Arthur Sherman - Chemical Vapor Deposition for Microelectronics (779637), страница 3
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We can then write the partial pressure of NO formed as(9)PNO--~MPComparing Equations (8) and (9), and recalling that Kp = Kp(T), we recognizethat the extent of the reaction (x) is independent of the total pressure p, anddepends only on the temperature through Kp .The behavior of a reaction when an inert gas or an excess of reactants areadded is of interest. We can understand this situation if we rewrite the partialpressures in Equation (1) as(10)where na/n is the mole fraction of species a. Then Equation (1 ) becomes(11 )6Chemical Vapor Deposition for Microelectronicswhen fj.v is positive (more molecules on right-hand side than left-hand side ofreaction equation), the addition of an inert gas, while keeping p the same, results in an increase in n which must be compensated by a corresponding decrease in na and nb.
In other words, the reaction will shift to the right. For anegative £iv, the reverse is true. Also, an excess of one of the reactants willhave the same effect. For example, consider the dissociation of PCl s .where fj.v = 1. If we add CI 2 or any inert gas to the PCl s and keep p constant,then n will increase and the reaction will shift toward the products. Said anotherway, if one has dilute phosphorus pentachloride (in an inert gas or chlorine mixture), then more of it will dissociate than if we had 100% PCl s . We will promote the dissociation reaction.When a heterogeneous reaction is considered, the partial pressures includedin the II Law of Mass Action, II Equation (1), are only those for the gaseousreactants.
For example, when silicon is deposited on a surface due to silanepyrolysis, we haveand(12)The partial pressure of Si does not appear because it is in equilibrium with thesol id, and at a given temperature is a constant regardless of pressure. Said another way, the activity of silicon is 1 when it is present as a solid in pure form.The equilibrium constant for each reaction, as a "function of ternperature,was originally determined experimentally.
However, the standard free energychange is related to the equil ibrium constant by!(13)Owhere R is the gas constant. The convention is that the fj. F value for elementsis zero, and for other compounds these values are tabulated as functions oftemperature. Adding the fj. F values for each compound in a reaction gives thevalue for the reaction as a whole. For example, consider the water gas reactionONow,6F CO-94.45 KCal/mole2Fundamentals of Thermal CVDliFOH270liFH20-54.65 KCal/mo1eliFCO-33.00 KCal/mo1eThen, ~ FO for the overall reaction is --33.00 + (-54.65) -- (-94.45), or 6.8 Kcalper mole. Therefore, the equilibrium constant for this reaction can be calculatedfrom Equation (13) as1.2.2 Reactions with Multiple SpeciesWhen there are many species involved, the problem becomes much morecomplex and species partial pressures have to be calculated by approximatenumerical techniques.
As an illustration, consider the Si-CI-H system 3 withonly eight gaseous species (H 2 , HCI, SiH 4 , SiH 3 CI, SiH 2 CI 2 , SiHCI 3 , SiCI 4 andSiCI 2 ) allowed, where the deposition of solid Si on the surface of the containermust be allowed for.It must also be made clear that the accuracy of this approach depends onthe skill with which the significant species are selected. For example, Si 2 H 6may be a significant species in this reaction, but it has been neglected.To treat th is problem along the Iines we have been discussing, we choosesix reactions as follows:Si (s) + 4HC1 (g) :.
SiC1 4 (g) + 2H 2 (g)(14)PSi C142PH2Si(s) + 3HC1(g) :. SiC1 H(g) + H (g)32(15)(16)Si (s) + 2HC1 (g) :. SiC1 H (g)2 28Chemical Vapor Deposition for MicroelectronicsSi(s)(17)+HC1(g)+H2 :. SiC1H 3(g)Si(s) + 2HC1(g) :. SiC1 2(g) + H2 (g)(18)(19)A seventh equation is derived from the fact that the sum of the partial pressures must equal the total pressure (in this example, 1 atmosphere).(20)A -Final (eighth) relationship is obtained when we specify the CI/H ratio.For this system, Si may leave the gas phase in going from one equil ibrium toanother (deposit as a thin film), but the CI and H will remain the same.
Therefore, we can write(21 )C1H4PSiC14+3PSiC13H + 2PSiC12H2 + 2PSiC12 + PSiC1H3+PHCl2PH2 + PSiC1 H + 2PSiC12H2 + 3PSiC1H3 + PHCl + 4PSiH43These eight equations [Equations (14) through (21)] can now be solved for theeight species partial pressures. Since they are highly nonl inear, they were solvedby a numerical iterative procedure on a high-speed digital computer. 3 The FreeEnergies of Formation were evaluated from available thermodynamic data. 4Additional sources of such data are generally available. sThe results of this calculation are presented in Table 1.
For each value ofCI/H and temperature, there will be a calculated value Si/CI. If the originalFundamentals of Thermalcva9value of Si/CI (at room temperature) is larger than this, then Si depositionwill occur in going from the initial state to the final one. If the original valueis less than the final state, etching of Si will occur. Clearly, since we are comparing equilibrium states, there is no way we can evaluate deposition/etchrates. For this, we will have to look at the kinetics of this situation.The Si/CI ratio can be calculated from(22)SiITTable 1: Equilibrium Partial Pressures of Species(H-CI-Si System, Atmospheres)3HC110001000100010001200120012001200140014001400140010°10- 110- 210- 310°10- 110- 210- 310°10- 110- 210- 36.289.399.909.986.09.189.839.985.378.819.£19.98xxxxxxxxxxxx10- 110- 110- 110- 110- 110- 110- 110- 110- 110- 110- 110- 11.52x10- 21.07 x 10- 25.40 x 10- 31 .67 x 10- 35.80 x 10- 23.97 x 10- 21 .501.981 .388.631.851 .9810- 210- 310- 110- 2x 10- 2x 10- 3xxxx2.743.031.741.552.612.444.301 .262.161.222.072.6610- 110- 210- 310- 510- 110- 210- 410- 710- 1x 10- 2x 10- 5x 10- 9xxxxxxxxx8.01 x 10-21.88 x 10- 22.26 x 10- 36.58 x 10- 57.36 x 10- 21 .54 x 10- 27.72 x 10- 41.74x10- 65.99 x 10- 28.89 x 10- 37.84 x 10- 59.56 x 10- 8S;C1 21000100010001000120012001200120014001400140014002.70 x 10- 31.34 x 10- 33.39 x 10- 43.22 x 10- 52.92 x 10- 31.37xl0- 31.94 x 10- 43.38 x 10- 62.63 x 10- 31.03xl0- 34.71 x 10- 55.44 x 10- 75.05 x 10- 55.33 x 10- 52.83 x 10- 58.78 x 10- 65.95 x 10- 56.23 x 10- 52.52 x 10- 53.37 x 10- 65.75 x 10- 55.88 x 10- 51.40 x 10- 51.53xl0- 62.555.706.346.453.438.029.229.493.649.771.211.25x 10- 710- 710- 710- 710- 710- 7xxxxxx 10- 7x 10- 7x 10- 7x 10- 7x 10- 6x 10- 61.99xl0- 46.62 x 10- 51 .59 x 10- 51.50 x 10- 65.01 x 10- 31.53 x 10- 32.03 x 10- 43.49 x 10- 64.62 x 10- 21.10x10- 24.52 x 10- 45.13 x 10- 6Reprinted by permission of the publisher, The Electrochemical Society, Inc.10Chemical Vapor Deposition for MicroelectronicsReferring to Table 1, we find at T = 1OOOoK and CI/H = 1 a value of Si/CI =0.263.
If we had started with dichlorosilane, Si H 2 C1 2 , we would have hadSi/CI = 0.500 initially, and deposition would have occurred. If instead we hadstarted with silicon tetrachloride, SiCI 4 + 2H 2 , the initial value would havebeen Si/CI = 0.250 and etching would have been the result. Then, if we addmuch more H 2 to the original mixture (i.e., SiCI 4 + 2000 H 2 ), we would haveCI/H = 10-3 as well as Si/CI = 0.250, and would find Si/CI = 0.0618. So, usingoSiCI 4 at 1000 K, we can go from etching to deposition, depending on howmuch H 2 we add to the mixture. Figure 3 shows the temperature variation ofthe deposition/etch regimes as a function of the partial pressure of silicon tetrachloride which is mixed with hydrogen.
Clearly, increasing the gas mixturetemperature can lead to etching rather than the desired deposition unless theSiCI 4 is heavily diluted with H 2 .160015001400C 13000-:EDEPOSITIONw~1200110000.100.150.200.300.40PSiCI 4 (ATM.)Figure 3: Boundary between etching and deposition in a SiCI 4 and H 2 mixtureat one atmosphere.1.2.3 Minimization of Gibbs Free EnergyAs noted earlier, when chemical reactions become more complex (i.e., athigh temperatures, 20 or more species are important), it becomes increasinglymore difficult to calculate the equil ibrium species concentrations.
Initially, wehave discussed the equilibrium constant approach to such calculations. Thisrequired a priori knowledge of the significant species developing in the reaction(phases as well), and we were required to write out the appropriate reactionequations. In real ity, it should not be necessary to specify these reactions, sincechanging from one thermodynamic state to another should be independent ofreaction paths involved. Also, it is not always possible to correctly choose whichspecies will be significant or what phases will be formed.
To deal with thesemore complex situations, another approach has been developed to calculatenumber densities at thermodynamic equilibrium for arbitrarily large numbersof species.Fundamentals of Thermal CVD11As is well known from thermodynamics,6 a system will be in equilibriumwhen the Gibbs Free Energy is at a minimum. Therefore, all that is necessaryis to express the Gibbs Free Energy in terms of the degree of completion of thereaction, and then minimize that function.
The Gibbs Free Energy can be expressed as follows, 7G(23)n. (c) ~ F~ (c),;wheremsndg)nj(c)N(g)L\ FOf . (g)=L\ F fj (c)=0PTR1number of gaseous speciesnumber of solid phase speciesnumber of moles of gaseous species inumber of moles of condensed species itotal number of moles of gaseous speciesfree energy of formation of gaseous species ifree energy of fornlation of sol id species itotal pressuretemperaturegas constantand values of the nj's have to be found that minimize G, subject to the massbalance constraint.
That is, the number of atoms of a particular element mustbe conserved. First, let us define aij as the formula numbers specifying thenumber of atoms of element j in a molecule of species i. For example, for thespecies CH 4 , we define i = 1 and let j = 1 represent C and j = 2 represent H, sothat all = 1 and a12 = 4 for this case. Then the constraint can be written as qequations:smL;=1a;j(g) n;(g) +L1=1aih(c) n; (c)bj(24)(j = 1, 2, ... , q)where bj is the number of moles of element j in the original mixture, and thereare q elements in the system.Then the solution to the problem of determining the equilibrium state atthermodynamic equilibrium reduces to one of finding the minimum in thefunction G subject to the constraints of Equation (24).
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