 2009 - 04:32 PM
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Use a Voltage Transformer???
Safer than use resistor divider network, isn't it?

Or use the resistor in series like Jim said, and in the end use optocoupler, then your AVR is isolated.

Brunomusw

Regards,

Bruno Muswieck

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ka7ehk
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Posted by ka7ehk: Fri. Apr 3, 2009 - 05:14 PM
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Whoa -

Transformers only work on AC. VanDerGraf is NOT AC.

Optocoupler not so good because it is really an on/off device and not linear nor stable.

You want to run the resistor string from the top of the generator to ground. So long as you connect the ground of the AVR circuit to the same point that the ground end of the divider connects to, then there is no problem. For safety, I would connect a diode between processor Vcc and the analog input (cathode to Vcc). This way, if there are any "accidents", nothing will get fried without wiping out the diode. A good Schottky diode of the same general type used for switch mode power supplies would be ideal.

Jim

Jim Wagner Oregon Research Electronics, Consulting Div. Tangent, OR, USA http://www.orelectronics.net

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brunomusw
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Posted by brunomusw: Fri. Apr 3, 2009 - 05:31 PM
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:oops:
Always learning... thanks Jim..

Quote:
For safety, I would connect a diode between processor Vcc and the analog input (cathode to Vcc). This way, if there are any "accidents", nothing will get fried without wiping out the diode.

But before wiping out the diode, all the current goes to Vcc, isn't? Will not blow the power supply?

Brunomusw

Regards,

Bruno Muswieck

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polecat
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Posted by polecat: Fri. Apr 3, 2009 - 05:33 PM
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There are several types of high value resistors rated for thousands of volts. Be warned that they also have a very high price tag when you can find them.
might look at Vishay ROX series.

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ossi
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Posted by ossi: Fri. Apr 3, 2009 - 07:17 PM
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And i think the higher the voltage gets along
the resistor-chain the more danger of
corona discharges into the air might be a problem.

The geometrical construction with avoiding sharp edges
and thin wires may be a real challenge.

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Kleinstein
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Posted by Kleinstein: Fri. Apr 3, 2009 - 09:47 PM
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For very high impedance, you may get away with a field mill. Its the elektronic, more quatitative analog to the elektroskope. Its just a groundet kinde of fan infront of an elektrode that is connected to a sensitive AC current measurent amplifier. Usually sycronized recitification is used. The only problem is, that you have to calcibrate it before use. For calibration you may need a few 100 V.

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tpappano
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Posted by tpappano: Fri. Apr 3, 2009 - 10:54 PM
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Only 50 kv? Is it one of the World's tiniest Van de Graff generators? Your voltage divider will need to be really high impedance, it is just static, after all. Why not just measure the spark length? That would probably be more accurate because any resistive load would likely drag the output down significantly.

Tom Pappano
Tulsa, Oklahoma

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ka7ehk
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Posted by ka7ehk: Sat. Apr 4, 2009 - 01:13 AM
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On Bruno's last question, No, it won't wipe out the power supply because the available current is so low, and the series resistance is so high.

Lets do a reality check, here. Suppose that you use the largest easily available resistor value, which ought to be larger than 1Meg (maybe 10Meg). So, you have 20 of these, in series. Makes 20Meg. From a 50KV source, the current will be 0.4ma. That is a LOT of current for a Van Der Graff generator!. Even at 10Meg each, the current will be 0.04ma = 40ua. That is still a lot, though maybe usable.

How many 10Meg resistors can you afford to get?

Jim

Jim Wagner Oregon Research Electronics, Consulting Div. Tangent, OR, USA http://www.orelectronics.net

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DocJC
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Posted by DocJC: Sat. Apr 4, 2009 - 11:46 AM
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I'm with Tom.

If the drive motor RPM is constant, then the output voltage is dependant upon the design. A spark will jump about 1 inch / 10 KV. Use your spark gap to measure the voltage at that RPM. Remove the spark gap and do your experiment. If the RPM and humidity don't change much then neither will the voltage.

How accurate do you need to be?

Jc

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brandon
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Posted by brandon: Sat. Apr 4, 2009 - 03:15 PM
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DocJC wrote:
A spark will jump about 1 inch / 10 KV.

More like 1 inch / 84 KV !! Usually quoted as 33 kV / cm (1 inch = 2.54 cm, even in USA). Do be careful!

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DocJC
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Posted by DocJC: Sun. Apr 5, 2009 - 12:53 AM
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HI Brandon,

Oh Man, you are Correct!

I don't know where I pulled that figure from, but it wasn't even close :oops: :!:

Thank you for correcting me!

JC

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DocJC
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Posted by DocJC: Sun. Apr 5, 2009 - 01:17 AM
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I'm having a technically challenged day... My camera won't focus correctly, even with a newly charged battery...

Anyways: Here is my version of Jim's suggestion, above. It was for a project a few years ago. I needed a 50 ohm load and a 10KV peak, much less than discussed above, but the concept is the same.

Input voltage on the left, an array of 30 resistors to keep the voltage across any single one within specs, and an additional tap for the ADC on the right side.

JC

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Dren
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Posted by Dren: Wed. Apr 8, 2009 - 08:59 AM
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@DocJS That's a neat little board :) For this app I'd probably lay them out in a line (down the side of the drive belt) inside a rubber tube.

I guessed at 50Kv, as it can get sparks jumping about a cm, but maybe 100Kv would be more realistic. It's not a big generator, just a small desktop unit to demo to students. I suppose that there is a health and safety reason why it can't be too impressive these days.

Using a pot divider will take a lot of resistors, OK they are cheap but several hundred of them will take time to build and be a pretty long tube.

I like the idea of an electronic mill spinning away, but that's the 'impressive' demo of electronic wind so the teacher would probably want to save that for the grand finale (or is that when a student gets zapped?).

I think I might have a play with the electroscope/diode/phototrans idea. Yep, I've also seen that gold leaf shoot across the room :S ! So maybe I'll just have a couple of metal plates and a tiny hinge. If bounce a beam off it then I don't need much mechanical deflection.

I suppose that I could do something with a sting of resistors/neon bulbs ? There's also vacuum discharge tubes on Ebay but they need too much current to drive.

This project is only so that the teacher can show the charge gradually building and a rough ballpark measurement of the volts. So any ideas welcome :)

<º))))><

I am only one lab accident away from becoming a super villain.

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DocJC
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Posted by DocJC: Wed. Apr 8, 2009 - 04:55 PM
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Perhaps in a darkened room you could also watch a nearby fluorescent tube glow? (Haven't tried this...).

You could, or course, automate it with a photocell taped to the tube to measure it's luminance and add a digital display...

JC

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