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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 11.1 11.2 and 11.3 Due April 16th, 2001. – Note change!
Sheath a,
Sa
Electrode
a
Sheath b,
Sb
Electrode
b
L
PLASMA
Area a,
Aa
Area b,
Ab
Basic setup.
First, we will start with the simplest model of a CCP system. To do this requires a few
simplifying assumptions.
1) The frequency as which the electric field is applied is such that
 ne 2 
ω 2pi << ω 2 << ω 2pe  =

 ε0m 
ω 2 ≥ ν m2
Effectively this implies that the ions don’t respond to the instantaneous electric field but
the electrons do respond. Further, the electrons don’t collide very often in a cycle of the
applied field. Thus the electrons can carry the power put into the plasma from the E.
2) The areas of the electrodes are equal, Aa = Ab, and large enough so that the E and J are
only directed normal to the electrodes. This implies no transverse fields, which is strictly
not correct but in some systems a reasonable approximation. When taken into account
with (1) this implies that the total current, free charge current and displacement current is
constant across the discharge.
3) The electron density is zero in the sheath – implying that the time-average sheath width,
S , is large compared to the Debye length. Further that Child’s Law is a good
approximation for the sheaths on each side.
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
4) The sheaths on both sides expand and contract with the applied field but the sum of the
sheath widths is a constant. 2 S = Sa (t ) + Sb (t ) = const
5) Finally, we will assume that the ion density is uniform across the discharge, including the
sheath and the electrons are uniform except for the sheaths.
Now, we can start looking at the total current that flows through the plasma. At sheath a (this
also applies to sheath b) we find
e
∂ x E = ni x ≤ Sa (t )
ε0
e
∂ x E = − ni x ≥ Sb (t )
ε0
We can integrate and use the boundary condition E( S, t ) ≈ 0 to show that
e
E( x, t ) = ni ( x − Sa ) x ≤ Sa (t )
ε0
e
E( x, t ) = ni ( Sb − x ) x ≥ Sb (t )
ε0
Now, we can get the displacement current in the sheaths.
Idisplace (t ) = Aa / b jdisplace (t ) = ε 0 Aa / b ∂t E(t )
⇓
Idisplace (t ) = eAa ni ∂t ( x − Sa )
a
x ≤ Sa (t )
Idisplace (t ) = eAb ni i ∂t ( Sb − x ) x ≥ Sb (t )
b
Now at the sheath boundary, the current changes from displacement current to free current in
terms of electron motion. We already know that the free current is the form of
I free (t ) = Aa / b j free
ƒ iωt
= −eAa / b ne Re ve
= −eAa / b ne v0 cos(ωt )
Thus,
Idisplace (t ) = I free (t )
a
a
= −eAa ne v0 cos(ωt )
= eAa ni ∂t ( x − Sa )
x ≤ Sa (t )
⇓
}
n 
∂t Sa = − e  v0 cos(ωt )
 ni 
≈1
⇓
Sa = Sa − Sa 0 sin(ωt )
and
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
Idisplace (t ) = I free (t )
b
b
= −eAb ne v0 cos(ωt )
= eAb ni ∂t ( Sb − x )
x ≥ Sb (t )
⇓
≈1
}
n 
∂t Sb = − e  v0 cos(ωt )
 ni 
⇓
Sb = Sb + Sb 0 sin(ωt )
v
where Sa / b 0 = 0 and the phase difference between sheath edges. Further we note that
ω
Sa + Sb = const . Now, we can get the sheath potential directly from the electric field.
Sa
V (t ) a = ∫ E( x, t )dx
0
Sa
e
1
= ni  x 2 − Sa x + const 

ε0  2
0
e
ni Sa2
2ε 0
where const = 0 is determined by making V ( x = 0, t ) = 0 . Likewise,
=−
l
V (t ) b = ∫ E( x, t )dx
Sb
l
e
1
= − ni  x 2 − Sb x + const 

ε0  2
S
b
e
ni Sb2
2ε 0
1
where const = − l 2 + Sbl from V ( x = l, t ) = 0 . Both sheath voltages are nonlinear in time, as
2
can be easily determined from the above equations.
e
V (t ) a = −
ni Sa2
2ε 0
=−
=−
2
e
ni ( Sa − Sa 0 sin(ωt ))
2ε 0
=−
e
ni Sa
2ε 0
(
2
)
+ Sa20 sin 2 (ωt ) − 2 Sa Sa 0 sin(ωt )
likewise
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
V (t ) b = −
=−
e
ni Sb2
2ε 0
2
e
ni ( Sb + Sb 0 sin(ωt ))
2ε 0
(
)
e
2
ni Sb + Sb20 sin 2 (ωt ) + 2 Sb Sb 0 sin(ωt )
2ε 0
Like the other electrode, the bias across the sheath is non-linear with the applied frequency.
Now the voltage between the electrodes is given by
V (t ) plasma = V (t ) a − V (t ) b 2
=−
(
=−
e
ni Sb
2ε 0
+
e
ni Sa
2ε 0
(
)
2
+ Sb20 sin 2 (ωt ) + 2 Sb Sb 0 sin(ωt )
2
+ Sa20 sin 2 (ωt ) − 2 Sa Sa 0 sin(ωt )
)
2e
ni Sb Sb 0 sin(ωt )
ε0
which is linear in the voltage response.
=−
Now let us consider a system in which the areas of the two electrodes are unequal.
1) For the purposes of our example we will assume that Aa > Ab .
2) As before, we will assume that the density of the ions is constant across the plasma.
(Diffusion of the ions will insure that this is not far form the truth.) Because of this we
find that the current densities at each electrode are the same. We will further assume that
the current density is constant across the face of each of the electrodes. Thus we find
that ja ≈ jb .
3) We will assume that the electrons are excluded from the sheath regions. This implies that
we can assume that the sheaths can be adequately modeled with Child’s Law. Thus,
1/ 2
1/ 4
3 j
M
(V )3 / 4 =  i   i  S
2 ε
2e
1/ 4
3  M j2 
=  i i2  S
2  2eε 
We can rearrange this to show that
4 2
e1/ 2ε
ji =
(V )3 / 2 1 / 2 2
9
Mi S
(Note our sign change on the potential.)
4) Each sheath region has a capacitance associated with it that is proportional to the
A
electrode area divided by the sheath width. C ∝
S
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
5) The rf voltage is capacitively divided between the two sheaths. Thus
Va Cb
= .
Vb Ca
This make sense as the charge, CV, on the ‘sheath/capacitor’ is the same on each sheath.
Combining the last two assumptions we find that
Va Ab Sa
=
Vb Sb Aa
Further using 2 and 3 we find,
1/ 2
4 2
3/ 2 e ε
ja =
Va )
(
9
Mi1/ 2 Sa2
≈ jb
=
1/ 2
4 2
3/ 2 e ε
Vb )
(
9
Mi1/ 2 Sb2
⇓
1
1
a
b
(Va )3 / 2 S 2 = (Vb )3 / 2 S 2
Now we can combine all of the pieces to get
4
Vb  Aa 
= 
Va  Ab 
This means that the smaller electrodes result in larger voltages. This effect is very strong – 4th
power! Further a substantial voltage can be generated on even grounded surfaces! (This
potential is of course relative to the plasma potential.)
Plasma Heating
Now we want to know how power is deposited into the plasma. Clearly, we are using the
electric field to accelerate and decelerate the electrons. However, to ‘permanently’ deposit the
energy, we need some mechanism with which to extract the energy from the electrons, so that
they do not give the energy back to the electric field. The simplest mechanism is via ionizing
collisions with the neutral gas. Such collisions are reminiscent of resistance to current flow in a
wire. This resistance leads naturally to Ohmic heating of the wire. There are two known method
from which the electrons gain energy, the first is the general sloshing of the electrons back and
forth between the electrodes. This is in effect a bulk motion of the electrons in response to the
push and pull of the electric fields. The second method of heating the electrons is known as
‘stochastic heating.’ Stochastic heating is not unlike hitting a ping pong ball with a paddle.
Initially, the paddle is moving at a much higher velocity than the ball but after the collision
occurs the ball’s velocity greatly exceeds that of the paddle. We will examine Ohmic heating
initially.
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
Ohmic heating
It can be shown that the time-averaged power obtained via Ohmic heating can be described as
1 T
pave = ∫ J total (t ) ∑E(t )dt
T 0
1
= Re J total (t ) ∑E* (t )
2
To get at this, we need to know the electric field and the current density.
[
]
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 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
∇ ∧ 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
− or −
p
]
= iωε p Eƒ = iωε 0κ p Eƒ
where ε p is know as the plasma dielectric constant.
Now, we can determine the power deposited into the plasma.
1 T
pave = ∫ J total (t ) ∑E(t )dt
T 0
1
= Re J total (t ) ∑E* (t )
2
[
]

 2
ω 2ps
1 
= Re iωε 0 1 − 2
 Eƒ 
2 
ω
−
ων
i
(
)

m 
 

ε 0ω 2ps  ƒ2 
1 
= Re  iωε 0 +
E 
2  
(iω + ν m )  
=
ε ω 2 (ν − iω )  ƒ2 
1 
Re  iωε 0 + 0 ps 2 m 2
E 
2  
ω
+ ν m )  
(


(
)  Eƒ 
2
ε 0 (ω 2 + ν m2 ) − 1
1   ε 0ω psν m
+ iω
= Re  2
2   (ω + ν m2 )
(ω 2 + ν m2 )

2



2
1 ε 0ω psν m 2
E
2 (ω 2 + ν m2 )
Or we can calculate this in terms of the current density rather than in terms of the electric field to
get
1 νm 2
pave =
J .
2 ε 0ω 2ps
=
Now, we know the current density field from our model. Thus we find that the average power
deposited into the plasma is
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
l
Pave = ∫ pave dx
0
1 νm 2
J dx
0 2 ε ω2
0 ps
=∫
=
l
1 νm 2 l
J dx
2 ε 0ω 2ps ∫0
1 νm 2
J l
2 ε 0ω 2ps
This of course assumes that the current density is constant across the plasma. Plugging in for the
plasma frequency gives
=
l
Pave = ∫ pave dx
0
1 νm 2
J dx
0 2 ε ω2
0 ps
=∫
=
l
1 νm 2 l
J dx
2 ε 0ω 2ps ∫0
1 mν m 2
J l
2 n0 e 2
(Note that the way we have done this could also apply to ion motion.)
=
The final thing that we need to look at is stochastic heating.
Stochastic Heating
To understand stochastic heating, we have to examine two different reference frames, the lab
frame and the stationary sheath edge frame. First, let us draw pictures of these frames.
Lab Frame
Sheath Frame
-ve
-ve + vsheath
vsheath
v’e
-(-ve + vsheath)
Assuming that the sheath has infinite ‘mass,’ than in the sheath frame, the electron must retain all
of its energy. Thus in the sheath frame the final velocity of the electron must be –(-ve + vsheath).
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
Transforming back to the lab frame we find that the final velocity of the electron must be
ve′ = − ve + 2vsheath .
This shows us that in the lab frame, the electron has more energy after colliding with the sheath.
(The lab frame is what matters as the neutral gas is assumed to be at rest in the lab frame.)
Now we need to find out how this affects the electron energy on average – if one electron is
energized while a second losses an equal amount of energy than there is no net heating of the
plasma.
To start, we first need to determine how many electrons of a velocity between ve and ve + dve
will collide with the sheath in time dt . This is best understood in the sheath frame.
Sheath Frame
D=(-ve + vsheath)dt
Then the number per unit area is simply
# D#
=
A Vol
= Df (ve , t )dve
= ( − ve + vsheath ) f (ve , t )dve dt
The change in energy of each of these electrons with a velocity between ve and ve + dve is
1
∆Energy = m(ve′ 2 − ve2 )
2
Integrating over all velocities we find the energy deposited
∞ 1
∆SStoch = ∫
m(ve′ 2 − ve2 )( − ve + vsheath ) f (ve , t )dve dt
−∞ 2
BUT THIS IS NOT QUITE CORRECT AS SOME ELECTRONS OUTRUN THE SHEATH
SO…
∆energy
∆SStoch =
A dt
v sheath 1
=∫
m(ve′ 2 − ve2 )( − ve + vsheath ) f (ve , t )dve
−∞
2
v sheath 1
2
=∫
− ve2 )( − ve + vsheath ) f (ve , t )dve
m(ve2 − 4ve vsheath + 4vsheath
−∞
2
=∫
v sheath
=∫
v sheath
−∞
−∞
2
m( − ve vsheath + vsheath
)(−ve + vsheath ) f (ve , t )dve
2 mvsheath ( − ve + vsheath ) f (ve , t )dve
2
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
While this is ‘true’ at any point in time in general, again it does not matter if we put energy in to
the electrons just to take it out in the next half cycle – so we now average over a cycle of the rf.
We do this by noting that the sheath velocity is simply
vsheath = vsh 0 cos(ωt )
plugging this into the above equation and averaging over a cycle gives
1 T
SStoch = ∫ ∆SStoch dt
T 0
1 T v sheath
2
= ∫ ∫
2 mvsh 0 cos(ωt )( − ve + vsh 0 cos(ωt )) f (ve , t )dve dt
−∞
0
T
1 T v sheath
= ∫ ∫
2 mvsh 0 (ve2 cos(ωt ) − 2ve vsh 0 cos2 (ωt ) + vsh2 0 cos3 (ωt )) f (ve , t )dve dt
−∞
0
T
= −∫
v sheath
−∞
=∫
∞
− v sheath
2 mve vsh2 0 f (ve , t )dve
2 mve vsh2 0 f (ve , t )dve
Now in general the sheath velocity is slow compared so we can let the sheath velocity go to zero.
Further we will assume that the electrons have a Maxwellian distribution. Thus
∞
SStoch ≈ ∫ 2 mve vsh2 0 f (ve , t )dve
0
∞
= 2 mvsh2 0 ∫ ve f (ve , t )dve
0
∞
= mvsh2 0 ∫ ve f (ve , t )dve
−∞
 1  8kT  1/ 2 
e
= mv  ne 
 
 4  πme  
2
sh 0
Page 10