Chapter 5 Kinetic Theory And Vacuum. Vern Lindberg

Chapter 5 Kinetic Theory And Vacuum Vern Lindberg April 26, 2012

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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 3

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 number 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 quantities 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. î, ĵ, 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 elastically: 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 different 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 6 N 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 î m v î = 2 m v î (5.1) This is equal to the impulse of the wall on the molecule. In a time t the total impulse on all 1 6N molecules will be F t = p = 1 ) ( 2m 6 N v î (5.2) where F is the force of the wall on the molecules. Thus F = 1 N m v 2 î (5.3) 3 d is the average force of the molecules on the wall, and the average pressure, P, on the wall of area A is F P = A = 2 3 N 1 2 m v2 A d (5.4) But Ad=V, the volume, and ( 1 2 m v2) = 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 N KE (5.5) 3 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 pressure. Surprisingly, the detailed result is exactly the same as the above, with KE being 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 KE 3 (5.7) Here k, sometimes written k B, is Boltzmann 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 n mol being 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 average of the speed. The rms (root mean square) speed is defined by v rms = The average speed is: v 2 3k T N = m (5.10) v 8k T 8R T v ave = v = N = π m = 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 temperature, T = 298 K, the molecules have average 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 average 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 vacuums? First: Pressure is an average quantity close measurements would reveal fluctuations around an average. Second: Pressure is dependent on temperature, 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 differently 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 Definition 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 Figure 5.2: Open-tube manometer. If the liquid is mercury, the pressure difference is h Torr. surface, we can specify the pressure by specifying the height of the column of liquid, assumed to be mercury unless otherwise stated. The units are: Table 5.2: Pressure Units Based on Atmosphere 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 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 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 pressure 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 pressure difference, P. P = ρ g h (5.12) Since ρ and g are constant at the earth s 1 But not always: in hospitals, water is sometimes used as the manometer fluid. 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 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 confuse 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 altogether 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 conversion: 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 pressure range we are discussing. Some conversions 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 described 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 place in medium vacuums of 1 mtorr to a few hundred mtorr. Evaporation of materials 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 conditions 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. Frequently these gauges are marked in units that show change from atmospheric pressure this is called gauge pressure. Most physical phenomena depend on absolute pressure. To get the absolute pressure, add atmospheric pressure to gauge pressure. The pressure gauges on tanks of pressurized gas are marked in gauge pressure. One abbreviation is psig, pounds per square inch, gauge In physics applications three pressure regimes are common. Sputtering, plasma etching and other plasma processes take 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 Theory In addition to the gas law and the expressions 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 pressure will be given, but frequently the temperature 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 occupies 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 P Torr cm ( 3 at 25 C. = 3.24 10 16 1 Torr cm 3 T ) P. (5.13) The number density at standard temperature 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, N Av 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 Å. More careful calculations show that the diameter, 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 Å. 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 H 2 O 0.47 number density, V = 1/n. From this we can determine the average distance, D in centimeters, between gas molecules to be D (cm) = 1 ( T = 4.70 10 7 n1/3 P T in K, P in Torr = 3.14 10 6 P 1/3 ) 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 distance 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, v. λ = v τ (5.16) We can estimate the mean free path as follows. We will use an atom diameter (air) of d 0 = 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. Assume regularly spaced atoms (gray) separated by D. Each atom has size d 0. If another atom (black) is to collide, it must be within the square area shown by solid line, of size 2d 0. 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 probability of collision as the ratio of the two areas, 4d 2 0 /D2, 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 = 5 10 3 P n in cm 3, d 0 in cm 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/ v = 0.1 seconds. Exercise: Suppose a chamber is a cube of side 1 m at a pressure of 1 10 6 Torr. Estimate (order of magnitude) the number 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 follows. 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 v, the collisions per second with the walls is [ n a 3 ] [ v a ] = n a 2 v (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 n v (5.19) 6 A correct derivation results in a similar expression, 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 centimeter of a solid surface. In coating a piece of glass we want to ensure that every atom on the piece of glass is coated by a deposited atom. We thus will compute the atomic collision flux, the number of gas atoms per second that hit one surface atom. Since each atom in a solid occupies 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 formation time. This is the time needed to form a single layer of molecules on a surface assuming that every molecule that strikes the surface sticks to it, and that the molecules form a layer one atom high. While these assumptions are not precisely true, we do get an estimate of the actual monolayer formation 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. and condense into the liquid or solid. Eventually an dynamic equilibrium will occur, when equal fluxes of molecules vaporize and condense. Thus an equilibrium pressure, a vapor pressure, occurs. 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 vaporize 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 sufficient 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 Figure 5.4: Establishing an equilibrium vapor pressure. On the left is a nonequilibrium state. At equilibrium the flux of particles evaporating equals the flux of particles 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 8k T 8R T v cm/s π m = 46600 π M 3k T v rms cm/s 50600 m 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 (d 0 in cm) 0.225 n d 2 0 Z atom/cm 2 /s 1 4 n v = 3.51 1022 P M T Zd 2 0 atom/atom/s Monolayer formation time (s) 5 10 3 P 3.77 10 20 P 5.16 105 P 1.94 10 6 P low ( ) H P = P 0 exp 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 relatively high vapor pressures at room temperature. If we hope to have a UHV system we must avoid these items. Even at moderate 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 outgassing 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 turbulent flow, similar to the flow of water in a white-water rapids. The criterion for turbulent 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 exert 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 direction. 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 nothing to do with the pressure gradient. This is illustrated in Figure 5.5. Figure 5.5: Cosine Emission Law for Molecular 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 moving 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 distance 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 pressure. 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 regions. In the molecular flow region the molecules collide most often with the walls rather than each other. Viscosity becomes unimportant, since the molecules rarely collide. At the same time, however, the fluctuations 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 pressure. The behavior of vacuum systems will depend on which of these three types of gas flow occur. Two dimensionless numbers are used to describe gas flow. At higher pressures we consider the Reynolds Number, defined as R = U ρ d (5.25) η 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 molecular 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 chambers 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 Conductivity At high vacuum the mean free path becomes long and we tend to have molecular flow. This means that there are very few collisions between molecules in the gas: instead the molecules bounce from one surface to another in straight-line motion. The molecules cannot exchange energy with one another directly, and thus the thermal conductivity of the gas becomes small. The kinetic theory model for this is similar to the model for viscosity in the above section. Here however the molecules transport energy from one surface to another. At high pressures the thermal conductivity can be shown to become constant for a given temperature and type of gas. At lower pressures the thermal conductivity decreases linearly with pressure, and becomes very small. At low pressures radiation heat transfer dominates. Thermocouple and Pirani gauges use the dependence of conductivity 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 temperature. We typically say that the temperature 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 evaporation source or substrate heaters. A second vapor species such as the evaporated material may be present in the system and may have a different temperature than that of the background gas. The average kinetic energy, and thus the temperature of the evaporated material will be much higher than the average kinetic energy or temperature of the background gas. This non-equilibrium state of two temperatures will persist for quite a long time, much longer than the processing time of the deposition. 5.8 Typical Vacuum System In order to get a high vacuum, we usually need two types of vacuum pumps. The mechanical 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 pressure gauges, and IG is an ionization pressure gauge. The diagram on the right uses standard US symbols. European symbols are different. 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 evacuation 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 (5.27) dt When applied to vacuums, this poses a problem. As the pressure decreases, the number of molecules in a given volume decreases, and so the rate of pressure change is not constant.

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, consisting 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 inserted at the 1/4 point, pulled down ejecting all molecules below it. with 3N/4 molecules the pressure is 3P 0 /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 second. 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 temperature is fixed the pressure is just proportional 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, suggesting an exponential decrease. Semi-log paper is useful in any case where the pressure 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 entering the system and negative for molecules leaving the system. By the ideal gas law, P V = NkT. So dn dt = d(p V/kT ) dt Usually, we keep T = constant. Thus dn dt d (P V) dt (5.28) (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 already discussed this temperature remains approximately fixed throughout the operation. d (P V) Q = (5.30) dt and is measured in such units as Torrliters/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 = P dv dt + V dp dt (5.32) You may be tempted to make the first term zero, since the volume of the chamber is constant. 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 system. 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 example. Diffusion of gases through a spongy surface layer: you must clean the surface, perhaps by periodically sandblasting 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 + V dp dt (5.33) Consider two extremes for a vacuum system. 1. The pressure has reached an ultimate, steady-state value, P ult (also called base pressure) where every molecule that enters from Q leaks is 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 determine 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 Equation 5.33 we find: or V dp dt dp P = S V = P S (5.34) 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) = P 0 exp ( SV ) 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. Nonconstant speed will lead to a deviation from exponential pressure decrease that will usually increase the time needed to pump down to a value longer than that predicted. Also, the final pressure, P ult, 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 pressure as a function of time in the case of constant speed is 2 P (t) = P 0 exp ( SV ) t + P ult (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 t + P ult (5.39) V The rate of rise of pressure in a vacuum system 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. Common ones are Torr-l/s and Pa-l/s. Another common unit is the standard cc per minute or SCCM. Standard refers to standard atmosphere, and the volume is measured in cubic centimeters. Throughput has units of 2 To be totally correct, the term multiplying the exponential should be (P 0 P ult )so that at t = 0, P = P 0. In practice, P ult P 0 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 regular graph paper. The rate of rise of pressure is constant. power, and can be expressed in Watts. Conversions 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 entering the pump is given by dn dt = Z A = 1 n v A (5.40) 4 From the ideal gas law we have N = P V/ (k T ) and n = P/kT. Assuming constant temperature, Equation 5.40 becomes 1 d (P V ) = P kt dt kt v A (5.41) 4 and this allows us to cancel k T and introduce the throughput, Q = d (P V ) /dt Q = 1 P v A P S (5.42) 4 Hence we have an expression for the maximum 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 v A (5.43) 4 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 temperature 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 evacuation. 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 theoretically. You may well ask why the pipe affects the throughput. In the viscous flow region, the gas sticks to the side of the container somewhat, reducing the rate of flow. In the molecular flow region, the pipe separates the chamber from the pump. In order to be removed, 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 developed, 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. 3 S Dushman, The Scientific Foundations of Vacuum Techniques, 2nd edition, John Wiley, 1962 4 J.F. O Hanlon, A User s Guide to Vacuum Technology, (Third edition), John Wiley & Sons, 2003.

22 5.12 Effect of Conductance on Pumping Speed As an example of the formulae consider a cylindrical pipe of length l and diameter D (both in cm), with l D. For viscous conductance of air at 20 C, where the average pressure in the pipe is P, In the high vacuum regions of my vacuum system where the mean-free-path is large, in order to make the conductance of my pipes large I will want the diameters to be large. U(L/s) = 182 D4 l For molecular conductance of air at 20 C, U(L/s) = 12.1 D3 l eg. 3 If I reduce the length to 1/3 of the original, how does the conductance P (5.45) change? P in Torr, D and l in cm U 1/l (for D = constant) U 2 = U 1 (l 1 /l 2 ) = U 1 (1/.33) = 3U1 The conductance is inversely proportional to the length. D and l in cm (5.46) Viscous conductance will be present at higher pressures, and molecular flow at lower pressures. The switch occurs when D is approximately the mean free path λ, and is in the range of a mtorr (alias µ) for typical pipes of D = 5 cm. (See the earlier discussion of the Knudsen number.) eg. 1 A hose 30 inches long and 1 inch in diameter connects a vacuum pump to a vacuum vessel. What is the conductance in the molecular flow region? l= 30 in = 76.2 cm D = 1 in = 2.54 cm U = 12.1D 3 /l = 2.60L/s eg. 2 Suppose I choose a hose of the same length, but 1 1/4 inches in diameter. By what factor will conductance change? U D 3 (for l = constant) U 2 = U 1 (D 2 /D 1 ) 3 = U 1 (1.25/1) 3 = 1.95 U 1 Notice that a 25% change in the diameter almost doubles the conductance! For pipes of other shapes, the same qualitative effects hold true: conductance is strongly dependent on the cross-sectional area ( A 3/2 ) and inversely proportional to the length. For a given area, a circular cross-section will be best. 5.12 Effect of Conductance on Pumping Speed Given the connection of Figure 5.11, we want to know what the effective speed of the pump is at the chamber when the pump of speed S p is connected to the chamber through a pipe of conductance U. Call this S eff. We can compute the throughput of the pipe at either end in terms of the speeds and pressures. At the chamber: At the pump: Q = S p P p Q = S eff P c These are the same, since the number of molecules entering the pump equals the number leaving. Also from the definition of conductance,

5.13 Conductances in Parallel and Series 23 What is the effective pumping speed at the chamber? l = 6 in = 15.2 cm D = 5.08 cm U = 12.1D 3 /l = 104 L/s Then S eff = S p /(1 + S p /U) = 300/(1 + 300/104) L/s = 77 L/s Thus only a small fraction of the pump capabilities are being used. The tube should be made larger. Figure 5.11: Pipe conductance U has an effect on the pressures and effective speeds at various places in the vacuum system. In steady-tate, the throughput is the same everywhere. Q = U (P c P p ) (5.47) Combining these (eliminating the pressures) ( Q Q = U Q ) S eff S p (5.48) 1 S eff = 1 S p + 1 U (5.49) eg. 5 A new tube is used in the above example that is 4 inches in diameter (twice the previous diameter). What is the new effective pumping speed? U D 3 So doubling D, octuples U. U = 832 L/s and S eff = S/(1 + S/U) = 220 L/s This is a great improvement! Looking at Equation 5.50 we can make some generalizations. Low Conductance ] If U S p, S eff S p /(S p /U) = U and the size of the pump is immaterial. In order to use at least half of the capabilities of the pump, we must have U = S p. High Conductance For U S, S eff S p. We can also write this as S eff = S p 1 + S p /U (5.50) The rate of evacuation is exponentially related to the speed. The larger the effective speed, the quicker we will pump down a system. This equation is one of the most important equations used in designing a vacuum system. Consider the following examples. eg. 4 A pump of speed 300 L/s operates in the molecular flow region. A tube 6 inches long and 2 inches in diameter connects the pump to the chamber. 5.13 Conductances in Parallel and Series Suppose we have several pipes, valves, etc., and we know the conductance of each

24 5.14 A Practical Example U 1, U 2, U 3,... If we place the sections in parallel, the total combined conductance is approximately U parallel = U 1 + U 2 + U 3 +... (5.51) This makes intuitive sense if we think of the parallel sections effectively increasing the diameter in Equations 5.45 and 5.46. For sections connected in series, the effective length is increased, so the conduction should decrease. In fact approximately 1 U series = 1 U 1 + 1 U 2 + 1 U 3 +... (5.52) The reciprocal of conductance (in a pipe) is defined as impedance of a pipe: W = 1/U in units of s/l. The rules for pipe impedances in series and parallel are the same as the rules for electrical impedances in series and parallel. These rules, unlike those in the electrical case, are only approximate in the molecular flow regime. A series connection of elements is the most common, so how careful must we be in choosing elements to combine. One of the elements will have the minimum conductance, U min, and for this 1/U min is a maximum. Clearly, none of the other elements can increase my total conductance they only make it smaller. To improve the overall conductance I must improve the conductance of the element having the minimum conductance. eg 5 Consider a three element series system: a valve of conductance 20 L/s and two pipes of conductance 50 L/s and 100 L/s. What is the conductance of the system? What should we do to improve the conductance of the system? The total conductance is 1/U series = 1/50 + 1/20 + 1/100 = 0.08 U series = 12.5 L/s If I want to increase the conductance, I should spend most of my efforts on the valve that has minimum conductance, and less time (and money) on the piping. 5.14 A Practical Example How does this all come together? Here let us look at the type of calculations that are possible for a vacuum system. Assume that we have a vacuum chamber of volume V = 530 cubic feet which is connected to a diffusion pump of speed 5300 L/s. Between this pump and the chamber are a series of a valve of conductance 5000 L/s, a pipe of conductance 6000 L/s, and a cold trap of conductance 4200 L/s. The diffusion pump is backed up with a roughing pump of speed 120 cubic feet/minute. We expect a base pressure of 8.0 10 7 Torr. These are plausible numbers for a 10 inch diffusion pump system. 1. (a) How long will it take to pump the chamber from atmosphere to 70 mtorr using the rotary pump? (b) What is the total conductance between the diffusion pump and the chamber? (c) What is the effective speed of the diffusion pump as measured at the chamber?

5.14 A Practical Example 25 (d) How long will it take to pump the chamber from 70 mtorr to 2 10 6 Torr using the diffusion pump? Next consider the moment when the chamber is at a pressure of 1 mtorr. (e) When the pressure in the chamber is 1 mtorr, what is the pressure in the diffusion pump? (f) When the pressure in the chamber is 1 mtorr, what is the pressure in the rotary pump? Finally, consider the chamber when it is at its base pressure, 8.0 10 7 Torr. (g) What is the leakage throughput, Q leaks, into the chamber? (h) If the hi-vac valve were suddenly closed, how long would it take for the pressure in the system to rise to 8.0 10 6 Torr? Let s first define symbols and convert to metric values. Volume V = 530 cubic feet 28.3168 L/cubic feet = 15 000 L Diffusion pump speed S DP = 5300 L/s Conductances U 1 = 5000 L/s U 2 = 6000 L/s U 3 = 4200 L/s Roughing pump speed S RP = 120 cfm 0.47195 L/s / cfm = 56.63 L/s Base pressure P ult = 8.0 10 7 Torr Solution 1. (a) How long will it take to pump the chamber from atmosphere to 70 mtorr using the rotary pump? Since both pressures are well above the ultimate pressure we can use Equation 5.37. The final pressure is 70 mtorr = 0.070 Torr and we start at 760 Torr. t = (V/S) ln (P 0 /P ) = (15000/56.63) ln (760/.070) = 2461 sec = 41 minutes. (b) What is the total conductance between the diffusion pump and the chamber? We combine the conductances to get the total conductance U tot according to Equation 5.52. 1/U tot = 1/5000 + 1/6000 + 1/4200 = 0.0006046 U tot = 1/0.0006046 = 1654 L/s. (c) What is the effective speed of the diffusion pump as measured at the chamber? We use Equation 5.50 to get the effective speed, S eff. S eff = 5300/[1 + 5300/1654] = 1261 L/s (d) How long will it take to pump the chamber from 70 mtorr to 2 10 6 Torr using the diffusion pump? Now the final pressure is close to ultimate so we use Equation 5.37. P P ult = 2 10 6 8 10 7 = 1.2 10 6 Torr. We use the effective speed at the chamber that we just calculated.

26 5.14 A Practical Example t = 15000 ( ) 70 10 3 1261 ln 1.2 10 6 = 131 sec = 2 minutes 10 sec. In practice the time would be quite a bit longer due to outgassing. (e) When the pressure in the chamber is 1 mtorr, what is the pressure in the diffusion pump? Here we use the fact that the throughput, Q = P S, is the same at all points in the chamber. Thus: to 8.0 10 6 Torr? Finally we use Equation 5.39 for the rate of rise. P = (Q leaks /V )t + P ult. 8 10 6 Torr = 8 10 7 Torr + [( 1.01 10 3 Torr L/s ) /15000 L ] t. 7.2 10 6 Torr = ( 6.73 10 8 Torr/sec ) t t= 107 sec Q = P chamber S eff = P DP S DP ( 1 10 3 Torr ) (1261 L/sec) = P DP (5300 L/sec) P DP = 0.24 10 3 Torr (f) When the pressure in the chamber is 1 mtorr, what is the pressure in the rotary pump? Again use the throughput. ( 1 10 3 ) (1261) Torr L/sec = P RP (56.4 L/sec) P RP = 22.4 10 3 Torr = 22.4 mtorr (g) What is the leakage throughput, Q leaks, into the chamber? We compute throughput at the ultimate pressure. Q leaks = ( 8 10 7 Torr ) (1261 L/sec) = 1.01 10 3 Torr L/sec = 0.079 sccm. (h) If the hi-vac valve were suddenly closed, how long would it take for the pressure in the system to rise