DIY HV Probe

[Alberto]

Hello.

I´m going to make a HV Prove 1.000:1 using resistors. But I see a lot of people using great values (1.000 Mohm). What is the problme if I use 10 resistors of 10 Mohm and one of 100Kohm?

I´ll use it for flybacks and that kindo of things, below 50 KV. I want to use a cheap DMM for this and its internal resistance is 1 Mohm, so I need tu ose low values.


Thank you.

Kind regards

[davekni]

Don't forget to include meter resistance in the bottom 100k.  In other words, you need a 111k resistor so that it becomes 100k when in parallel with the meter's 1meg input resistance.  Or, what I do is use the meter's 1meg as the bottom resistor of the divider, making it a 100:1 probe.  (Actually 101:1, but the resistors and cheap meter aren't likely accurate enough for that to matter.)

50kV across 10 resistors is 5kV per resistor.  You'll need something like Vishay's VR68000 family of resistors (rated for 10kV).  Typical 1/4 watt resistors will be non-linear by 1kV and arc over before 5kV.  I've used standard resistors in 100-long strings, and VR68000 parts in shorter strings.

[Alberto]

Don't forget to include meter resistance in the bottom 100k.  In other words, you need a 111k resistor so that it becomes 100k when in parallel with the meter's 1meg input resistance.  Or, what I do is use the meter's 1meg as the bottom resistor of the divider, making it a 100:1 probe.  (Actually 101:1, but the resistors and cheap meter aren't likely accurate enough for that to matter.)

50kV across 10 resistors is 5kV per resistor.  You'll need something like Vishay's VR68000 family of resistors (rated for 10kV).  Typical 1/4 watt resistors will be non-linear by 1kV and arc over before 5kV.  I've used standard resistors in 100-long strings, and VR68000 parts in shorter strings.

Thank you for your answer.

I don´t understand the thing about the 111K resistance. Can you show me how I can calculate it?

And yes, I was thinking about Vishaiy´s 68000 for the big value resistors. I have found 51, 56 and 62 Mohn and 1 watt. So the bottom resistance must be around 50-60Kohm.

[davekni]

The meter behaves as a 1-meg resistor from +input to -input (ground).  That 1-meg resistance is in parallel with the bottom resistor of your divider.  If you look up the formula for parallel resistors, it's typically stated as R_parallel = R1 * R2 / (R1 + R2).  Plugging in 1-meg for R1 and 111k for R2 gives 99.9k, so roughly 100k.

Perhaps a more intuitive way to look at it is by conductance, or 1/resistance.  Conductivities add for parallel resistors, just like resistances add for series resistors.  The parallel formula then becomes 1/R_parallel = 1/R1 + 1/R2.  This makes the algebra easier for calculating R2 given desired R_parallel and R1:  1/R2 = 1/R_parallel - 1/R1 = 1/100k - 1/1meg ~= 1/111k.

For ten 62meg resistors in series, total is 620meg.  Thus you want 620k for the bottom of the divider.  Use the above formula to figure out what to use in parallel with the meter.  Or, just use 16 resistors instead of 10.  16 * 62meg = 992meg.  That's close enough to 999meg, so will work with no bottom resistor other than the meter.  That's the way my dividers work.  Load on the voltage being measured is a bit lower that way too.

[Alberto]

The meter behaves as a 1-meg resistor from +input to -input (ground).  That 1-meg resistance is in parallel with the bottom resistor of your divider.  If you look up the formula for parallel resistors, it's typically stated as R_parallel = R1 * R2 / (R1 + R2).  Plugging in 1-meg for R1 and 111k for R2 gives 99.9k, so roughly 100k.

Perhaps a more intuitive way to look at it is by conductance, or 1/resistance.  Conductivities add for parallel resistors, just like resistances add for series resistors.  The parallel formula then becomes 1/R_parallel = 1/R1 + 1/R2.  This makes the algebra easier for calculating R2 given desired R_parallel and R1:  1/R2 = 1/R_parallel - 1/R1 = 1/100k - 1/1meg ~= 1/111k.

For ten 62meg resistors in series, total is 620meg.  Thus you want 620k for the bottom of the divider.  Use the above formula to figure out what to use in parallel with the meter.  Or, just use 16 resistors instead of 10.  16 * 62meg = 992meg.  That's close enough to 999meg, so will work with no bottom resistor other than the meter.  That's the way my dividers work.  Load on the voltage being measured is a bit lower that way too.

Thank you for your answer.

Yessss, now I understand. I knew the formula, but I didn´t understand why I needed to apply. Now all is clear.

So, if I use 10 x 62 Mohm resistors, the bottom resistor has to be 62Kohm = 1.000.000R/(1.000.000+62.000)   R = 66,1 Kohm.

Or the wayyou propose.

But now my question is. Why the people use very high value resistors if it can be done with these resistors?

The other thing is, I have only found the Vishay's in 1 watt Is that enought?

Thank you

[davekni]

Since you know the parallel-resistor formula, you likely know the power formula too.  Figure out what voltage across a 62meg resistor produces 1 watt.  10 resistors allows for 10 times that voltage.

62k load resistance will make a 10000:1 probe.  Is that what you want?  Remember, your input resistance is 10 * 62meg = 620meg.

If you do want 10000:1 and the corresponding 62k, you came up with the correct 66.1k resistor.  Formula has a typo, however:
"62Kohm = 1.000.000R/(1.000.000+62.000)   R = 66,1 Kohm"
Should be:
62kohm = 1000000 * R / (1000000 + R).  R = 66,1k.

I suspect long strings are used for two reasons.  One is to get higher total resistance, thus loading the measured voltage less.  For example, my CW voltage multiplier for static-electricity demonstrations has ~400meg output impedance.  (That's intentional for safety, as I use it to charge people - hair standing on end etc.)  Measuring that accurately would require a much higher impedance resistor string.  The other reason is, if you have more time than money, generic low-voltage resistors are orders-of-magnitude cheaper than high-voltage resistors.  Soldering a very long string of small resistors is cheaper if the cost of time spent isn't counted.

[Alberto]

Since you know the parallel-resistor formula, you likely know the power formula too.  Figure out what voltage across a 62meg resistor produces 1 watt.  10 resistors allows for 10 times that voltage.

62k load resistance will make a 10000:1 probe.  Is that what you want?  Remember, your input resistance is 10 * 62meg = 620meg.

If you do want 10000:1 and the corresponding 62k, you came up with the correct 66.1k resistor.  Formula has a typo, however:
"62Kohm = 1.000.000R/(1.000.000+62.000)   R = 66,1 Kohm"
Should be:
62kohm = 1000000 * R / (1000000 + R).  R = 66,1k.

I suspect long strings are used for two reasons.  One is to get higher total resistance, thus loading the measured voltage less.  For example, my CW voltage multiplier for static-electricity demonstrations has ~400meg output impedance.  (That's intentional for safety, as I use it to charge people - hair standing on end etc.)  Measuring that accurately would require a much higher impedance resistor string.  The other reason is, if you have more time than money, generic low-voltage resistors are orders-of-magnitude cheaper than high-voltage resistors.  Soldering a very long string of small resistors is cheaper if the cost of time spent isn't counted.

Ok thank you. Yes, I make a mistake with a 0.

I´ll try and I´ll tell.

[Hydron]

If you're going to go ahead with something like this, then please don't hold the probe or even meter when using it - set it up and stand back before turning on the power. That way even if there is a mistake you're not at risk - just your cheap meter. I'd apply this rule even if it works the first time - unless you're an expert in HV design it's not worth the risk that it will fail when you're measuring something dangerous (e.g. a big HV cap or something, though even a flyback can be dangerous).

[Alberto]

Thank ypu for the advice.

Yes yes, this prove is non-touch. I´ll apply it in the source when is discconected, and later I´ll stard the source.

[Alberto]

Ok, I have finally recieved the components.

I´m not going to touch anything while is conected, but I´m thinking on covering everithing in parafin, and later cover with Heat-shrink tubing to avoid arcing.

But I think it will be difficult. Another solution? Maybe teflon tape (From plumbing)

Thank you

[davekni]

If you are using VR68000 family resistors, I haven't had trouble with those arcing from one lead to the other even without any encasement.  To keep the entire string insulated and rigid, I slid mine into a section of polyethylene tube (from hardware store - normally used as water supply line).

[Alberto]

Ok, thank you so much

[Alberto]

I have made it.



It works very well. I have put it 5 layers of heat-shrink tubing.

I know I don´t have to touch it, but if I put it inside PVC pipes (one inside another) and I create a 10 or 15 mm thick wall. Wouldn´t it be enought? For example 20 KV DC?

[davekni]

First, any safety advice is just my thoughts.  All decisions and associated risks are yours alone.  Back in high-school and college, 4+ decades ago, I managed to shock myself every week or so, usually by touching 120V 60Hz line voltage while trying to measure or fix something without being cautious enough to unplug it.  If I had any heart issues, I'd likely not be alive now.

For me, the degree of safeguards depends on how lethal a mistake is likely to be.  A typical TV flyback transformer is unlikely to generate enough current to be lethal for most people, unless you have an undetected heart problem or implanted medical electronics.  Would make an uncomfortable burn though.  For something like that I'd happily hold a probe.  On the other hand, my coin-shrinker uses a TV flyback transformer to charge a 14uF 20kV capacitor (2.8kJ).  I don't get close to that when charged.  One mistake would likely be my last ever.  (It sits in the corner of my garage.  I charge and trigger it from inside my house.)

[Alberto]

Ok ok. Thank you for your answer.

Yes, Of course is not the same a MOT than a small flyback.

One question that I have allways had it. If I touch with my bare hand one terminal of a flyback and I´m not wearing shoes. I know is very dangeros, because the current flow acroos my body (Includin heart). But I undertood that the current needs a path to back to the source. A circuit. Why the current of the flyback go across my body if theres is not a close circuit?

Thank you

[davekni]

Yes, current needs a closed circuit.  That is one reason I survived my youth - knew enough to use one hand at a time.  Most shocks were just from one part of a hand to another part of the same hand.

For high-frequency high-voltage such as a Tesla coil, the return path may be capacitance between you and ground.

For DC such as a typical TV flyback transformer, the return path needs to be DC-coupled, so no capacitors.  (Even standing on an insulating stool, you would get a bit of shock as your body charges to 30kV or whatever the flyback is generating, much like typical static-electricity shocks.)  If the floor you are standing on insulates well enough, you actually could touch a DC flyback output with only the initial capacitive charging shock.  However concrete and even wood (depending on moisture content) will conduct enough at 30kV to make a return path to ground.  The flyback return (-) lead is typically also grounded, completing the loop.  If not explicitly, 30kV is enough to break down insulation from the flyback return lead to the flyback primary, then back through the power supply to ground.  If you ran a flyback on isolated battery power well insulated from ground, then again you could touch it even if you are grounded.

[Alberto]

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Great answer!! Thank you very much, it is the first time someboy give a real good andwer to that question.