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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 8 Plasma Sheaths
New homework problems:
Lieberman 6.1, 6.3 and 6.5 Due March 19th, 2001.
Wednesday’s class will be at 1 PM in MP 2.228 - get other make up time for Thursday.
We know from the previous section that the electrons and ions must level a plasma at the same
time and position averaged rate. (This implies that we can have the electrons leaving at one
location and the ions leaving at another location. It also implies that the ions might leave at a
certain time while the electrons leave at a different time – provided the time difference is not
great.) We want to examine plasmas in terms of this requirement of charge neutrality.
Our fundamental fluid equations are as before:
∂n 

f c =  ∇ r ∑(n v ) + 

∂t 
∂ v

+ v ∑∇ r v  = ∆M c −
mn
 ∂t
 123
momentum
lost via
collisions
mv fc
12
4 4
3
− ∇ r ∑P + qn(E + v ∧ B)
momentum change
via particle gain / loss
Let us examine what each of these equations mean.
First integrating over the continuity equation
∂n
f c dτ − ∫∫∫ dτ = ∫∫∫ ∇ r ∑(n v )dτ
∫∫∫
∂t
Vol
Vol
Vol
=
∫∫ n v
∑ds
∂Vol
=
∫∫ Γ ∑ds
∂Vol
where Γ is the flux of the particles.
Likewise we can rewrite of energy equation, here integrating along the E field to give
m
 ∂v 1

1

1

∫  m ∂t + 2 m∇r ∑vv ∑dl = ∫  n ∆M c − n v f c  ∑dl − ∫  n ∇r ∑P ∑dl + ∫ q(E + v ∧ B) ∑dl
These are in general very difficult problems to solve. Thus we will make some approximations.
First, we will let the collision terms go to zero. Then we will let the pressure differential go to
zero. Both of these approximations are reasonable for many plasma systems. Typically they
have very thin sheaths compared to the mean-free path and are in low-pressure systems. Thus
our equations simplify to:
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
− ∫∫∫
Vol

∫  m
∂n
dτ = ∫∫ Γ ∑ds
∂t
∂Vol
∂v 1

+ m∇ r ∑vv ∑dl = ∫ q(E + v ∧ B) ∑dl

2
∂t
We can further simplify our model by assuming time independence and no magnetic field. This
eliminates a lot of real sheaths that we might run across but it makes the model understandable.
Thus, we are dealing with the following set of equations
∫∫ Γ ∑ds = 0
∂Vol
⇒ Γ = nv = const
1
∫  2 m∇
r
∑vv ∑dl = ∫ qE ∑dl = − ∫ q∇φ ∑dl

⇒
1
m∆v 2 = − q∆φ
2
We will use these equations to give us the ion density and velocity at different points in the
sheath. (The above equations are general and we had not applied them to any species.) The only
thing that we need to know is what is the potential as a function of position. Still, we can
determine the velocity and the density as a function of the potential. Thus we have
v
ni = is nis
vi
mi vi2 = 12 mi vis2 + e(φ s − φ )
where ‘s’ stands for the value at the sheath edge. Combining the equations gives
2
 nis 
1
vis  = 12 mi vis2 + e(φ s − φ )
2 mi 
 ni 
1
2
mi nis2 vis2 = ni2 ( mi vis2 + 2e(φ s − φ ))
1/ 2


2e
φ − φ )
nis = ni 1 +
2 ( s
 mi vis

or
−1 / 2


2e
φ − φ )
ni = nis 1 +
2 ( s
 mi vis

This is slightly different than what is in the book as I have included the fact that the plasma has a
potential relative to other items, labeled, φ s .
To get the potential, we can use Poisson’s equation.
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
ρ e
= (ne − ni )
ε ε
Again we are missing a piece of the puzzle, but now it is the electron density. There are a couple
of readily useful options. 1) There are no electrons in the sheath. 2) The election density is set
by Boltzmann’s equation. The second option is the one used most commonly, while the first is
appropriate only for very special situations.
∇ 2φ = −
Initially, let us assume that the electrons follow Boltzmann’s equation. Thus,
ne = n0 e e∆φ / kTe
where n0 is the electron density at the point at which the potential is zero. Or we can write this
in terms of the potential at the sheath edge and get
e φ − φ / kT
ne = nes e ( s ) e
This and our equation for the ion density can be plugged into Poisson’s equation to give
−1 / 2

 
e
e
− e (φ s − φ ) / kTe
2
∇ φ =  nes e
− nis 1 + 1
φ − φ ) 
2 ( s
ε
 2 mi vis
 
.
−1 / 2




e
e
∇ 2 Φ =  nes e eΦ / kTe − nis 1 −
( Φ ) 
Ε
ε
is

 
where Φ = φ − φ s < 0 and Ε is = 12 mi vis2 .
This equation can be integrated once analytically. This is done in the following way. First
multiply by ∇Φ. This gives
−1 / 2

 
e
e
2
eΦ / kt
∇Φ(∇ Φ) = ∇Φ  nes e
− nis 1 −
( Φ ) 
ε
 Ε is
 
Now integrate over dl
−1 / 2
l
l
 eΦ  
e
eΦ / kt e
2
∫0 ∇Φ(∇ Φ) ∑dl = ∫0 ∇Φ ε  nese − nis 1 − Ε is   ∑dl
∫
l
0
1
2
1/ 2
Ε is  eΦ  
e  kTe eΦ / kt
+ 2 nis
e
1−
∇(∇Φ) ∑dl = ∫ ∇ nes
 ∑dl
0ε
e
e  Ε is  

2
l
where 0 is the sheath edge and l is a position in the sheath. Now, ∇ =
first principle of calculus and get
Page 3
d
so that we can use the
dl
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
0
1
2
1/ 2
le d 
d
kTe eΦ / kTe
Ε is  eΦ  
2
e
+ 2 nis
(∇Φ) ∑dl = ∫0
1 −
  ∑dl
 nes
dl
e
e 
Ε is  
ε dl 
l
1
2
(∇Φ)
2 l
l=0
1/ 2
e  kTe eΦ / kTe
Ε is  eΦ  
e
=  nes
+ 2 nis
1 −
 
e
e 
Ε is  
ε
l=0

=0
}


}


e  kTe  eΦ ( l ) / kTe
Ε is
2
e Φ ( l ) / kTe 
1 

=  nes
−e
2 (∇Φ( l)) −  ∇ Φ( 0 )
e
 + 2 nis
e 
e


  ε 



2
=0
1
2
Ε
e  kT
(∇Φ(l)) =  nes e eeΦ( l ) / kTe − 1 + 2nis is
e
e
ε
[
2
]
1/ 2
=0

}

 
1/ 2
 eΦ(l) 
 e Φ ( l )  
 − 1 −
1 −

Ε is 
Ε is  


 

 eΦ(l)  1/ 2 
1 −
 − 1
Ε is 



2
=0
=0
}

 e  kT 
}


Ε is
2
eΦ ( l ) / kTe
e Φ ( l ) / kTe 
e
1 
 =  nes
−e
2 (∇Φ( l)) −  ∇ Φ( 0 )
e
 + 2 nis
e 
e


  ε 



1/ 2
=0

}

 
1/ 2
 eΦ(l) 
 e Φ ( l )  
 − 1 −
1 −

Ε is 
Ε is  


 

or
2e  kTe eΦ ( l ) / kTe
Ε
e
− 1 + 2 nis is
 nes
e
e
ε 
 eΦ(l)  1/ 2 
∇Φ(l) =
1 −
 − 1
Ε is 


There is no known way to further analytically integrate this equation. However we can still tell
that the term inside the square root cannot be negative – or else the potential is imaginary. We
can learn something from this requirement.
1/ 2
 kT

Ε is  eΦ(l) 
eΦ ( l ) / kTe
e
−
e
− 1 + 2 nis
1
 ≥ 0
1 −
 nes

Ε is 
e
e 


For small potentials, we can expand the exponential term to get
1/ 2
 kT  eΦ(l)  

Ε is  eΦ(l) 
e
−
−
+
−
n
1
+
1
2
n
1
1
 ≥ 0

 es

 


is
Ε
e
kT
e






e
is




[
[
]
]
1/ 2


Ε is  eΦ(l) 
l
Φ
n
2
n
1
1
−
+
−
(
)
 ≥ 0

 es


is
Ε is 
e 


Further we can expand the square root term to get
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

Ε is
 nes Φ(l) + 2 nis e

 1 eΦ(l)  
1 −
 − 1 ≥ 0
Ε
2

 
is


Ε is
 nes Φ(l) − 2 nis e

 1 eΦ(l) 
 2 Ε  ≥ 0
is 

(nes Φ(l) − nis Φ(l)) ≥ 0
nes ≤ nis
− remember that Φ(l) < 0!
This means that the electron density must be less than the ion density but not by much. This is
not surprising as we have already said that the sheath occurs because the electrons would
otherwise leave faster than the ions. This makes it such that electron density is slightly smaller
than the ion density. Now, if we assume that the electron and ion densities are approximately the
same, so that nes ≈ nis = ns as we typically assume in the plasma. This is certainly an option.
We can now go back and examine the right hand side of our inequality again.
1/ 2
 kT

Ε is  eΦ(l) 
eΦ ( l ) / kTe
e
−
e
− 1 + 2 ns
1
 ≥ 0
1 −
 ns

e
e 
Ε is 


[
]
1/ 2
 kT

Ε is  eΦ(l) 
eΦ ( l ) / kTe
e
−1 + 2
e
1 −

 − 1 ≥ 0
e 
Ε is 

 e
Now we will expand the equation to the second order so that
 kTe  eΦ(l) e 2 Φ 2 (l)  
Ε is  eΦ(l) e 2 Φ 2 (l)  
0≤
+
−
1 −
1 +
 − 1 + 2
 − 1
2 k 2 Te2  
2Ε is
8Ε is2  
kTe
e 
 e 
[
]

eΦ 2 (l) 
eΦ 2 (l)  
Φ
l
0 ≤   Φ( l ) +
−
+
(
)
2 kTe  
4Ε is 

  eΦ 2 (l)   eΦ 2 (l) 
0 ≤ 

−
  2 kTe   4Ε is 
kTe ≤ 2Ε is
1/ 2
 kT 
vis ≥ vB =  e  = ion acoustic / Bohm velocity
 Mi 
This is known as the Bohm Sheath criteria. In effect, it says that the ions have to be traveling at
the ion acoustic velocity before they enter the sheath. This makes sense as the if the velocities
were zero at the edge of the sheath, then the flux of the ions into the sheath would be zero (unless
the density was infinite!). This means that we have to have a region in which the ions are
accelerated from 0 to the acoustic velocity. This region is known as the ‘presheath.’
What does this mean? First to accelerate the ions to the Bohm velocity we need to have a
potential drop such that energy is conserved. Thus, ignoring collisions we find that
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
1
m∆v 2 = − q∆φ
2
⇓
≈0
67
8
1  2
2
Mi  vB − v plasma  = − q φ B − φ plasma
2 

(
)
1
kT
Mi e = − qΦ B
2
Mi
1
qΦ B = − kTe
2
Likewise we know that we can determine the electron density at the edge of the sheath from
Boltzmann’s equation.
ns = n plasma e − qΦ B kTe
= n plasma e −1 2
= 0.61n plasma
The above discussion implies that there has to be some sort of presheath region which
accelerates the ions from zero to the Bohm velocity. We can get some handle on the subject if
we assume a few simple things. First that the ion and electron densities are equal throughout the
presheath. Second, that the electron density follows Boltzmann’s equation. Third and final, that
the ions enter the presheath at zero velocity and leave on the other side at the Bohm velocity.
Thus we have the following picture.
Φ, n
n0 = ni = ne , v0 = 0
Φ=0
plasma
ni = ne = n0 e eΦ / kTe
v = vB =
kTe
Mi
presheath
sheath
X or r
To understand the presheath, we need to look at the ion flux through the presheath. This is
simply
Γi = ni v i .
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
Taking the derivative of both sides gives
∇ ∑Γi = ∇ ∑(ni v i )
= ni ∇ ∑(v i ) + v i ∑∇(ni )
now dividing through by the flux, we get
1
1
1
∇ ∑Γi = ∇ ∑(v i ) + ∇(ni ) .
Γi
vi
ni
Plugging in Boltzmann’s equation gives
1
1
1
∇ ∑Γi = ∇ ∑(v i ) + ∇(ni )
Γi
vi
ni
(
)
=
1
1
eΦ / kTe
∇ ∑(v i ) +
eΦ / kTe ∇ n0 e
vi
n0 e
=
1
1
qΦ / kTe e
∇ ∑(v i ) +
∇(Φ)
qΦ / kTe n0 e
vi
n0 e
kTe
=
1
e
∇ ∑(v i ) +
∇(Φ)
vi
kTe
What does this mean?
Well we can get some handle on the subject if we examine what happens to the ion energy as it
passes through the pre sheath. The only simple example is conserved energy. (Obviously there
are ways in which energy can also be added or subtracted!)
1
If energy is conserved than m∆v 2 = −e∆φ . This implies for our case that
2
1
−2eΦ
Mi v 2 = −eΦ or v =
2
Mi
thus
−2eΦ
∇vi = ∇
Mi
=
−2e
∇ Φ
Mi
=
1 −2e 1
∇Φ
2 Mi Φ
=
1 −2eΦ −1
Φ ∇Φ
Mi
2
=
1 −1
vΦ ∇Φ
2
⇓
1
1
∇vi = Φ −1∇Φ
vi
2
and at the sheath edge
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
1
kT
Mi e = −eΦ
2
Mi
1 kTe
2 e
Plugging this into our flux equation we find
1
1
e
∇ ∑Γi = ∇ ∑(v i ) +
∇Φ
kTe
vi
Γi
Φ=−
=
1 −1
e
∇Φ
Φ ∇Φ +
2
kTe
=
1 2e
e
∇Φ +
∇Φ − at the sheath edge
2 kTe
kTe
=0
This implies that
∇ ∑Γi = 0
in the presheath if the energy is conserved. Other solutions typically have to be found via
computer simulations.
Classified Sheaths
There are a few important standard sheaths. These are sheaths to a floating wall, sheaths to a
grounded wall and sheaths to a driven wall. We will start with the sheath potential at a floating
wall.
If a wall or other object is inserted into a plasma and there is no way for the charge to drain from
the surface, we have a situation in which the wall is electrically floating. This gives rise to a
‘floating sheath’. In this case, the object will charge up until the flux of electrons and ions
match. Thus to find floating potential of the object we simply need to determine the potential at
which the fluxes match.
The flux of ions to the wall is given by
Γi = const
= n s vB
kTe
Mi
This flux is matched by the flux of electrons to the surface. The electron flux is those electron
headed toward the surface which also have enough energy to overcome the potential. Thus,
= ns
Γe = ∫
∞
v min
vx f (v)dvx dvy dvz
where
2eΦ w
me
is set by energy requirements
Thus
vmin =
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
 me 
Γe = ne 

 2 πkTe 
3/ 2
 me 
= ne 

 2 πkTe 
1/ 2
∞
∞
∫ ∫ ∫
−∞ −∞
∫
1  2 kTe 
= ne 

2  πme 
1/ 2
1  2 kTe 
= ne 

2  πme 
1/ 2
∞
2 eΦ w
me
∫
∞
∫
∞
2 eΦ w
me
2 eΦ w
me
∞
2 eΦ w
me
(
 − me vx2 + vy2 + vz2
vx exp 
2 kTe

 − me (vx2 ) 
vx exp 
 dvx
 2 kTe 
) dv dv dv

x
y
z
 − me (vx2 )   me vx2 
exp 
d 

 2 kTe   2 kTe 
exp[ −U ]d (U )
average


speed
 −eΦ w 
 −eΦ w  
 kTe 
1 }

= ne 
v exp 
= ne
 exp 


4
 2 πme 
 kTe 
 kTe  



Now the flux of the ions and the electrons must come into balance so that
1/ 2
 kTe 
ne 

 2 πme 
1/ 2
 −eΦ w 
kTe
exp 
= ns

Mi
 kTe 
= ne
kTe
Mi
⇓
Φw = −
kTe  2 πme 
ln
e  Mi 
Grounded Sheath
In this case the electrode is electrically connected such that we can add or subtract any necessary
charge to provide balance. As before the flux of ions to the surface is
Γi = n i ( x )vi ( x ) = const
(Note that while I am doing this for flux, you really want to do this for current. The distinction is
that the ions might be doubly – or greater – charged. Thus the flux and the current are not
equivalent. You really need to know how much charge you are carrying to the surface.)
To keep things as simple as possible, we will assume that the energy is conserved. Thus
1
Mi ∆v 2 = −e∆Φ .
2
If we assume further that the initial kinetic energy and potential is zero – hence ignore the
presheath – we find
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
1
Mi (v 2 − vB2 ) = −e(Φ − Φ S ) or
2
v = vB2 −
2e(Φ − Φ s )
Mi
but
vB2 =
kTe
Mi
ΦS = −
1 kTe
2 e
thus
v=
kTe 2e(Φ) kTe
−
−
Mi
Mi
Mi
= −
2e(Φ)
Mi
and
Γi = n i ( x )vi ( x ) = const
= ni ( x) −
2eΦ
Mi
⇓
− Mi
2eΦ( x )
This means that in the sheath, the ion density can be determined if we know the potential. We
can find the potential from Poisson’s equation
ρ e
∇ 2φ = − = (ne − ni )
ε ε
Plugging in our ion density variation electron density from the Boltzmann’s equation
ne = n0 e eΦ / kTe
we find
ρ e
− Mi 
∇ 2φ = − =  n0 e eΦ / kTe − Γi
ε ε
2eΦ( x ) 
As we did before, we can integrate this by multiplying by ∇φ and then integrating. Thus
n i ( x ) = Γi
Page 10
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
∇ 2 Φ∇Φ =

e  eΦ / kTe
Mi
n0 e
∇Φ + Γi
∇( − Φ)

ε
2e( − Φ)

⇓
∫
r
0
∫
r
0
∇ 2 Φ∇Φ ∑dr =

e r  eΦ / kTe
Mi
Φ
Γ
Φ
∇
+
∇
−
n
e
(
)
0
i
 ∑dr
ε ∫0 
2e( − Φ)

 Mi ( − Φ)  
1
e r  kT
2
∑dr
∇(∇Φ) ∑dr = ∫  n0 e ∇ e eΦ / kTe + Γi ∇ 2
2
2e  
e
ε 0

(
)
⇓
r
r
 − Mi Φ  
e  kT
1
2
∇
Φ
( ) = n0 e eeΦ / kTe + Γi  2

2
2e   0
ε e

0
(
)
r
 − Mi Φ  
e  kT
=  e ne (r) + Γi  2

2e   0
ε e

Now assume the following. 1) ∇Φ(r = 0) = 0 , 2) Φ(r = 0) = 0 , AND 3) ne (r) = 0 . The first two
assumptions are reasonable, while the third assumption is an important approximation that
allows us to obtain an analytic solution. With these assumptions we find
1
e  − Mi Φ 
(∇Φ)2 = Γi  2

2
2e 
ε
⇓
eΓ
−∇Φ = 2 i 
 ε 
1/ 2
 − Mi ( − Φ) 



2e 
1/ 2
 Mi 
 2e 
1/ 4
⇓
∇( − Φ)
 eΓi 
1/ 4 = 2
 ε 
( − Φ)
Integrating again
1/ 4
Page 11
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
r
∇( − Φ)
 eΓ 
d
r
=
1
/
4
∫0 (−Φ)
∫0 2 ε i 
r
1/ 2
 Mi 
 2e 
1/ 4
dr
⇓
∫
r
0
4
eΓ
3/ 4
∇( − Φ) (r)dr = 2 i 
 ε 
3
1/ 2
 Mi 
 2e 
1/ 4
 Mi 
 2e 
1/ 4
∫
r
0
dr
⇓
4
eΓ
(− Φ)3 / 4 (r) = 2 i 
3
ε
1/ 2
r
⇓
M
3 eΓ
(− Φ) =  i   i  r
2 ε
2e
This is known as the Child’s Law sheath. It is the only known analytic description, albeit an
approximate solution, to the sheath equation.
1/ 2
1/ 4
3/ 4
Ion Matrix Sheath
The ion matrix sheath occurs when the potential on an object changes very rapidly. In this case,
the electric field change is such that the electrons leave the immediate region while the ions
remain fixed for a small instance. (The heavy ions can not respond as fast as the light electrons.)
Then there is an electric field that occurs through a uniform distribution of ions. How thick is
such a sheath and what is the potential structure? From Poisson’s equation we know:
= const = n 0
=0
} 
ρ e }
2
∇ Φ = − =  ne − ni 
ε ε

=
−e
n0 = const
ε
⇓ integrating once gives
−e
∇Φ =
n0 r ( = − E)
ε
⇓ integrating again gives
−e 2
n0 r
Φ=
2ε
⇓ inverting
Page 12
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
r = −Φ
2ε
- Thus at the wall
en0
Swall = − Φ Wall
= λ Debye
2ε
en0
−2Φ Wall
εkTe
, λ Debye =
kTe
en0
Now we can start to add back parts to our model of the sheath. One of the important things that
we left out was collisions. How do we add these back into our model?
Page 13