Transcript KAIST EQM - UCLA Henry Samueli School of Engineering

A new theory of gas discharges
based on experiments
Francis F. Chen, University of California
KAIST, Daejeon, S. Korea, April 2011
Two problems with gas discharges
1.
Anomalous skin depth in ICPs
2.
Electron diffusion across magnetic fields
12
10
10
-3
n (10 cm )
3 mTorr, 1.9 MHz
Prf(W)
8
800
240
200
6
4
2
0
-5
0
5
10
15
r (cm)
Problem 1: Density does not peak near the antenna (B = 0)
20
Problem 1: anomalous skin effect
3
n (10
11
-3
cm )
2
KTe (eV)
1
RF Bz field
skin depth
0
0
Data by John Evans
5
r (cm)
10
15
In the plane of the antenna, the density peaks well outside the
classical skin layer
First solution to skin depth problem
Consider the nonlinear effect of the Lorentz force
on the motion of an electron in an RF field Eq

dv
m
 e E  v  B
dt

FL
This force is in the radial direction, while E is in the q direction
An electron trajectory over four RF cycles
with and without the Lorentz force FL
without FL
RF phase
(degrees)
0
180
360
540
720
900
1080
1260
Skin depth
with FL
After 32,000 iterations
UCLA
Density profile in four sectors of equal area
1.2
Relative density
1.0
0.8
0.6
0.4
0.2
Points are data from Slide 5
0.0
0
5
r (cm)
10
15
Problem 2: Diffusion across B
Classical diffusion predicts slow electron diffusion across B
B
B
i
n
i
n
e
D 
D
1  (c2
/ )
2
e
ci  ce
Hence, one would expect the plasma to be
negative at the center relative to the edge.
Density profiles are never hollow
B
r
n
If ionization is near the boundary, the density
should peak at the edge. This is never observed.
Helicons are always more positive at center
1.2
24
n/n0
Vs
1.0
-3
cm )
Density (10
11
0.8
0.6
16
0.4
12
0.2
0.0
8
-15
-10
-5
0
r (cm)
5
10
15
Plasma potential (volts)
20
Resolution: discharges have finite length
The Simon short-circuit effect
e
+
+
SHEATH
APPARENT
ELECTRON FLOW
HIGH DENSITY
LOWER DENSITY
e
B
Electrons are Maxwellian along each field line, but not across lines.
A small adjustment of the sheath drop allows electrons to “cross the field”.
This results in a Maxwellian even ACROSS field lines.
The E-field moves ions to the center
e
+
+
SHEATH
e
+
-
E
ION DIFFUSION
HIGH DENSITY
LOWER
DENSITY
B
This is why density is not peaked at the edge
Hence, the Boltzmann relation holds even across B
n  n0ee / KTe  n0e 
Er
 ( KTe / en)(dn / dr )
As long as the electrons have a mechanism that
allows them to reach their most probable
distribution, they will be Maxwellian everywhere.
This is our basic assumption.
This radial electric field pushes ions from high density to low density,
filling in hollow profiles so that, in equilibrium, the density is always
peaked on axis.
An idealized model of a high-density discharge
rLi >> a
a
+
B
-
rLe << a
L
1.
B = 0 or B  0; it doesn’t matter.
2.
The discharge is cylindrical, with endplates.
3.
All quantities are uniform in z and θ (a 1-D problem).
4.
The ions are unmagnetized, with large Larmor radii.
5.
Ion temperature is negligible: Ti / Te = 0
The ion equations
Motion:
Mnv v  en(E  v  B )  KTi n  Mnv io
E  ,   e / KTe , and cs  (KTe / M )½
1-D:
Continuity:
vr
dv r
d
 cs2
 io v r  io  nn   v cx  nn Pc (r )
dr
dr
 (nv)  Q(r )  nnn Pi (r)
Pi (r )   v ion (r )
dv r
d (ln n) v r
 vr

 nn Pi (r )
dr
dr
r
Pi is the ionization
probability
Pc is the collision
probability
We keep the two highlighted equations
We now have three equations
dv r
d
 cs2
 io v r
dr
dr
Ion equation of
motion:
vr
Ion equation of
continuity:
dv r
d (ln n) v r
 vr

 nn Pi (r )
dr
dr
r
Electron
Boltzmann relation:
(which comes from)
Er
 (KTe / en)(dn / dr)
n  n0ee / KTe  n0e
3 equations for 3 unknowns:
vr(r) (r) n(r)
Eliminate unknowns (r) and n(r)
This yields an ODE for the ion radial fluid velocity:

cs2  v v 2
dv
 2 2   2 nn Pc (r )  nn Pi (r )   0 (v  vr )
dr cs  v  r cs

Note that dv/dr   at v = cs (the Bohm condition,
giving an automatic match to the sheath
We next define dimensionless variables
u  v / cs
to obtain…
k  Pc / Pi   v cx /   v ion
Reminder: Bohm sheath criterion
ne = ni = n
ns
ni
n
PRESHEATH
ne
v = cs
PLASMA
+
SHEATH
xs
x
The universal equation
du
1  nn Pi
u
2

1  ku    0
2 
dr 1  u  cs
r

Note that the coefficient of
(1 + ku2) has the
dimensions of 1/r, so we
can define
This yields

  (nn Pi / cs )r
du
1

d  1 u2

u
2
1

ku






Except for the nonlinear term ku2, this is a universal equation giving
the n(r), Te(r), and (r) profiles for any discharge and satisfies the
Bohm condition at the sheath edge automatically.
1.0
1.0
0.8
0.8
a
0.6
a
a
0.4
n / n0
V / Cs
Solutions for different values of k = Pc / Pi
0.6
p (mTorr)
1
10
100
0.4
0.2
0.2
0.0
0.0
0.5
1.0

1.5
2.0
0.0
0.0
0.2
0.4
0.6
r/a
We renormalize the curves, setting a in each case
to r/a, where a is the discharge radius.
Once v is known, all the other profiles can be
calculated from the three equations.
We then find that there’s a universal profile, the
same for all discharges!
0.8
1.0
A universal profile for constant k
1.0
0.0
-0.2
n/n0
0.8
-0.4
0.6
-0.6
0.4
-0.8
v/cs
eV/KTe
0.2
-1.0
0.0
-1.2
0.0
0.2
0.4
0.6
0.8
1.0
r/a
This is independent of magnetic field!
A program EQM was devised by Curreli
The universal profile is good for k = constant, where k is the ratio Pc/Pi.
However, Pc depends on v(r) and Pi depends on Te. But we know v(r) from
our solution, so we can calculate Pc(r) and use it. Te, however, depends
on ionization balance and will be treated next.
Ph.D. student Davide Curreli from the University of Padua, Italy,
wrote a program to solve all the equations with radial profiles.
1
These are density profiles
for different values of
constant Te, but with selfconsistent Pc.
n / n0
0.8
KTe (eV)
0.6
2
3
4
0.4
0.2
0
0
0.2
0.4
0.6
r/a
0.8
1
It is no longer necessary
to calculate a collisional
presheath and match it to
the collisionless sheath.
It’s all included.
Ionization balance
Consider the number of ions in a cylindrical shell dr of unit length
dnT (r )
 2 rdr  n(r )nn (r )  Pi (Te ) input
dt

dr
output
dnT (r )
1 d
 2 rdr  [n(r )v(r )]  2 rdr 
 rn(r )v(r )]
dt
r dr
Result:
1 d
 rnv   nn Pi (Te )
nr dr
Previous equation (u = v/cs)
du
1

dr 1  u 2
 u nn 2

u Pc  Pi   0
 
 r cs



Assume nn (or pressure po) to be uniform, and solve these equations
simultaneously.
Ionization balance result
For given tube radius a, there is only one Te that puts the sheath edge there.
1.0
10
Tube radius (cm)
8
KTe (eV)
V / Cs
0.8
0.6
0.4
KT e (eV)
3.00
3.49
4.00
0.2
2.5
5
10
6
4
2
0
0.0
0
1
2
3
r [cm]
4
5
1
10
p0 (mTorr)
100
This gives the familiar result that Te varies inversely with po,
But more accurately, since the profiles of v(r) and n(r) are taken into account.
The p-Te relation depends on the tube radius a.
Neutral depletion
What about nn(r)? This is more difficult, since the neutrals’
motion depends on the exact geometry of the discharge.
D2nn  nnnPi
Eqn. of continuity
for neutrals
We have to assume that the mfp’s are short, and the neutrals are
collision-dominated, with a diffusion constant D. The come in and
out uniformly at the surface. Inside the plasma they are ionized at a
rate proportional to electron density n and ionization rate Pi(r).
Results for neutral depletion
1.2
45
1.2
Density (10
40
KT e (eV)
p0 (mTorr)
1
5
10
0.4
cm )
1.0
30
0.8
25
0.6
20
15
p (mTorr)
n (10 12 cm -3 ) & p / p 0
35
0.6
-3
1.0
2.5
5.0
1
0.8
12
0.4
10
0.2
0.2
5
0
0
0
0.5
1
1.5
2
2.5
r (cm)
Neutral density profiles for various
pressures in a 5-cm diam tube.
The ne profiles (----) are peaked at
1012 cm-3.
0.0
0.0
0.5
1.0
1.5
2.0
2.5
r (cm)
Same profiles as peak density is
varied. Input pressure is 1 mTorr.
The corresponding Te profiles (----)
are shown, but these are not realistic
because radiation losses have not
yet been accounted for.
Energy balance: helicon discharges
To implement energy balance requires specifying the type of
discharge. The HELIC program for helicons and ICPs can calculate
the power deposition Pin(r) for given n(r), Te(r) and nn(r) for various
discharge lengths, antenna types, and gases. However, B(z) and
n(z) must be uniform. The power lost is given by
Pout  Wi  We  Wr
Energy balance: the Vahedi curve
The particle losses Wi and We are simple. The radiation loss Wr,
however, has to be calculated from the Vahedi curve, known for argon.
This gives the energy needed to create each ionization, including all the
radiation that is lost before that happens.
1000
E c (eV)
1.61
Ec » 23exp(3.68/ TeV
)
100
10
1
2
KTe (eV)
5
10
Energy balance gives us the data to calculate Te(r)
Iteration between EQM and HELIC
The final step is the tedious process of iterating between EQM and HELIC.
Starting with constant Te, EQM calculates n(r) and nn(r). These are put into
HELIC to obtain Te(r). This is then put into EQM to get new n(r) and nn(r).
The process is repeated until it converges.
The right Te(r) is extremely important, since ionization varies
exponentially with Te.
Here are three cases, before iteration
1.2
1200
Case 1
Case 2
Case 3
1000
1.0
0.8
n/n 0
2
P r ( /m )
800
600
0.6
400
0.4
200
0.2
0
0.0
0
0.5
1
1.5
r (cm)
2
2.5
Case1
Case 2
Case 3
0
0.5
1
1.5
r (cm)
2
2.5
Two cases with iteration
8
KT e (eV)
-3
cm )
11
3
2
2
2
14.6
2
1
1
0
1.0
r (cm)
1.5
2.0
2.5
-3
12
11
n
Pr
2.5
6
10
5
27.12 MHz
120G, 1000W
15.0
8
4
Te (eV)
p (mTorr)
14.8
15.2
3
14.6
2
14.4
2
1
14.2
0
0
6
4
0
0.5
2.0
12
4
0.0
1.5
2
8
1.0
1.0
1.5
r (cm)
2.0
2.5
14.0
0.0
0.5
1.0
1.5
r (cm)
2.0
2.5
p (mTorr)
27.12 MHz
120G, 1000W
0.5
r (cm)
Pr (k/m )
cm )
16
14.4
0.0
KT e (eV)
0.5
20
n (10
0
0
0.0
14.8
3
p (mTorr)
4
15.0
KTe (eV)
p (mTorr)
4
Pr (K/m )
4
n
Pr
15.2
13.56 MHz
65G, 400W
5
5
13.56 MHz
65G, 400W
6
n (10
6
6
One case of absolute agreement
12
cm -3 )
Measured
Calc. L=20
Calc. L=25
Calc. L=30
10
8
11
3
Density (10
P(z) (arb. units)
4
2
1
6
4
2
0
0
0
5
10
15
z (cm)
20
25
30
0
100
200
300
400
RF power (watts)
The unfinished task is to predict the Big Blue Mode, which has
complete ionization on axis. But that involves higher ionization states.
Title here
Title here