Will voltage breakdown occur in
the Sem Fusor?
Paschen's law is
an equation that gives the breakdown
voltage,
that is, the voltage necessary
to start a discharge or electric
arc,
between two electrodes in a gas as a function of pressure and gap
length.
See:
wiki/Paschen law
Suppose we made a good vacuum (1.10^{6} Torr)
and filled the vacuum chamber again with deuterium gas up to a pressure
of 30 microns (3.10^{2} Torr).
Droom11.16
Suppose the distance between two rings/spheres is 30 cm, and the voltage
difference is 200 kV.
We take H2 gas, 30 cm x
3.10^{2} Torr
=0.9 cm Torr
According to figure 1 the breakdown voltage is a lot higher than
10^{4
}
Volt (curve is rapidly increasing but ends..)
200 kV = 2.10^{5
}
V
If
you interpolate the curve for H2, you will come close to this
2.10^{5
}
V
So I think that in the Sem Fusor one must be aware of the possibility of
voltage breakdown.
According to figure 3, the voltage breakdown for D2 gas is a bit higher
than for H2 gas.
In our Sem Fusor we use D2 gas, so this is favourable.
But one possible configuration of the Sem Fusor is only with two
negatively charged rings:
Droom11.21
Then we do not have a voltage difference between rings...
Fig 1.
Paschen curves obtained for various gasses.
(According to
wiki/Paschen_law)
Fig. 2. (https://commons.wikimedia.org/wiki/File:Paschen_Curves.PNG)
Fig. 3.(
https://www.researchgate.net/figure/PaschencurveDirectcurrentbreakdowntensionVbatRTofseveralgasesversus_fig2_344464964
)
