Re: [volt-nuts] Bohnenberger electrometer DANGER

[ed breya]

Here's a simplistic view that may be sufficient. Some energy (in the 
form of charge redistribution, which includes current flow) has to come 
from the capacitor, and some from the input signal, to do the work 
needed to push the leaf against gravity. When the input signal is 
removed, some of the energy (charge) stored on the leaf is returned to 
the cap as gravity restores the initial position - roughly the same 
amount of work, depending on leakage and mechanical loss, and heating of 
the protective series resistor.


With a quick review of electrostatics, you could estimate up a simple 
model and the field equations to get a more satisfying, detailed answer. 
It may be more straightforward to look at it from a circuit perspective. 
Picture it as as a very small, non-linear capacitor (the leaf structure) 
in series with a much much larger regular capacitor charged up to a 
constant DC voltage. Any actual series resistance is just resistance, 
and the mechanical loss can be represented as more resistance added in 
series. Presuming the leaf never actually touches or emits particles* or 
arcs to the reference capacitor node, it's basically a capacitive 
voltage divider, and the applied signal may be considered to be 
transient, or even AC - it steps to the applied voltage, then returns to 
zero (or open), then the cycle may be repeated. Each experiment is 
adding or subtracting charge, then reversing the process. Ideally, the 
cap would never lose its DC bias if there were no losses.


The problem is that figuring out all these details may not be trivial. 
It may be more fun to just try some experiments and see how it goes - 
you'll get some idea of how the real thing holds up, and figure what 
amount of C is OK.


*You shouldn't have to worry too much about corona discharge if the 
maximum voltage on anything is below 3 kV or so. Beyond that, it could 
cause problems.


Ed
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Re: [volt-nuts] Bohnenberger electrometer DANGER

[Dr. David Kirkby]

On 8 March 2018 at 07:19, Andre  wrote:

> Hi, re. capacitors it might be worth mentioning that the normal equation
> assumes charge and discharge through a constant current.
>

What 'normal equation' do you mean?


> Don't forget that the equation includes a non linear term so you'll need
> to take that into account (Q=CV2 iirc) where Q is Coulombs, C is
> capacitance.
>

I am puzzled by CV2.  The energy (joules) stored in a capacitor is 1/2 C
V^2, where C is the capacitance and V the voltage. I don't know if that's
what you mean.


> If this is done using something like an LM317T in CC mode or even a string
> of them (my idea) with anti-overload circuitry added externally then this
> may well work.
> Any series resistance will cause problems so you'd need quite a lot of
> regulators but there are ways to use JFETs selected by hand if you really
> wanted to
> make a test setup.
>

I am totally lost here!

If anyone has an explanation of whether any of the energy to move the leaf
comes from the battery/capacitor, or does it all come from the charge
applied to the unit, I would like to know.  If no energy (apart from
leakage) comes from the device applying the electric field, a small
capacitor is suitable, and very safe. If at least some of the energy
required to move the leaf comes from the voltage supplying the electric
field, then a small value capacitor will be no use.

This is a fairly low priority task for me at the minute, as I need to do
some real work until Friday evening. But over the weekend I will play with
this.

Dave
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