Chapter 5 Kinetic Theory And Vacuum

Vern Lindberg

April 26, 2012

2

Contents

5.1 A Brief Introduction to Kinetic Theory . . . . 4

5.2 Units of Pressure . . . . 6

5.3 Gauge and Absolute Pressure . . . . 7

5.4 More Kinetic Theory . . . . 8

5.4.1 Number Density . . . . 8

5.4.2 Sizes and spacing of atoms . . . . 8

5.4.3 Mean free path, mean time between collisions . . . . 9

5.4.4 Collision Flux, Monolayer formation time . . . . 10

5.5 Vapor Pressure . . . . 11

5.6 Viscosity, Turbulence, Molecular Flow . . . . 13

5.7 Thermal Conductivity . . . . 15

5.8 Typical Vacuum System . . . . 16

5.9 Pumping Speed, Throughput . . . . 16

5.10 Ideal Pump Speed . . . . 20

5.11 Conductance . . . . 21

5.12 Effect of Conductance on Pumping Speed . . . . 22

5.13 Conductances in Parallel and Series . . . . 23

5.14 A Practical Example . . . . 24

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4 5.1 A Brief Introduction to Kinetic Theory

5.1 A Brief Introduction to Kinetic Theory

A vacuum is a region where the pressure is less than atmospheric—i.e. where the num- ber of molecules per unit volume is less than what is normally expected. The purpose of vacuum technology is to create regions of vacuum, maintain them, and measure them. To fully appreciate the difficulties, we need to know the connection between the microscopic description of a gas—mass of molecules, velocities of molecules, etc.—

and the macroscopic description—pressure, volume, temperature, number of moles, etc.

This is the subject of The Kinetic Theory of Gases, which will be described fully in thermodynamics texts. A simplified version of Kinetic Theory will allow us to determine the functional dependence of various quan- tities without evaluating integrals.

Consider a gas consisting of a large number, N , of hard molecules. These molecules will each have mass m and speed v. They will also be allowed to have only kinetic energy.

The molecules will be constrained to move along one of the 3 axes - i.e. ˆi, ˆj, or ˆk, and initially will collide only with the walls of a container. The walls will be rigid so that in any collision the molecules will collide elas- tically: they simply reverse their direction, but continue at the same speed.

Consider the box shown in Figure 5.1. What
is the average force exerted by the molecules
on the wall? The number of molecules
moving toward the wall will be ^{1}_{6}N . The
molecules will be spread out uniformly in
the box and will strike the walls at dif-
ferent times. We can say that all of the

Figure 5.1: Box containing an ideal gas. In a simplified model, all molecules move at same speed and along one of the axes.

1

6N molecules will hit the wall in a time

∆t = d/v . As each molecule bounces off, its momentum (a vector) will change.

∆~p = m v

−ˆi

− m v ˆi = −2 m v ˆi (5.1)

This is equal to the impulse of the wall on
the molecule. In a time ∆t the total impulse
on all ^{1}_{6}N molecules will be

F ∆t = ∆~~ p = 1 6N

−2m v ˆi

(5.2) where ~F is the force of the wall on the molecules. Thus

F =~ 1 3

N m v^{2}

d ˆi (5.3)

is the average force of the molecules on the wall, and the average pressure, P , on the wall of area A is

P = F~

A = 2

3N

1
2m v^{2}

A d (5.4)

But Ad = V , the volume, and ^{1}_{2}m v^{2} = KE
is the kinetic energy of a molecule (for real

5.1 A Brief Introduction to Kinetic Theory 5

gas this will be the average kinetic energy.) So

P V = 2

3N KE (5.5)

The model chosen is oversimplified. A real kinetic model would allow molecules to have a range of speeds, and allow motion in any direction. Then integrals must be evaluated to obtain the final result for the average pres- sure. Surprisingly, the detailed result is ex- actly the same as the above, with KE be- ing the average kinetic energy of a molecule.

The equation above is known as the Ideal Gas Law, with the average kinetic energy measured by absolute temperature, T .

P V = N k T (5.6)

with

k T = 2

3KE (5.7)

Here k, sometimes written k_{B}, is Boltz-
mann’s constant, k = 1.38 × 10^{−23}
J/K.

In chemistry we usually write

P V = n_{mol}R T (5.8)
with R being the gas constant and nmol be-
ing the number of moles of gas. You can
easily show that R = N_{Av}k, where N_{Av} is
Avogadro’s number. We will not use moles
in this text. Instead we will define n = N/V
as the number density of particles in such
units as molecules per cm^{3}. Do not confuse
the two meanings of n! The Ideal Gas Law
can thus be written as

P = n k T (5.9)

A real kinetic model uses Maxwell- Boltzmann distribution functions to

allow us to determine various average speeds.

There are different ways to compute the av- erage of the speed. The rms (root mean square) speed is defined by

v_{rms} =r P v^{2}

N =

r3k T

m (5.10)

The average speed is:

v_{ave} = v = P v

N =

r8k T π m =

r8R T π M

(5.11) where m is the mass of one molecule in kg and M is the molecular mass (g/mol). For air (average M = 29 g/mol) at room temper- ature, T = 298 K, the molecules have aver- age speed of 466 m/s and an rms speed of 506 m/s. At the same temperature molecules with a smaller mass will have larger aver- age speeds. This was one of the methods used to separate isotopes of uranium in the Manhatten project.

What do we need to know about the Kinetic Theory that will apply to our study of vac- uums?

First: Pressure is an average quantity — close measurements would reveal fluc- tuations around an average.

Second: Pressure is dependent on temper- ature, but not on the type of molecule.

Two containers filled with different gases, but at the same temperature, can have the same pressure. However, some of the pressure gauges will respond dif- ferently to different gases even if they are at the same pressure.

6 5.2 Units of Pressure

5.2 Units of Pressure

The most direct units of pressure are based on its definition, Table 5.1:

Table 5.1: Pressure Units Based on Defini- tion

SI 1 N/m^{2} = 1 Pascal, 1 Pa
cgs 1 dyne/cm^{2} = 0.1 Pa
British 1 psi (pound/in^{2}) = 6895.3 Pa
Other units are based on the atmospheric
pressure that is always around us on earth.

In Europe the bar is the most widespread unit, although it is being replaced with the Pascal. See Table 5.2

Table 5.2: Pressure Units Based on Atmo- sphere

1 atmosphere (atm, A.P.) = 1.013 × 10^{5} Pa
1 bar (just smaller than 1 atm)

= 1 × 10^{6} dynes/cm^{2} = 10^{5} Pa = 100 kPa

The third set of units, Table 5.3, and the
most common in the USA, is based on the
manometer, a U-tube that shows a pres-
sure difference between the two sides of the
tube as a difference in height of a liquid in
the tube—usually mercury^{1}. An open-tube
manometer is shown in Figure 5.2 From the
local gravitational field strength, g, and the
density of the liquid, ρ, we calculate a pres-
sure difference, ∆P .

∆P = ρ g h (5.12) Since ρ and g are constant at the earth’s

1But not always: in hospitals, water is sometimes used as the manometer fluid.

Figure 5.2: Open-tube manometer. If the liquid is mercury, the pressure difference is h Torr.

surface, we can specify the pressure by spec- ifying the height of the column of liquid, assumed to be mercury unless otherwise stated. The units are:

Table 5.3: Pressure Units Based on Manometer

1 mm-of-Hg = 1 Torr 1 in-of-Hg = 25.4 Torr

1 µm-of-Hg = 1 µ-of-Hg = 1 µ = 1 mTorr

CAUTION: The mm-of-Hg, etc. are not to be canceled with the units of length such as mm. Thus 1 mm-of-Hg/ 5 mm 6= 0.20 This should be clear if the “-of-Hg” is kept with the unit. Unfortunately, people in a hurry tend to say “x mm of pressure” or

“the pressure was y microns”. If you con- fuse the pressure-mm with the length-mm, you will be flogged with a piece of vacuum hose! It is best to avoid confusion by using the Torr.

Lest you find these units comprehensible, be
warned that people often omit the unit alto-
gether and talk of “pressure of 10^{−6}.” Usu-

5.3 Gauge and Absolute Pressure 7

ally, if a physicist from the USA or Canada is talking this means the units are Torr.

There is one convenient approximate conver-
sion: 1 Torr ≈ 1 mbar. If a journal reports
a pressure of 2 × 10^{−4} mbar we can mentally
convert to 2×10^{−4}Torr and know what pres-
sure range we are discussing. Some conver-
sions are given in Table 5.4

Table 5.4: Pressure Conversions 1 std. atm. = 760 Torr = 760 mm-of-Hg

= 1.01323 bar

= 1.01323 × 10^{5} Pa

= 29.92 in-of-Hg

= 14.69 lbs/in^{2}

1 Torr = 1000 mTorr = 1000 µ

= 0.0013 atm

= 133.32 Pa

= 1.34 mbar 1 millibar = 0.745 mm-of-Hg

= 0.745 Torr

= 100 Pa

1 Pa = 7.45 × 10^{−3} Torr

= 10 bar.

Different degrees of vacuum may be de- scribed qualitatively by the terms in Table 5.5.

Table 5.5: Levels of Vacuum (in Torr)

Rough vacuum 760 to 1

Medium vacuum 1 to 10^{−3}
High vacuum (HV) 10^{−3} to 10^{−8}
Ultra-High vac (UHV) 10^{−8} and lower

In physics applications three pressure regimes are common. Sputtering, plasma etching and other plasma processes take

place in medium vacuums of 1 mTorr to a
few hundred mTorr. Evaporation of mate-
rials takes place in high vacuum conditions,
typically 1 × 10^{−6} Torr. This is a pressure
typical of low earth orbit where the space
shuttles operate. Surface analysis equipment
and particle accelerators require UHV con-
ditions of 1 × 10^{−9} Torr or less.

As a final note, remember that 1 atmosphere
of pressure is about 1 Ton/ft^{2}. This means
that vacuum structures are subject to large
stresses!

5.3 Gauge and Absolute Pressure

Some vacuum gauges (such as the open tube manometer) read a pressure difference. Fre- quently these gauges are marked in units that show change from atmospheric pressure

— this is called gauge pressure. Most phys- ical phenomena depend on absolute pres- sure. To get the absolute pressure, add at- mospheric pressure to gauge pressure. The pressure gauges on tanks of pressurized gas are marked in gauge pressure. One abbre- viation is psig, ”pounds per square inch, gauge”

e.g. My tire pressure is 25 lbs/in^{2}. What
is absolute pressure?

P = 25 + 14.7 = 39.7 lbs/in^{2}.

8 5.4 More Kinetic Theory

5.4 More Kinetic The- ory

In addition to the gas law and the expres-
sions for averages of speeds, 7.1.3 and 7.1.4,
we need several other expressions. We will
not provide rigorous proof for the results,
but will give estimates of the results followed
by the more accurate equations. The full
equations involving temperature and pres-
sure will be given, but frequently the tem-
perature is constant at 25^{◦}C and the gas
chosen is air. These equations will also be
given.

5.4.1 Number Density

Number density, n, is the number of atoms
or molecules per unit volume, usually in
units of molecules/cm^{3}. Clearly n = N/V .
We know that 1 mole of an ideal gas occu-
pies 22.4 liters at STP (0^{◦}C and 760 Torr).

From this it is easy to derive
n atoms/cm^{3} = 9.66 × 10^{18}

K

Torr cm^{3}

P
T
at 25^{◦}C. = 3.24 × 10^{16}

1

Torr cm^{3}

P

. (5.13)

The number density at standard tempera-
ture and pressure, 2.5 × 10^{19}atoms per cm^{3},
is known as Loschmidt’s Number.

Equation 5.13 was written in clear form with the units applied to the constant as is usually done in physics. It is common in engineering and other disciplines to present the equation as follows:

n = 9.66 × 10^{18}P

T (5.14)

with P in Torr, T in Kelvin,
n in molecules/cm^{3}.

This engineering convention will be followed in the remainder of the equations.

At room temperature and a pressure typical
of evaporation, 1 × 10^{−6} Torr we find the
number density of 3.24×10^{10}molecules/cm^{3}.
At all pressures attainable in the laboratory
we will have huge number densities, and so
the principles arising from the ideal gas law
remain true.

5.4.2 Sizes and spacing of atoms

Sizes of atoms and distances between them.

We will make a crude estimate of the size of atoms as follows. In the solid and liquid states, molecules basically touch each other.

If we know the density and molecular weight we can find the size of the molecule.

Consider water with a density of ρ = 1.0
g/cm^{3}, and a molecular mass of M = 18
g/mol. Using these numbers and Avogadro’s
number, NAv we find a volume per water
molecule of 3 × 10^{−23} cm^{3}. If we consider
the molecules as cubes, the side of the cube
is the cube root of the volume, about 0.3 nm

= 3 ˚A.

More careful calculations show that the di-
ameter, d_{0}, of atoms and small molecules
ranges between about 0.2 and 1.0 nm. For
helium d_{0} is 0.22 nm and for water vapor d_{0}
is 0.47 nm. For air we use an average value
of 3.7 × 10^{−8} cm = 0.37 nm = 3.7 ˚A.

The volume of space available to each molecule in the gas is just the inverse of the

5.4 More Kinetic Theory 9

Table 5.6: Molecular diameters Gas diameter (nm)

air 0.37

He 0.22

H2O 0.47

number density, V = 1/n. From this we can determine the average distance, D in cen- timeters, between gas molecules to be

D (cm) = 1

n^{1/3} = 4.70 × 10^{−7} T
P

1/3

T in K, P in Torr

= 3.14 × 10^{−6}P^{−1/3}

at 25^{◦}C. (5.15)

5.4.3 Mean free path, mean time between collisions

Mean-free-path, mfp or λ is the average dis- tance traveled between collisions, and mean time between collisions, τ is the average time between collisions of molecules in the gas.

They are related by the average speed of molecules in the gas, hvi.

λ = hvi τ (5.16)

We can estimate the mean free path as fol-
lows. We will use an atom diameter (air) of
d0 = 0.37 nm, and a separation of atoms at
STP of D = 3.4 nm. In order to hit, the
atoms must pass within 2d_{0} = 0.74 nm of
each other, as is shown in Figure 5.3.

The collision cross section is thus about
4 d^{2}_{0} = (0.74)^{2} = 0.55 nm^{2} compared with
the area occupied by an atom of D^{2} =

Figure 5.3: Estimating mean free path. As-
sume regularly spaced atoms (gray) sepa-
rated by D. Each atom has size d_{0}. If an-
other atom (black) is to collide, it must be
within the square area shown by solid line,
of size 2d0. The chance of collision in one
atomic layer is the ratio of areas of the solid
square to the dotted square.

(3.4)^{2} = 11.6 nm^{2}. We estimate the prob-
ability of collision as the ratio of the two ar-
eas, 4d^{2}_{0}/D^{2}, and expect a 0.55/11.6 = 0.087
chance of a collision when an atom travels
3.4 nm. By dividing the separation of layers
by the probability of collision in one layer
we estimate the mean free path at STP to
be 3.4 nm/0.087 = 39 nm. If you put in the
symbols this gives λ = 0.25/ n d^{2}_{0}. A more
realistic derivation results in

λ (cm) = 0.225

n d^{2}_{0} n in cm^{−3}, d0 in cm

= 5 × 10^{−3}

P at 25^{◦}C. (5.17)
For air at STP the mean free path becomes
66 nm, which compares well with our crude
estimate.

At a typical evaporation pressure of 1 × 10^{−6}

10 5.4 More Kinetic Theory

Torr the mean free path is 5000 cm = 50 m!

Recall that the number density of particles
at this pressure is huge, n = 3.24 × 10^{10}
molecules/cm^{3}, but even so the molecules
rarely meet each other. For air at room
temperature the average speed is about 480
m/s, so the mean time between collisions
at 1 × 10^{−6} Torr is τ = λ/hvi = 0.1 sec-
onds.

Exercise: Suppose a chamber is a cube of
side 1 m at a pressure of 1 × 10^{−6} Torr.

Estimate (order of magnitude) the num- ber of collisions that a molecule makes with the walls in the mean time between collisions.

5.4.4 Collision Flux, Monolayer formation time

We may also compute the arrival rate of molecules, Z, also known as collision flux.

This is the number of molecules per unit area
per unit time that arrive at a surface. We
can get an estimate for this quantity as fol-
lows. Consider a cube of length a. The total
area of the cube faces is 6a^{2} and the volume
is a^{3}. With a number density n molecules
moving at average speed hvi, the collisions
per second with the walls is

n a^{3} hvi
a

= n a^{2}hvi (5.18)
and the number of collisions per unit area
per second is this quantity divided by the
area of the cube faces,

Z = 1

6nhvi (5.19)

A correct derivation results in a similar ex- pression, expressed in terms of P, M , and

T ..

Z molecules/cm^{2}/s = 3.51 × 10^{22} P

√ M T P (Torr) T (K), and M (g/mol)

= 3.77 × 10^{20}P
for air at 25^{◦}C
.
(5.20)
For air we use a weighted average M =
29.

At an evaporation pressure 1×10^{−6}Torr and
room temperature, Z is 3.8 × 10^{14} molecules
per cm^{2} per second. While this is a large
number, we must also remember that there
are a large number of atoms per square cen-
timeter of a solid surface.

In coating a piece of glass we want to en-
sure that every atom on the piece of glass is
coated by a deposited atom. We thus will
compute the atomic collision flux, the num-
ber of gas atoms per second that hit one sur-
face atom. Since each atom in a solid occu-
pies about d^{2}_{0} of area, the atomic collision
flux in molecules per atom per sec is just
Z·d^{2}_{0}. For air at room temperature hitting a
surface where the atomic spacing is the same
as solid air,

Zatoms/atom/s) = 5.16 × 10^{5}P (5.21)
A related quantity is the monolayer forma-
tion time. This is the time needed to form
a single layer of molecules on a surface as-
suming that every molecule that strikes the
surface sticks to it, and that the molecules
form a layer one atom high. While these as-
sumptions are not precisely true, we do get
an estimate of the actual monolayer forma-
tion time. This is just the inverse of the

5.5 Vapor Pressure 11

Table 5.7: (a) Kinetic Theory Quantities. Air at 25^{◦}C

Pressure n λ τ Monolayer time

(Torr) (atoms/cm^{3}) cm (sec) (sec)

760 2.5 × 10^{19} 6.7 × 10^{−6} 1.4 × 10^{−10} 2.6 × 10^{−9}
1 3.2 × 10^{16} 5.1 × 10^{−3} 1.1 × 10^{−7} 1.9 × 10^{−6}

10^{−3} 3.2 × 10^{13} 5.1 1.1 × 10^{−4} 1.9 × 10^{−3}

10^{−6} 3.2 × 10^{10} 5.1 × 10^{3} 1.1 × 10^{−1} 0.19

10^{−9} 3.2 × 10^{7} 5.1 × 10^{6} 110 1.9 × 10^{3}

Pressure Arrival Rate Atomic Collision Flux
(Torr) (Atoms/cm^{2}/sec) (Atoms/atom/sec)

760 2.9 × 10^{23} 3.9 × 10^{8}

1 3.8 × 10^{20} 5.2 × 10^{5}

10^{−3} 3.8 × 10^{17} 5.2 × 10^{2}

10^{−6} 3.8 × 10^{14} 0.52

10^{−9} 3.8 × 10^{11} 5.2 × 10^{−4}

atomic collision flux.

τ = 1.94 × 10^{−6}

P (5.22)

Values for the various quantities are shown
in Table 5.7 for air at 25^{◦}C. A summary of
formulas is in Table 5.8.

5.5 Vapor Pressure

Even if we produce a perfect vacuum, we will be hard put to contain it. Any walls on the vacuum system—solid or liquid—will vapor- ize to some extent. This occurs even though the walls are not at their boiling point. The Kinetic Theory explanation of vaporization is that some molecules in the solid or liquid gain energy (randomly) and this will be suf- ficient to let some molecules vaporize. Thus a sealed container will fill with vapor. The molecules in the vapor randomly collide, and occasionally some molecules will lose energy

and condense into the liquid or solid. Even- tually an dynamic equilibrium will occur, when equal fluxes of molecules vaporize and condense. Thus an equilibrium pressure, a vapor pressure, occurs.

Figure 5.4: Establishing an equilibrium vapor pressure. On the left is a non- equilibrium state. At equilibrium the flux of particles evaporating equals the flux of par- ticles condensing.

The equilibrium vapor pressure is very temperature dependent. The Clausius- Clapeyron equation from thermodynamics says that the vapor pressure should fol-

12 5.5 Vapor Pressure

Table 5.8: Important formulae from kinetic theory. P in Torr, T in K, m in kg, M in amu

Quantity General For Air at 25^{◦}C

hvi cm/s

r8k T π m =

r8R T

π M 46600

v_{rms} cm/s

r3k T

m 50600

n atoms/cm^{3} 9.66 × 10^{18}P

T 3.24 × 10^{16}P

d_{0} nm 0.37

D cm 4.70 × 10^{−7} T

P

1/3

3.14 × 10^{−6}P^{−1/3}

λ cm (d0 in cm) 0.225

n d^{2}_{0}

5 × 10^{−3}
P
Z atom/cm^{2}/s ^{1}_{4}n hvi = 3.51 × 10^{22} P

√M T 3.77 × 10^{20}P

Zd^{2}_{0} atom/atom/s 5.16 × 10^{5}P

Monolayer formation time (s) 1.94 × 10^{−6}

P

low

P = P_{0}exp −∆H
k T

(5.23) where ∆H is the latent heat of vaporization (enthalpy change) and is nearly temperature independent. Plots of the vapor pressure versus temperature will be given later.

Some typical vapor pressures are in Table 5.9.

From even these few data several things should become clear. Water vapor, oils (such as fingerprints), and greases have rel- atively high vapor pressures at room tem- perature. If we hope to have a UHV system we must avoid these items. Even at mod-

erate high vacuum this outgassing will be a problem.

Solids have very low vapor pressures at room temperature, or even elevated temperatures.

This is true with a few exceptions such as zinc.

In a vacuum system we have two sources of gas: the original gas in the chamber and gas that has stuck to the surfaces of the chamber and which will leave that surface by a process of vaporization generally called outgassing. Frequently the gas load on our pumping system will be dominated by the water vapor in the chamber and by the out- gassing from the surface.

5.6 Viscosity, Turbulence, Molecular Flow 13

Table 5.9: Vapor Pressures of Common Materials

Material T (C) P (Torr) Material T (C) P (Torr)

Water 100 760 Ice -10 1.95

30 32 -50 0.030

0 4.6 -78 0.56 × 10^{−3}

Mercury 100 0.27 Vacuum Grease

20 1.2 × 10^{−3} 20 10^{−6} to 10^{−8}

-78 3 × 10^{−9}
Copper 1000 10^{−9}

500 10^{−11}
20 10^{−35}
Pump Oils

Roughing 40 10^{−4} Diffusion 100 10^{−4}

100 10^{−2} 200 1

5.6 Viscosity, Turbulence, Molecular Flow

Pumping on a vacuum system means that the gases will be in motion. If the gas flow is relatively rapid, the gas will be in turbu- lent flow, similar to the flow of water in a white-water rapids. The criterion for turbu- lent as opposed to laminar (non-turbulent, viscous) flow depends on the dimensions of the pipes containing flow, the average speed of the molecules, the density of the gas, and the viscosity of the gas. At normal pressures we have viscous flow of liquids (molecules ex- ert forces on each other), however at very low pressures the flow changes to molecular flow (molecules interact only with walls.)

To understand the flow of molecules we must examine the interaction of a molecule with a surface. When a molecule hits a surface it generally sticks to the surface for a short time. It then can be re-emitted in any direc- tion. The distribution of directions is usually

assumed to be a “cosine distribution”: the chance of emission in a direction θ from the normal is proportional to cos θ.

This means that the molecule is most likely to be emitted normal to the wall, and has an equal chance of going left or right at equal angles. The direction of emission has noth- ing to do with the pressure gradient. This is illustrated in Figure 5.5.

Figure 5.5: Cosine Emission Law for Molec-
ular Flow. The probability of emission in a
range of angles θ to θ + dθ is ^{1}_{2}cos θdθ, where
θ is measured from the normal.

A viscous force is a force between molecules.

In the Kinetic Model mentioned previously, the molecules had only kinetic energy; since

14 5.6 Viscosity, Turbulence, Molecular Flow

there was no potential energy, there was no force between the molecules except during collisions. Real molecules will have some force and therefore some viscosity.

Figure 5.6: Definition of Viscosity. The bottom plate is at rest while the top plate moves. If we divide the space into regions of size equal to the mean free path λ, then the molecules next to a plate stick to it, the next layer interacts with these molecules etc.

Consider two plates with the top one mov- ing relative to the bottom one. The layer of molecules next to a plate ”sticks” slightly to the wall and the next layer of molecules sticks to that, etc. Thus the molecules near the bottom plate have an average velocity of zero, while the molecules near the top plate have the same average velocity as that of the top plate. This is shown in Figure 5.6. The thickness of the layers is the mean free path.

The result is molecule drift speeds ranging from zero next to the bottom plate up to a large value near the top plate. The viscous force arises from the transfer of momentum between layers in the gas. Viscosity limits the speed with which gas can be pumped out of a vacuum system. At relatively high pressures the viscous force on a plate of area A separated by a stationary plate by a dis- tance d and moving with velocity ~v is

F = −η~ v~_{0}A

d (5.24)

The viscosity of a gas is denoted η, and has
units N · s/m^{2} = 10 poise (non-SI unit).

Some values are given in Table 5.10. At higher pressures the viscosity depends only on the temperature and the size and mass of the molecules. It is independent of the pres- sure. At lower pressures, where the mean free path is about the same size as the size of our vacuum tube, the viscous force is linearly proportional to pressure. This fact is used in viscometer vacuum gauges that are used to calibrate other high vacuum gauges.

Molecular Flow refers to low-pressure re- gions. In the molecular flow region the molecules collide most often with the walls rather than each other. Viscosity becomes unimportant, since the molecules rarely col- lide. At the same time, however, the fluctu- ations in pressure, and density, become more of a factor. Since the emission of gas from a surface follows the cosine law, we are equally likely to find a molecule moving to a region of higher pressure as to one of lower pres- sure.

The behavior of vacuum systems will depend on which of these three types of gas flow oc- cur. Two dimensionless numbers are used to describe gas flow. At higher pressures we consider the Reynolds Number, defined as

R = U ρ d

η (5.25)

where U is the speed of gas flow, ρ is the gas density, η is the gas viscosity, and d is the pipe diameter. Laminar flow exists for R < 1200, turbulent flow for R > 2200, and between these values either type of flow may exist.

At lower pressures the Knudsen Number is

5.7 Thermal Conductivity 15

Table 5.10: Viscosities of some materials

Material η (N/(m s)) η cP (centiPoise)
Chocolate Syrup, 20^{◦}C 15 15000

Motor Oil SAE60 1 1000

Motor Oil SAE20 0.065 65

Water 20^{◦}C 0.001 1.0

Water 99^{◦}C 0.00028 0.28

Air STP 1.8 × 10^{−5} 0.018

defined as

Kn = λ

d (5.26)

where λ is the mean-free-path and d is the pipe diameter. For Kn < 0.01 we have laminar flow, for Kn > 1 we have molec- ular flow, and in between we have transition flow.

At a pressure of 1 × 10^{−6} Torr, the mean free
path for air at room temperature is about
50 meters. In a pipe of diameter less than
50 meters we have molecular flow. In a
pipe larger than 5000 meters we have viscous
laminar flow. Thus hi-vac vacuum cham-
bers usually operate in the molecular flow
region.

In the roughing lines in a vacuum chamber you have pressures on the order of 0.05 Torr.

At this pressure for air at room temperature we have molecular flow for pipes of diameter smaller than 1 mm and viscous laminar flow in pipes larger than 10 cm. In pipes between these sizes we have transition flow, which is a bit of both kinds, and complicated to deal with.

5.7 Thermal Conductiv- ity

At high vacuum the mean free path becomes long and we tend to have molecular flow.

This means that there are very few colli- sions between molecules in the gas: instead the molecules bounce from one surface to an- other in straight-line motion. The molecules cannot exchange energy with one another di- rectly, and thus the thermal conductivity of the gas becomes small. The kinetic theory model for this is similar to the model for vis- cosity in the above section. Here however the molecules transport energy from one surface to another.

At high pressures the thermal conductiv- ity can be shown to become constant for a given temperature and type of gas. At lower pressures the thermal conductivity de- creases linearly with pressure, and becomes very small. At low pressures radiation heat transfer dominates. Thermocouple and Pi- rani gauges use the dependence of conduc- tivity on pressure for their operation.

In most practical systems the walls of the chamber are kept near room temperature, while a smaller hot source may be inside the chamber. Since the molecules will have a

16 5.9 Pumping Speed, Throughput

greater chance colliding with the larger area of the cooler surface, the temperature of the gas in the system will remain near room tem- perature. We typically say that the temper- ature of a gas in a vacuum remains constant at its initial value even when there is a heat source in the chamber such as an evapora- tion source or substrate heaters.

A second vapor species such as the evapo- rated material may be present in the sys- tem and may have a different temperature than that of the background gas. The av- erage kinetic energy, and thus the temper- ature of the evaporated material will be much higher than the average kinetic en- ergy or temperature of the background gas.

This non-equilibrium state of two tempera- tures will persist for quite a long time, much longer than the processing time of the depo- sition.

5.8 Typical Vacuum Sys- tem

In order to get a high vacuum, we usually need two types of vacuum pumps. The me- chanical roughing pump evacuates the vessel via the roughing line down to a pressure of about 70 mTorr. The diffusion pump then pumps the vessel via the hi-vac valve down to a very high vacuum.

In Figure 5.7, V are valves, with V5 being the hi-vac valve, TC are thermocouple pres- sure gauges, and IG is an ionization pressure gauge. The diagram on the right uses stan- dard US symbols. European symbols are dif- ferent.

During the roughing operation valves V1,

V3, V4, and V5 are closed. Valve V2 is opened. When the vacuum is about 70 mTorr, valve V2 is closed, V1 is opened and then V5 is opened. The diffusion pump then will reduce the pressure to low values. Valves V3 and V4 are needed to vent the system to atmospheric pressure.

We will need to discuss the various elements of this typical system, including:

Piping its effect on pumping speeds Vacuum Pumps pumping speed, ultimate

pressure, etc.

Gauges ranges, use, accuracy Valves types

Leaks finding and fixing

5.9 Pumping Speed, Throughput

We would like to know how rapidly we can evacuate a chamber. The rate of evacua- tion is a complicated function of the pump, the pressure, and the piping in the system.

Pump speed, S, is defined as the rate at which a volume of a substance is transferred out of a system, and is measured in units such as liter/sec or cubic feet per minute (CFM).

S = dV

dt (5.27)

When applied to vacuums, this poses a prob- lem. As the pressure decreases, the num- ber of molecules in a given volume decreases, and so the rate of pressure change is not con- stant.

5.9 Pumping Speed, Throughput 17

Figure 5.7: A typical vacuum system using diffusion and rotary pumps. Valves are denoted V.

Suppose we initially have 1,000,000 molecules in a chamber of volume 4 cubic inches and a pressure of 2.00 Torr.

We imagine a simple pump, Figure 5.8, con- sisting of a metal plate that we insert into the chamber isolating 1/4 of the volume.

This contains 1/4 of the molecules.

Figure 5.8: A simple pump. A plate is in- serted at the 1/4 point, pulled down ejecting all molecules below it. with 3N/4 molecules the pressure is 3P0/4

Now I pull the plate downwards and eject

all the molecules in the small portion of the chamber. I repeat this process once per sec- ond. The gas behind the plate expands to uniformly fill the volume. Thus after each cycle I end up with 3/4 of the molecules that I began the cycle with.

The pumping speed is just 1 cubic inch per second. The number of molecules remaining varies as shown in Table 5.11. If the tem- perature is fixed the pressure is just propor- tional to the number of molecules.

Figure 5.9 shows two plots of these data.

When plotted on regular graph paper, a curve results. If we use semi-logarithmic graph paper the data fit a straight line, sug- gesting an exponential decrease. Semi-log paper is useful in any case where the pres- sure varies over a wide range of powers of 10.

We can get the same results analytically.

18 5.9 Pumping Speed, Throughput

Table 5.11: Pumping With a Simple Pump Time Number of molecules Pressure

(sec) of Molecules (Torr)

0 1 000 000 2.000

1 750 000 1.500

2 562 500 1.125

3 421 875 0.844

4 316 406 0.633

5 237 305 0.475

6 177 979 0.356

7 133 484 0.267

8 100 113 0.200

9 75 085 0.150

10 56 314 0.112

We would like to know how many molecules per second are entering a chamber—call this dN/dt. This is positive for molecules enter- ing the system and negative for molecules leaving the system. By the ideal gas law, P V = N kT . So

dN

dt = d(P V/kT )

dt (5.28)

Usually, we keep T = constant. Thus dN

dt ∝d (P V)

dt (5.29)

A measurement of P V at a fixed T is thus proportional to a measure of N . The value d(P V)/dt is defined as the throughput, Q, of the system. The temperature to be used is room temperature, for as we have al- ready discussed this temperature remains approximately fixed throughout the opera- tion.

Q =d (P V)

dt (5.30)

and is measured in such units as Torr- liters/sec. A pump of speed S and inlet

Figure 5.9: Linear and Semi-logarithmic plots of the data in Table 5.11

pressure P connected to the system has a throughput of

Q = −P S (5.31)

with the negative sign indicating the removal of molecules from the system. Equation 5.30 can be expanded

Q = PdV

dt + VdP

dt (5.32)

You may be tempted to make the first term zero, since the volume of the chamber is con- stant. However the volume is actually the volume of gas in the system, and this can

5.9 Pumping Speed, Throughput 19

change even when the chamber volume and pressure are constant. Even while a pump removes molecules from the chamber, several mechanisms allow molecules to enter the sys- tem. Some of these mechanisms are:

Outgassing of molecules stuck to the walls while the chamber was vented.

Real leaks connecting the chamber to the outside atmosphere.

Virtual leaks caused by small volumes of trapped air, in screw threads for exam- ple.

Diffusion of gases through a spongy sur- face layer: you must clean the sur- face, perhaps by periodically sandblast- ing the chamber.

Backstreaming of molecules from the pump back into the chamber.

We will refer to all of these additions to
a chamber by a throughput of molecules
into the system Q_{leaks}. The second term
represents the change in the number of
molecules due to pumping. Equation 5.32
becomes

Q = Q_{leaks}+ VdP

dt (5.33)

Consider two extremes for a vacuum sys- tem.

1. The pressure has reached an ultimate,
steady-state value, P_{ult}(also called base
pressure) where every molecule that en-
ters from Qleaksis removed by the pump
and P is constant: dP/dt = 0. Then
Q = Q_{leaks}= |−P_{ult}S|.

If we know Q_{leaks}and P_{ult}we can deter-
mine the pump speed at this pressure.

2. Q_{leaks} = 0: The pump dominates and
Q = V (dP/dt).

In the second extreme, negligible leaks, we can determine the pressure as a function of time. Connect a vacuum chamber of volume V to a pump of constant speed S. As we pump down, the throughput is Q = −P S, the throughput of the pump (negative since dP/dt is negative.) Putting this into Equa- tion 5.33 we find:

VdP

dt = −P S (5.34)

or dP

P = −S

V dt (5.35)

If the pump speed is constant and does not
depend on pressure, and if P = P_{0} at t = 0
we integrate to get

P (t) = P0exp

−S V t

(5.36)

In general S is a function of pressure as
will be discussed in the next chapter. Many
pumps operate in a pressure range where
their speed is approximately constant and
Equation 5.36 is approximately true. Non-
constant speed will lead to a deviation from
exponential pressure decrease that will usu-
ally increase the time needed to pump down
to a value longer than that predicted. Also,
the final pressure, Pult, will not be zero, but
will be limited by the value of Q_{leaks}.
Knowledge of the leakage rate is important
in system maintenance. If the value of P_{ult}
increases suddenly we know that there is
a problem that has occurred—a leak has
sprung, the system is too dirty and needs
cleaning, or the substrates are outgassing

20 5.10 Ideal Pump Speed

too much. Our full expression for the pres-
sure as a function of time in the case of con-
stant speed is^{2}

P (t) = P0exp

−S V t

+ Pult (5.37)

We can also look at the system when we close the valve leading to the pump so that the throughput of the pump is zero. We expect the pressure to rise as outgassing and leaks occur. We can equate the two terms of the throughput to give

V dP

dt = Q_{leaks} (5.38)
that is easily integrated to give

P (t) = Q_{leaks}

V t + P_{ult} (5.39)
The rate of rise of pressure in a vacuum sys-
tem is linear with time so long as the Q_{leaks}
is constant. Figure 5.10 shows the expected
graphs of pressure versus time for the cases
of pump down and leak up. The pump is
assumed to have a constant speed. When
graphing the pumpdown graph on semi-log
paper, it is convenient to plot (P − P_{ult}) on
the ordinate. This allows use of more of the
data in determining the slope.

Throughput may have various units. Com- mon ones are Torr-l/s and Pa-l/s. Another common unit is the standard cc per minute or SCCM. Standard refers to standard at- mosphere, and the volume is measured in cubic centimeters. Throughput has units of

2To be totally correct, the term multiplying the exponential should be (P0− Pult)so that at t = 0, P = P0. In practice, Pult P0 and the equation is essentially correct as presented in the text.

Figure 5.10: Pressure changes for a vacuum system (a) Pump down curve for an ideal pumping system of constant speed. On a semi-log plot this is linear. (b) Pressure rise for an isolated system. This plot is on regu- lar graph paper. The rate of rise of pressure is constant.

power, and can be expressed in Watts. Con- versions between the units are given in Table 5.12.

5.10 Ideal Pump Speed

Many different types of hi-vac pumps exist, as we will discuss later. We can compare all of them to an ideal hi-vac pump, one that removes every molecule that enters it. Con-

5.11 Conductance 21

Table 5.12: Throughput Conversions 1 sccm = 1.68872 Pa· L/s

= 1.68872 mW

= 12.667 mT · L/s 1 W = 1000 Pa· L/s

sider an ideal pump with an inlet of area A.

The number of molecules per second enter- ing the pump is given by

dN

dt = Z A = 1

4n hvi A (5.40) From the ideal gas law we have N = P V / (k T ) and n = P/kT . Assuming constant temperature, Equation 5.40 be- comes

1 kT

d (P V ) dt = P

kT hviA

4 (5.41)

and this allows us to cancel k T and intro- duce the throughput, Q = d (P V ) /dt

Q = 1

4P hvi A ≡ P S (5.42) Hence we have an expression for the maxi- mum speed of a hi-vac pump that depends only on the area of the pump inlet and the speed of the molecues,

S_{ideal}= 1

4hvi A (5.43) Diffusion pumps approach this ideal state.

For example a Varian VHS-6 Diffusion
Pump has an inlet diameter of d = 7.88
inches = 200.2 mm. The area of the inlet is
πd^{2}/4 = 0.03148m^{2}and for air at room tem-
perature the average speed is about 480 m/s
so the ideal speed should be 3.780 m^{3}/sec =
3780 liters/sec. The measured speed of this
pump for air is 2400 liters/sec, which is 63%

of ideal.

5.11 Conductance

The pipes, valves, and other plumbing that
connect the pump to the vacuum chamber
also have their effect on the rate of evacua-
tion. In order to evacuate the chamber, the
pressure at the pump inlet, P_{p}, must be less
than the pressure in the chamber, P_{c}. The
throughput of the pipe, Q, will depend on
the size of the pressure difference (P_{c}− P_{p}),
and on the geometry of the pipe. We write
in general

Q = U (P_{c}− P_{p}) (5.44)
The quantity U is the conductance of
the pipe, with units such as (liters/sec).

It depends on the geometry of the pipe, and is rather difficult to compute theoreti- cally.

You may well ask why the pipe affects the throughput. In the viscous flow region, the gas sticks to the side of the container some- what, reducing the rate of flow. In the molecular flow region, the pipe separates the chamber from the pump. In order to be re- moved, a molecule must travel a long way in one direction to get to the pump. In either region, the length of the pipe is a hindrance to the evacuation.

Formulae for conductances have been devel-
oped, notably by Dushman^{3}. These are not
always correct, as is discussed by O’Hanlon^{4},
but do give a feeling for the effect of various
parameters on the conductance.

3S Dushman, The Scientific Foundations of Vac- uum Techniques, 2nd edition, John Wiley, 1962

4J.F. O’Hanlon, “A User’s Guide to Vacuum Technology”, (Third edition), John Wiley & Sons, 2003.