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## Solid States Devices => solid state devices => Topic started by: nix85 on February 18, 2020, 02:14:11 PM

Title: Wimshurst - let's clear this up
Post by: nix85 on February 18, 2020, 02:14:11 PM
How exactly does Wimshurst produce HV? I understand how charge induction happens on opposite discs. Charge is picked up from two pairs of oppositelly charged sectors (two on each side), but these two pairs alone surely don't produce HV.

Are all the sectors somehow adding up in voltage? That sure does NOT seem to be the case, at least i don't see how that would work.

"The available voltage gain can be understood by noting that the charge density on oppositely charged sectors, between the neutralizer bars, is nearly uniform across the sectors, and thus at low voltage, while the charge density on same charged sectors, approaching the collector combs, peaks near the sector edges, at a consequently high voltage relative to the opposite collector combs."

Wikipedia explanation that sharpness of the combs makes charge density on the adjecent sector higher and thus provides higher voltage makes sense, but i still don't see how many dozens of kilovolts would be generated that way, especially by a single sector pair.

I watched MIT video and this one too https://youtu.be/nA4aCd5qFWs?t=239

Anyone?

Title: Re: Wimshurst - let's clear this up
Post by: nix85 on February 19, 2021, 01:09:16 AM
To clarify, as i understood later, it works similar to Van de Graaff generator, simultaneously adding and removing charge from opposite plates of a Leyden jar (cap).
Title: Re: Wimshurst - let's clear this up
Post by: antijon on February 19, 2021, 08:09:34 PM
As I understand it a static generator, like a van de graaff, operates by moving and depositing charge and functions as a current source. I’m not too familiar with a wimshurst, but I have a small van de graaff, and I know that the charge is able to build up because of the domed top. If you apply charge to the inside of a cup or sphere, it moves to the outside, so you can continue to add charges of lower potential as the outside develops a high potential.

Anyway, a current source has a near infinite internal resistance, unlike batteries or transformer windings, so the voltage increases to a value necessary to push current across a resistance. Theoretically a van de graaff could be used for free energy if it generated a high enough current because no matter the load resistance, as long as it’s lower than the internal resistance, the voltage would continue to increase to match ohms law - IXR=E. For example, if a 120V bulb had a resistance of 120ohms, it would require 1A to light up. If you had a current source of 1A, you could power thousands of bulbs in series. What people don’t realize is that with a current source power increases to match the load. The problem is making a static machine that generates a decent current.
Title: Re: Wimshurst - let's clear this up
Post by: nix85 on March 07, 2021, 07:28:01 PM
As I understand it a static generator, like a van de graaff, operates by moving and depositing charge and functions as a current source. I’m not too familiar with a wimshurst, but I have a small van de graaff, and I know that the charge is able to build up because of the domed top. If you apply charge to the inside of a cup or sphere, it moves to the outside, so you can continue to add charges of lower potential as the outside develops a high potential.

It works simply by electrostatic induction creating small voltages that charge the caps. Now this is a tricky part.

Indeed, adding charge to the center of a conductor allows to build up surface charge to any level cause the source does not "see" the built up charge.

And it does not have to be a dome

I long accepted this as solution for Wimshurst, but was lately skeptical cause Leyden jar is afterall a classical capacitor, not a VDG dome.

And we all know we can never charge a cap to 10V with 1V source.

But apparently this is bypassed in Wimshurst, exactly how no one explained clearly.

Quote
Anyway, a current source has a near infinite internal resistance, unlike batteries or transformer windings, so the voltage increases to a value necessary to push current across a resistance. Theoretically a van de graaff could be used for free energy if it generated a high enough current because no matter the load resistance, as long as it’s lower than the internal resistance, the voltage would continue to increase to match ohms law - IXR=E. For example, if a 120V bulb had a resistance of 120ohms, it would require 1A to light up. If you had a current source of 1A, you could power thousands of bulbs in series. What people don’t realize is that with a current source power increases to match the load. The problem is making a static machine that generates a decent current.

Max voltage obtainable by VDG is limited by volume of the sphere and insulation properties of surrounding medium, usually 450kV for 30cm sphere, but much more is achievable in closed ones.

Sadly author ignored my question below if adding external capacitance adds to max voltage, he clearly has no idea if it does.

As for the latter part of your post, i completely disagree, you are misinterpreting things. It is possible to extract energy from subtler energy forms ala Don Smith, but not just with electrostatic field, you need a dynamic component.
Title: Re: Wimshurst - let's clear this up
Post by: nix85 on March 08, 2021, 11:00:13 AM
Is anyone gonna explain how Leyden Jars get charged to dozens or hundreds of kilovolts with tiny voltages on the rotating discs.

It has to be the adding charge to center of conductor principle as in VDG, but again, we are talking Leyden Jars, that is, a capacitor with two plates and a dielectric.

This is confusing cause, again, you cannot charge a cap above the source voltage.

I guess when you are charging a cap by electrostatic induction instead of direct application of voltage, you can charge it to any level even tho the source voltage is tiny.

But then again, electrostatic induction is direct application of voltage, i don't see a difference.

Hopefully someone will clear this up.
Title: Re: Wimshurst - let's clear this up
Post by: antijon on March 08, 2021, 05:18:48 PM
Hey Nix, I doubt I can clear it up, but I can speculate. The action of charge "induction" may be the reason for the potential. I don't know much about induction, but I know it's different than conduction. e.g. my vdg has a built in light bulb by the lower brush to decrease humidity, because conductive atmosphere would kill the effect.

I'm guessing induction is related to the surface area of the plate, the potential charge of the plate, and it's distance to the pointed electrode. The electric field of the plate converges on the point of the electrode.

Hard to surmise, but in my simple mind it means the area of the plate reduces to the electrode, so the charge must have an equal increase in potential. Or, the area of the plate produces a combined action on the ions of the air causing the potential of a few ions to equal that of the charged plate. Humidity must form a return path to short out the plate which causes a drop in potential.

But what I said about infinite power, that's related to an ideal current source.
From wiki,
Quote
The voltage across an ideal current source is completely determined by the circuit it is connected to. When connected to a short circuit, there is zero voltage and thus zero power delivered. When connected to a load resistance, the voltage across the source approaches infinity as the load resistance approaches infinity (an open circuit).

So for Ohm's
1A x 10 Ohm = 10V = 10W
1A x 100 Ohm = 100V = 100W

It's just an ideal current source is completely backwards to the way that we use power. Everything would have to be wired in series and designed for the current output of the source.
Title: Re: Wimshurst - let's clear this up
Post by: nix85 on March 08, 2021, 06:11:09 PM
To start with ideal current source, that is just abstraction to make a point, battery is considered an ideal voltage source (almost) yet it does not provide infinite anything.

Now that we put that away, yes, pointed electrode increases the local electric field, i mentioned it in the first post, quote..

Quote
Are all the sectors somehow adding up in voltage? That sure does NOT seem to be the case, at least i don't see how that would work.