Comparison of axial field solves for Reduced Problem
INPUTS
Gas is argon with L&K cross sections
R=1 cm, height h = 20 cm
Gas temperature = 273K
Gas pressure = 2.83 mTorr
Iin = Input axial current = 15.2 mA
- Old axial field solve
In the previous axial field solve, the electron mobility,
µ = (e/m)/(ngas< v sigmaT(v)>),
(the average "< >" is taken over all electrons)
The axial field,
Ez = Iin h/(e µ Ne),
(Ne is the number of electrons)
XPDC1 OUTPUT:
Ne = 3.56e10
µ = 7404 m2/V-s
Ez = 72 V/m
However, there was a discrepancy between Iin = 15.2 mA,
the input axial current
used to determine Ez from µ, and the actual axial
current Iz found from integrating
the diagnostic jz(r).
Iz = 2 pi Int[ jz(r) r dr ] = 19.8 mA
Iz/Iin = 1.3
- New Axial Field Solve
In the new axial field solve, the mobility is given by,
µ = -4 pi e/(3 m ne ngas) Int[ (v³/(v sigmaT(v)) (dfe/dv) dv ],
(Eq. B.23 in the Lieberman and Lichtenberg text,
"Principles of Plasma Discharges and Materials Processing".)
Here, ne is the electron density, and fe(v) is the electron
velocity distribution function:
ne = 4 pi Int[ fe(v) v² dv ].
As before, The axial field, Ez = Iin h/(e µ Ne),
(Ne is the number of electrons)
XPDC1 OUTPUT:
Ne = 4.11e10
µ = 6503 m2/V-s
Ez = 71 V/m
The discrepancy between Iin = 15.2 mA and Iz
is even greater for the new method:
Iz = 2 pi Int[ jz(r) r dr ] = 22.2 mA
Iz/Iin = 1.46
Note that for both methods Ez is still around 72 V/m,
Eznew ~ Ezold
µold =7404 m²/V-s
µnew = 6503 m²/V-s
µold/µnew = 1.14
Neold = 3.56e10
Nenew = 4.11e10
Nenew/Neold = 1.15
For both methods Ez turns out to be about the same.
The smaller µ in the new method
is offset by the larger Ne in the new method.
Note that the discrepancy between the input axial
current Iin and the axial current Iz
calculated from integrating jz(r) increases in the new method.
(Iz/Iin)old = 1.3
(Iz/Iin)new = 1.46
Ideally, the axial field solve should generate an axial field which
produces the desired axial input current. In this respect, the
old method is superior.