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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.

Wikipedia says: (

"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


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.