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Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
Lecture 10 Capacitively Coupled Plasmas
New homework problems:
Lieberman 12.2 and 12.3 Due April 25th, 2001. – Last Set!
dB/dt
Wire/coil
Iwire
Dielectric plate
Ipla sma
PLASMA
Basic setup.
How do we find out how much current is being pushed about in the plasma? Well the place to
start is Maxwell’s equations.
∇ ∧ E = −∂ t B
∇ ∧ H = J free + ∂ t D
∇ ∑D = ρ free
∇ ∑B = 0
∇ ∑J = −∂ t ρ
If we take the curl of the first (second) equation we find
∇ ∧ [∇ ∧ E = −∂ t B]
∇ ∧ ∇ ∧ H = J free + ∂ t D
[
∇ ∧ ( ∇ ∧ E ) = −∂ t ∇ ∧ B
∇ ∧ (∇ ∧ E) = −∂ t µ∇ ∧ H
(
∇ ∧ (∇ ∧ E) = −∂ t µ J free + ∂ t D
]
∇ ∧ (∇ ∧ H) = ∇ ∧ J free + ∂ t ∇ ∧ D
)
∇ ∧ (∇ ∧ H) = ∇ ∧ σE + ∂ t ∇ ∧ εE
∇ ∧ (∇ ∧ H) = ∇ ∧ σE + ε∂ t ( −∂t B)
∇ ∧ (∇ ∧ E) = − µ∂ t J free − µ∂ t2 D
∇ ∧ (∇ ∧ H) = σ ( −∂t B) − ε∂t2 B
∇ ∧ (∇ ∧ E) = − µ∂ tσE − µ∂ t2εE
∇ ∧ (∇ ∧ H) = −σµ∂t H − εµ∂t2 H
∇ ∧ (∇ ∧ E) = − µσ∂ t E − µε∂ t2 E
∇ ∧ (∇ ∧ H) = −σµ∂t H − εµ∂t2 H
∇(∇ ∑E) − ∇ 2 E = − µσ∂ t E − µε∂ t2 E
∇(∇ ∑H) − ∇ 2 H = −σµ∂t H − εµ∂t2 H
(
)


∇ ρ free ε − ∇ 2 E = − µσ∂ t E − µε∂ t2 E
∇ µ ∇
∑B − ∇ 2 H = −σµ∂t H − εµ∂t2 H
14243
{
 =0 
assume = 0
∇ 2 E = µσ∂ t E + µε∂ t2 E
∇ 2 H = σµ∂t H + εµ∂t2 H
This leads directly to the general electromagnetic wave equation
Page 1
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
 E
 E
 E
∇ 2   = σµ∂ t   + εµ∂ t2  
 H
 H
 H
E =±
iµω
iµω 
H
; η=
=
=
η
γ
σ + iεω 

µ
if σ = 0
ε

sign det er min ed by growth / decay
( growth => −, decay => + )
β=
2π v ph
1
; v ph =
=
λ
ω
εµ
This leaves us with a need to know what ε and σ are…
First, we know that the plasma is driven with an rf electric field. We can model this field as a
sinusoidal variation,
ƒ iωt .
E = Re Ee
This field will accelerate the electrons
dv
m = qE − mν m v
dt
where ν m is the electron-neutral collision frequency. Hence the last term is simply the resistive
drag term that we need to have to transfer the power from the electrons to the neutrals and the
ions. Now assuming that the electron velocity is also sinusoidal, e.g. they follow the electric
field.
ƒ iωt
v = Re ve
Then we find that
ƒ iωt − mν Re ve
ƒ iωt = q Re Ee
ƒ iωt
miω Re ve
m
⇓
q
1
Eƒ
m (iω + ν m )
This is of course related to the free current density
2
1
ƒj = qnvƒ = nq
Eƒ
free
m (iω + ν m )
vƒ =
= ε 0ω 2ps
1
Eƒ
(iω + ν m )
= σ p Eƒ
Where σ p = ε 0ω 2ps
1
is the plasma conductivity.
(iω + ν m )
Likewise, we have the displacement current density
ƒj
ƒ
ƒ iωt
displace = ε 0 ∂ t E = iε 0ωEe .
Thus the total current density is simply
Page 2
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
∇ ∧ Hƒ = ƒjtotal = ƒj free + ƒjdisplace
 nq 2

1
+ iε 0ω  Eƒ
=
 m (iω + ν m )



ω 2ps
= iωε 0 1 − 2
 Eƒ
i
−
ω
ων
(
)

m 


ε 0ω 2ps  ƒ
= iωε 0 +
E
i
ω
ν
+
(
)
m


= iωε + σ Eƒ
[
0
p
− or −
]
= iωε p Eƒ = iωε 0κ p Eƒ
where ε p is know as the plasma dielectric constant.
Now let us go back to the wave equation
∇ 2 E = µσ∂ t E + µε∂ t2 E
ƒeiωt so that
Further, we will use E = Re E
∇ 2 E = iωµσE − ω 2 µεE
123 123
free current
disp current
= iωµ0σ p E − ω 2 µ0ε 0 E
(
)
= iωµ0 σ p + iωε 0 E
= −ω 2 µ0ε p E
= −ω 2 µ0ε 0κ p E
−ω 2
κ pE
c2
Now we can assume a one-dimensional problem. So that
∇2 E = ∂ z E
=
−ω 2
κ pE
c2
= α 2E
=
⇓
E = ( E+ e +αz + E− e −αz )
Because we cannot physically have the electric field grow as it goes to infinity than the first term
must be zero and
ω
1
α = i κ 1p/ 2 =
c
δ
Page 3
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
where δ is the skin depth. Further noting that α must be real (as well as ω and c!), we find
ω
α = Im κ 1p/ 2
c
So… what is κ 1p/ 2 ? From above
(
iωε 0κ p = iωε 0 + σ p
)


i
κ p = 1 −
σ p
 ωε 0 


1
i
= 1 −
ε 0ω 2ps
(iω + ν m ) 
 ωε 0
 ω 2ps

1
= 1 − 2
ω (1 − i ν m ω ) 

ω 2ps
1
2
ω (1 − i ν m ω )
where we have made the assumption that ω 2ps >> ω 2 . This assumption is reasonable as
frequencies above the plasma frequency will be cut off. Further the plasma frequency is
typically several GHz compared to our typical 13.56 MHz driving frequency. Thus
ω
1
κ 1p/ 2 ≈ i ps
1/ 2
ω (1 − i ν m ω )
Now we can examine the different cases
≈−
Collisionless case: ν m ω << 1
κ 1p/ 2 ≈ i
≈i
ω ps
1
1/ 2
ω (1 − i ν m ω )
ω ps
ω
⇓
1
δ
ω
= Im κ 1p/ 2
c
ω ω ps ω ps
=
=
c ω
c
α=
⇓
δ=
c
ω ps
Page 4
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
Collisional case: ν m ω >> 1
ω
1
κ 1p/ 2 ≈ i ps
1/ 2
ω (1 − i ν m ω )
≈i
ω ps
1
1/ 2
ω ( −i ν m ω )
=i
ω ps
1
1/ 2 3 / 2 1/ 2
ω i νm
=
ω
i ν ω
ps
1/ 2 1/ 2 1/ 2
m
⇓
1
δ
ω
= Im κ 1p/ 2
c
α=
=
ω

ω 
Im 1/ 2 1/ps2 1/ 2 
c  i νm ω 
=
ω 1/ 2ω ps  1 
Im 1/ 2
i 
ν 1m/ 2 c
⇓
δ=
ν 1m/ 2 c
Im(i1/ 2 )
1/ 2
ω ω ps
Now what is Im(i1/ 2 ) ?
[
Im(i1/ 2 ) = Im (eiπ / 2 )
[
= Im eiπ / 4
1/ 2
]
]
= Im[cos( π / 4) + i sin( π / 4)]
1 
 1
= Im 
+i
2 
 2
1
=
2
so
Page 5
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
ω 1/ 2ω ps
α = 1/ 2
νm c 2
and
δ=
ν 1m/ 2 c
ω 1/ 2ω ps 2
So now we need to ask how much power is being deposited into the discharge?
The power deposited is
1
Pave = ∫∫∫ J ∑E * dτ
2 Vol
For a planar system, this is very complex. If on the other hand, we where to assume that we have
a solenoidal coil, we end up with a simpler system. First, the both the current and electric field
are approximately azimuthal. Second, both the current and the electric field are approximately
constant. Thus,
1
Pave = ∫∫∫ J ∑E * dτ
2 Vol
1
≈ J ∑E * ∫∫∫ dτ
Vol
2
1
= Jφ ∑Eφ * ∫∫∫ dτ
Vol
2
R
1
l
≈ Jφ ∑Eφ * π(r 2 ) ( z ) 0
R −δ
2
1
2
= Jφ ∑Eφ * π R2 − ( R − δ ) l
2
1
= Jφ ∑Eφ * π(2 Rδ − δ 2 )l
2
1
≈ Jφ ∑Eφ * 2 πRδl
2
1
Now we can use the plasma conductivity, σ p = ε 0ω 2ps
, to get
(iω + ν m )
2 πRδl 2
Pave =
Jφ .
2σ p
For high pressure discharge ω << ν m , so that
1
σ p ≈ ε 0ω 2ps
νm
(
=
and
Pave =
)
e2 n
mν m
mν m πRδl 2
Jφ
e2 n
Page 6
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
Further noting that
Iφ = Jφ lδ
so that
mν πR
Pave = 2 m Iφ2
e nlδ
1
= Rp Iφ2
2
so
2 mν πR
Rp = 2 m
e nlδ
is the resistance. The other part of the picture is the inductance. By definition the induction is
the total magnetic flux Φ contained by the structure divided by the current I.
Lp = Φ / I
For a single turn of the coil, the flux is
Φ = πR2 Bz
= µ 0 πR2 Hz
further the magnetic field produced by the induced surface current is simply
Hz = Jφδ
If we assume that the source is a long solinoid then we can determine the magnetic field.
L
D
By symmetry in z and φ, the magnetic field is only dependent on the radial position.
B = B(r ) Further
B = Bz√
Then by Maxwell’s equations
µ0−1 ∫ B ∑dl = ∫∫ J ∑ds
With the exception of the curve shown above, the current through a given surface is zero. (Ok
you can pick some ‘fun’ surfaces that YOU can work with…) By letting D -> 0, with the coil
still passing through surface, we find
Page 7
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
µ0−1 ∫ B ∑dl = In′L
µ0−1 Bz L =
⇓
µ0 In′ inside
Bz = 
outside
 0
n
where ′ is the number of turns per length L.
The flux of the magnetic field from the coil (a) through a surface (b) is
Φ ab = ∫∫ Ba ∑ds b
Further the induction is given by
NΦ ab
Lab =
I
(Often L is used for self induction and M is used for mutual induction. Note that the mutual
induction is the such that Mab = Mba. This can be shown from
Φ ab = ∫∫ Ba ∑ds b
= ∫∫ (∇ ∧ A a ) ∑ds b
= ∫ A a ∑dl b
b
)
 µ 0 Ia 1 
dl a ∑dl b
= ∫
4 π ∫a r 
b
=
µ 0 Ia
4π
1
∫ ∫ r dl
a
∑dl b
b a
= Mab Ia
Now let us assume that the plasma is a single turn coil of radius R inside the power coil of radius
b. Thus,
area
ratio
}
πR2
Φ rf p = Φ rf rf
πb 2
µ NI
= 0 rf πR2
L
and
Page 8
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
Lrf
p
Lrf
rf
Lp
p
NΦ rf p
I
µ0 N 2 2
=
πR
L
NΦ rf rf
=
Irf
=
µ0 N 2 2
=
πb
L
=1
}
NΦp p
=
Ip
=
µ0 2
πR
L
Now, we need to consider the circuit, which looks like
Lrf
Lp
V rf
Vrf = iωLrf
Vp
Rp
I + iωLrf p I p
rf rf
Vp = Rp I p = iωLrf p Irf + iωLp p I p
Now the impedance of the source is
Page 9
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
Vrf
= Zrf
Irf
= iωLrf
rf
−
(R
ω 2 L2rf
p
p
− iωLp
p
)
µ02 N 2 2 4
π R
2
µ0 N
2
L
but Rp << ωLp
= iω
πb −
L
 2 mν m πR − iω µ0 πR2 
 e 2 nlδ

L
2
ω2
p
µ0 N 2 2 ωµ0 N 2 πR2
πb − i
L
L
2
µ N
= iω 0 π(b 2 − R2 )
L
= iωLs
or to get the real part
2 2
2 µ0 N
ω
π 2 R4
2
2
µ0 N
 2 mν m πR + iω µ0 πR2 
2
L
Zrf = iω
πb −
2
2 

L
L
  2 mν m πR   µ0 2   e 2 nLδ
R
ω
+
π
 2
 
 e nLδ   L
≈ iω
µ0 N 2 2 ωµ0 N 2 πR2 2 mν m µ0 N 2 πR
≈ iω
πb − i
−
L
L
e 2 nδL
= iωLs + Rs
Page 10