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Is it possible to divide an electrostatic voltage from a Van de Graff generator or electrostatic generator with very large resistors?

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  • \$\begingroup\$ yes. Voltage is voltage, and resistors are resistors, no matter the dimension of their values. I think you wouldn't be asking this question if you hadn't have a specific doubt, caused by anything: Could you elaborate on that? I'm certain the question "do fundamental laws of physics still apply for higher voltages" isn't what you meant to ask! \$\endgroup\$ – Marcus Müller Jul 18 at 19:00
  • \$\begingroup\$ As long as the resistor doesn't breakdown. Resistors have a max voltage withstand rating. \$\endgroup\$ – DKNguyen Jul 18 at 19:05
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    \$\begingroup\$ It won't be static any more. \$\endgroup\$ – Transistor Jul 18 at 19:09
  • \$\begingroup\$ Marcus Muller (the sleuth): Yes, the reason I ask is that lower value resistors would surely cause a typical electrostatic generator to deviate from normal function by 'asking' for too much current to flow or cause resistor thermal failure. Was hoping someone with practical experience might have a ballpark lower limit estimate for resistance that would keep your 'typical' EBay-bought (i.e. no technical data) electrostatic generator functioning, since I don't have an electrostatic voltage (charge) meter on hand and have few ways to experiment. \$\endgroup\$ – Ted Jackson Jul 19 at 15:18
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The comments so far on your post are good. I particularly like Transistor's:

It won't be static any more.

Let me expand on that...

An electrostatic voltage is a difference in charge caused (normally) by an excess or deficiency of electrons. One object (the negatively charged one) has extra electrons, and the other object (the positively charged one) has a relative deficiency. Both are insulated from each other, which keeps the charge from neutralizing.

When you use one or more resistors to connect the two objects, electrons flow between the objects as a current. They are no longer insulated from each other. Unless you are somehow regenerating the voltage (e.g., by a Van de Graff generator), the voltage difference between the objects will fall. At this point, all the rules you know from EE 101 apply as expected. The resistors will indeed divide the voltage, even while the overall voltage will be falling. The voltage and the rate at which it falls will both exponentially approach zero, following the natural response of a charged capacitor discharging through a resistor, because that's exactly what's happening. The two objects are equivalent to the two plates of a charged capacitor, and they're storing energy in the same way.

Now, in order to get this to work, the resistors can't break down or arc across the terminals, so your voltage has to be reasonable. Using physically longer resistors will decrease arcing through the medium around them. Also, because the system behaves as the RC circuit that it is, you'll want to use very high resistance resistors if you don't want to drain your source quickly.

Also, if you tap into the connection between the resistors to measure the voltage, take into account the current draw of your measuring tool. If you're using very large resistors, your voltmeter will probably draw (within an order of magnitude or two) as much current as the resistor, effectively changing the voltage ratio.

Blowing this up to infinity, if you somehow had two endless reservoirs of charge, your two resistors would certainly work as a voltage divider. This could be helpful, for instance, if the voltage difference between the two reservoirs was too large to measure with your voltmeter, so you needed to divide it down to something more manageable. Note, though, that a resistor divider would not be a good way to get a lower-voltage power source from the high-voltage static charge. The added current draw through the first resistor would change the voltage ratio, just like a voltmeter or any other load in parallel with the second resistor.

Of course, you could regenerate the source constantly... But then you'd just have a generator putting electricity through resistors, and the whole concept of electrostatic voltage is moot, making it much less interesting. ;)

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Yes, provided that the resistors are large enough in value so as to not "load down" the source. ALL voltage sources have some source resistance, which is VERY high for Van de Graffs, et. al. Also, as pointed out, the resistors must be rated for the voltage, i.e., they must be physically large enough so as not to arc-over.

As far as not being "static", that's just semantics - ALL "static" sources have some non-static behavior(albeit miniscule) except in a vacuum.

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  • \$\begingroup\$ What? I can't use a tiny surface mount resistor? Hee hee. But you're right, DP. The resistor must be beyond a certain length or nature herself will choose to ignore the poor resistor altogether. Zap! \$\endgroup\$ – Ted Jackson Jul 19 at 15:27
  • \$\begingroup\$ I would suggest wiring several large-ohm resistors in series to avoid any arc-over problem. \$\endgroup\$ – DontPanic Jul 20 at 13:58
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The voltage or electric potential is calculated as the total work required to move a positive test charge from the negatively charged electrode to the positively charged electrode against the electrostatic force generated by the effort to move the charge through an invisible electric field.

Applying the positive charge convention series resistors divide voltage and dissipate power when the voltage source performs electrical work to move some electric charge from the positive electrode to the negative electrode of the potential source. The current is calculated as the open circuit potential divided by the total series resistance including the measure or estimate of the internal series resistance of the source itself. So as DontPanic writes the external resistance can load down the source voltage if it is not large in comparison to the internal source resistance.

Link below has answers to a similar question stating that one should not attempt to make a voltage divider from resistors across a high voltage source due to issues of safety and limited comprehension of the geometry and properties of materials when dealing with high voltage sources:

https://www.quora.com/How-will-I-make-50k-volts-and-a-005A-circuit-using-resistors-with-input-voltage-400k-volts-What-is-the-actuality-of-resistor-since-I-can-t-understand-ohms-exactly

"I have been working with HV since the 1960s, and would not even attempt creating a resistive divider chain for a 400kV input."

The power dissipation in a resistor is voltage developed across the device times the current flow through the device. The current is voltage across the resistor divided by the resistance value. Special resistor materials and custom geometry may be necessary to create a safe and effective voltage divider in any high voltage application.

Ohmcraft makes high voltage axial resistors rated as high as 50kV and 10W power dissipation as shown under this link:

https://www.ohmcraft.com/leaded-resistors/new-high-voltage-axial-resistors-hva-series

This link is the ohmcraft paper outlining how to specify resistors:

https://www.ohmcraft.com/uploads/AN_SpecifyingResistorsTutorial.pdf

Maybe you can speak to some experts in ohmcraft or general high voltage industry to share their experience. I don't have direct high voltage experience and that's what I would do to learn from domain experts.

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  • \$\begingroup\$ You all make it clear that I should have been more concise in my question. Apologies. Any ballpark estimates for a lower limit on resistance for a 'typical' 12Vdc electrostatic generator module outputting 100,000V? Maybe the answer is, 'As big as you can find, stupid.' and as DontPanic makes clear 'as long'. \$\endgroup\$ – Ted Jackson Jul 19 at 15:24
  • \$\begingroup\$ Dry air breaks down at about 100kV/inch, so size your top-half divider chain accordingly. Some 10-meg, 1/8w resistors in my scrap box are about 1/4", so good for maybe ~25kV. Maybe some carbon-composite resistors would be larger (if you can find them). I assume you want to do division to measure voltage, if so, the lower-half of the divider should have a resistance ~10X less than the meter's input resistance. In any case, some experimentation may be required. \$\endgroup\$ – DontPanic Jul 20 at 14:17
  • \$\begingroup\$ As far as minimum divider resistance, I cannot say - it depends on generator, but certainly many, many megohms. You'll have to experiment. \$\endgroup\$ – DontPanic Jul 20 at 14:23
  • \$\begingroup\$ I RETRACT my previous suggestions (My Bad). As pointed out by @SystemTheory, you must also account for the power rating of the resistors. Assuming that N series resistors of value R1 ohms are used, and the input voltage is Vt, the voltage across one resistor is V1 = Vt/N. The power dissipated by that resistor is P1 = V1V1/R1 = VtVt/(NNR1). Rearranging, NxN=VtVt/R1P1. For R1=10Mohms @ 1/10W, that comes out to N=100. That's a LOT of resistors. \$\endgroup\$ – DontPanic Jul 26 at 21:31

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