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DIY High voltage probe

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radhoo
Mon Nov 29 2010, 06:43PM Print
radhoo Registered Member #1938 Joined: Sun Jan 25 2009, 12:44PM
Location: Romania
Posts: 699
Motivation/Short Intro
For my High voltage experiments, knowing the exact potential difference of a given supply value would be extremely useful. So I decided to build a high voltage probe, using a resistive divider. After doing some calculations, I decided to go for a 1:1000 divider ratio, for the many advantages including ease of reading the measured values (10kV -> 10V).
150x150Divider 2150x150
The High voltage goes as Vin, and the measurement is done at Vout.
I needed a 1GOhm resistor, so I purchased 5x 200MO Vishay Dale High voltage resistors from Ebay. Connecting them in series I get the desired value:
150x150150x150150x150

To insulate this resistor, I put it in a PVC pipe under oil:
150x150150x150150x150

If R1 is 1GOhm, and R2 is 1MOhm, the voltage drop across R2 , Vout will be aprox. 1/1000 * Vin . Instead of R2 I use a multimeter with 1MOhm impedance (However, to protect the multimeter, additional components are required). The results:
150x150150x150

The left picture shows a good result: for 30.3V in, the multimeter measures 32mA, very close to the 1/1000 ratio (actually the ratio is 1/1001).

Use the probe for HV
Exactly as Proud Mary suggested, I had to "throw away" the multimeter, and get a good old analog microAmpmeter.
-to be continued-

Old issues - fixed - please read the entire thread
However, trying to use this probe with my 50KV cascade supply, gives very poor results. The multimeter jumps between values, the biggest shown is 12 (12KV). Probably because of the high frequency currents.
Any idea on what am I missing? Do I need a capacitive divider? Do I need a combination of the two? Thanks.
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Proud Mary
Mon Nov 29 2010, 07:20PM
Proud Mary Registered Member #543 Joined: Tue Feb 20 2007, 04:26PM
Location: UK
Posts: 4992
Your '50kV cascade' supply seems to be a Cockcroft & Walton voltage multiplier, which has a DC output.

A voltage divider measurement network is suitable for use in DC and low frequency AC measurements, so this is not your problem.

Saying that your multimeter has an input impedance - Z - of 1M, is not at all the same as thinking of the meter as a simple 1M resistance - R - across the terminals. This is your problem, Radhu.

What will be a big problem is if your meter goes open circuit - when 50kV will appear at the bottom end of your 1G resistor.

Here is what you must do: throw away the meter with 1M input impedance and buy another with the standard Zin = 10M.

Connect a resistor of 1M1 from the bottom of the 1G resistor and the Earth. (Connect a neon or zener across this resistor for safety) Measure the voltage across this with your 10M input impedance meter and all will be well.


Avram Iancu - pehblendã / oxid de uraniu nativ smile




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radhoo
Mon Nov 29 2010, 09:35PM
radhoo Registered Member #1938 Joined: Sun Jan 25 2009, 12:44PM
Location: Romania
Posts: 699
Thanks for your reply and suggestions! I'll try to do the mods tomorrow and hope for better results.

Initially I planned to use two resistors R1-1GO and R2-1MO as per the schematics above. However I quickly learned that the multimeter's impedance will change the results. That's why I excluded R2 and simply used the multiplier with it's internal 1MO impedance. I would really need some clarification on why it works like this, when testing with a small voltage supply even if I'm mixing things (Rs with Zs)

I do have a better multimeter with 10MO impedance, and other features, but I don't want to put that anywhere close to the 50KV supply. I also have a digital 1MO impedance multimeter that I was planning to embed into this probe. Why do you think it is a must to use a 10MO multimeter?

About the Zener and the Neon: indeed a most welcomed protection, I already tried paralleling a Neon over the Multimeter's terminals but it was changing the measurements. I've seen this while testing the "probe" with small voltages. Probably my mistake, will try that again (since a neon glass lamp should be somewhere close to 200GO?)

"Avram Iancu - pehblendã / oxid de uraniu nativ"
Thanks for mentioning Avram Iancu over "Dracula" . Much appreciated.


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Proud Mary
Mon Nov 29 2010, 09:57PM
Proud Mary Registered Member #543 Joined: Tue Feb 20 2007, 04:26PM
Location: UK
Posts: 4992
radhoo wrote ...

Thanks for your reply and suggestions! I'll try to do the mods tomorrow and hope for better results.

Initially I planned to use two resistors R1-1GO and R2-1MO as per the schematics above. However I quickly learned that the multimeter's impedance will change the results. That's why I excluded R2 and simply used the multiplier with it's internal 1MO impedance. I would really need some clarification on why it works like this, when testing with a small voltage supply even if I'm mixing things (Rs with Zs)

I do have a better multimeter with 10MO impedance, and other features, but I don't want to put that anywhere close to the 50KV supply. I also have a digital 1MO impedance multimeter that I was planning to embed into this probe. Why do you think it is a must to use a 10MO multimeter?

About the Zener and the Neon: indeed a most welcomed protection, I already tried paralleling a Neon over the Multimeter's terminals but it was changing the measurements. I've seen this while testing the "probe" with small voltages. Probably my mistake, will try that again (since a neon glass lamp should be somewhere close to 200GO?)

"Avram Iancu - pehblendã / oxid de uraniu nativ"
Thanks for mentioning Avram Iancu over "Dracula" . Much appreciated.


Think of your two resistors in series: 1000M and 1M - a simple ratio of 1000:1.

Now, if I take a second resistor of 1M, and connect that in parallel with the first resistor of 1M, we have a resistance of 1/R = 1/R1 + 1/R2 which is 0.5M - 500K.

So we have a ratio of 1000M/0.5 = 2000:1 by using your 1M meter, which will be completely incorrect!

Now, if you take a 10M resistor (think of your meter), and connect that in parallel with your 1M, we have a combined resistance of 910K - so here your voltage divider would have a ratio of 1000:0.95 - no good to us!

But if we make the bottom resistor in the potential divider 1.111Meg-ohms (or 1M111 as we write it) we find that when we put this in parallel with your 10M resistor we have exactly 1M - and the 1000:1 ratio has been preserved! For all ordinary purposes, you can use a resistor of 1M1 as the bottom resistor, which we call the measuring resistor in this circuit.

Klar?
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radhoo
Tue Nov 30 2010, 09:23AM
radhoo Registered Member #1938 Joined: Sun Jan 25 2009, 12:44PM
Location: Romania
Posts: 699
yes, thanks for the examples, I did most of these scenarios on a piece of paper, however I still don't understand why the divider works without using R2, relying only on the multimeter's impedance (which is 1M):

Taking this another example:
Link2

The author speaks about R3, the internal resistance of the DMM. Before picking my multimeter I did some measurements like: Known voltage supply, measured by the multimeter in series with a 1MO resistor (Link2, where resulted that my multimeter can work in a divider as a factor of 1MO.

Moving further, since I still have the question of impedance vs. resistance (unfortunately I did a different school), I see the following on Wiki:
Electrical impedance, or simply impedance, describes a measure of opposition to alternating current (AC). Electrical impedance extends the concept of resistance to AC circuits, describing not only the relative amplitudes of the voltage and current, but also the relative phases. When the circuit is driven with direct current (DC) there is no distinction between impedance and resistance; the latter can be thought of as impedance with zero phase angle.

All my tests have been performed in DC (except measuring the 50KV supply that may still show some ripple), so the 1MO impedance multimeter performed as a 1MO resistor. So this is the answer to my question.

On repairfaq, the author further says:
While R2 is not strictly needed, it is recommended that it be included and approximately equal to the Z-in of the meter on the scale you will be using. The reason to include R2 is to insure that high voltage never can reach the meter.
I've seen this article before building my probe and probably this was another reason I dropped the resistor relying only on the multimeter's resistance.

However now I'm back to the start. Why doesn't this work?


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Proud Mary
Tue Nov 30 2010, 10:49AM
Proud Mary Registered Member #543 Joined: Tue Feb 20 2007, 04:26PM
Location: UK
Posts: 4992
Here is what to do.

Buy an old Russian moving coil 50μA panel meter and use this in series with your 1G resistor. You will have 50μA= 50kV. End of story.


1291114072 543 FT0 Russian 50ua Panel Meter


For safety's sake, you should connect a small neon bulb and a zener in parallel with the meter, so that if the meter fails and becomes open circuit - a very rare event, but possible - then the neon and zener will conduct away high voltage appearing at the end of the 1G resistor. If the zener should also fail, the neon will start to conduct.
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radhoo
Tue Nov 30 2010, 01:11PM
radhoo Registered Member #1938 Joined: Sun Jan 25 2009, 12:44PM
Location: Romania
Posts: 699
Thanks Proud Mary, for bearing with me on this one and for the practical advise, I'll look around for an old uA meter, but while I do that, I'm still intrigued about the poor results in my first attempt, using the DVM.

I need to understand the failure reasons. Did you (or others on the forum) perhaps tried to build a probe and maybe have a clue on this?
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Proud Mary
Tue Nov 30 2010, 01:23PM
Proud Mary Registered Member #543 Joined: Tue Feb 20 2007, 04:26PM
Location: UK
Posts: 4992
radhoo wrote ...

Thanks Proud Mary, for bearing with me on this one and for the practical advise, I'll look around for an old uA meter, but while I do that, I'm still intrigued about the poor results in my first attempt, using the DVM.

I need to understand the failure reasons. Did you (or others on the forum) perhaps tried to build a probe and maybe have a clue on this?

I think it's because the DMM input has some active devices in it - it's not just a resistance going straight to Earth - and perhaps the DC is 'dirty' with switching waveforms imposed upon it. I'd guess that if you used your C&W 'cascade' to charge a capacitor - say a Leyden Jar - and then measured the voltage across that, it would work properly. But do be careful if you try this, or you will be toast, my friend!

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Patrick
Tue Nov 30 2010, 02:52PM
Patrick Registered Member #2431 Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639
what material is the cork stuff in the ends of your oil filled pipe? how did you get a good leak proof seal?
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radhoo
Tue Nov 30 2010, 03:04PM
radhoo Registered Member #1938 Joined: Sun Jan 25 2009, 12:44PM
Location: Romania
Posts: 699
it's hot glue.

The first end (with the pipe empty of oil) was simply sealed with a perfectly matching cylinder of glue bar, and the resistor's connector inserted though it. Additional hot glue was added to ensure sealing.

The other end could not take a sealing bar since the air inside would have been compressed. So I carefully sealed it with hot glue, directly over the oil's surface. Not even one bubble of air left.
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