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TimeDomain CVD, Inc. |
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| Once we've gotten the required precursors into
the vapor phase we can ask whether they can
do anything useful: that is, could any chemical reactions occur to produce
a solid? The answer to this question has two pieces. The first, the thermodynamics,
examines what the eventual equilibrium state of the system is: can the
precursors and other gases and surfaces in the system react to form
anything else, or are they already as stable as they'll get? The second
piece, kinetics, is the examination of how long it would take to
actually get anything to take place, and whether the equilibrium state is
ever reached.
Thermodynamics, while hardly trivial, is much easier than kinetics to get results out of (although the basic data -- heat and entropy of formation--is surprisingly hard to find for many substances of interest in CVD).
Equilibrium StateIn our discussion of volatility, we used the fact that the equilibrium vapor pressure of a gas over its condensed phase is determined by minimization of the free enthalpy (Gibbs free energy). This result is not unique to evaporation; the equilibrium state of any chemical system -- the concentrations of reactants and products that will be present if you wait long enough -- occurs when the free enthalpy G has reached its minimum value. This is equivalent to saying that the change in free energy due to any reaction (involving a small number of moles) at the equiibrium concentration is 0 |
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For any single reaction like nA + mB => pC + rD we can define a free enthalpy of reaction for the reactants in their standard states. The standard state is typically 1 atmosphere partial pressure for a gas at 25 C. A pure liquid or solid is the standard state of the relevant substance. The free enthalpy for the reaction of n moles of A (and appropriate quantities of the other constituents) can be often be found from standard tables, since it is just the sum of the standard free enthalpy of formation for all the constituents A, B, etc. multiplied by the appropriate stoichiometric coefficients. |
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We can then find the equilibrium state of the reaction from the equilibrium constant. |
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| In more realistic situations, many overall
reactions, with numerous reactants and products, need to be considered. We
can treat such a situation by writing all the reactions we can think of
between reactants and products. We then search for the set of
concentrations which simultaneously satisfies the equilibrium constant for
each reaction (or equivalently minimizes the total system free enthalpy),
under the constraint that the number of each kind of atom summed over all
the constituents is equal to what we "put in" to begin with.
This is a problem in constrained minimization. It can be solved by
standard computational methods using Lagrange multipliers. A more elegant
approach is to employ element potential methods, originated by Powell and
Sarner (GE Report R59/FPD (1959)) and incorporated by William Reynolds of
Stanford into the program STANJAN. (As of 1998 a web implementation of
STANJAN was available at http://adam.caltech.edu/tcc/
.)
We can classify overall CVD reactions by examining the standard enthalpy of reaction. Reactions for which the change in free enthalpy is modest are potentially reversible: the actual process may be a balance beween etching and deposition, depending on the initial conditions. Reactions involving very large changes in free enthalpy have essentially no reactants left in equilibrium: they are completely irreversible and lead only to deposition. Let's look at a few examples: shown below are some important overall CVD reactions, with the standard free enthalpy of reaction per mole of product in KJoule/mole. The reactions are given in order of increasing free enthalpy. We see that the free enthalpy of reaction for the formation of silicon dioxide from silane is strongly negative. In terms of the equilibrium constant, the concentration of reactants is essentially zero in the equilibrium state: this is a highly irreversible reaction! The enthalpy release is larger when enough oxygen is available to completely oxidize the hydrogens into water. Silicon nitride formation from silane is less strongly driven, but still very much an irreversible reaction: one can't etch nitride with hydrogen. The decomposition ("pyrolysis") of silane to form solid silicon involves a modest release of free enthalpy; it is not unreasonable to think that the reaction could be reversed, and (as we'll examine in some more detail below), this can be easily accomplished by adding a bit of chlorine to the mixture. The deposition of TiN and Ti from titanium tetrachloride is actually not thermodynamically possible at room temperature! However, the entropy of these reactions is positive (because we're producing more high-entropy gas phase molecules than we consume). If we raise the temperature enough, the free enthalpy will become negative and the reaction will proceed: for titanium metal, temperatures in excess of 1000 C are required. This explains the commonplace observation that it is difficult to deposit titanium metal by CVD. Example: Selective Silicon DepositionLet's look in some detail at a more complex application of thermodynamic calculations to a deposition process. [after R. Madar and C. Bernard, J. Vac. Sci. Tech. A8 p. 1413 (1990)]
At 1000 K, for example, the equilibrium Si:Cl ratio is about 0.18. Thus if we use a feed of pure dichlorosilane, with a Si:Cl ratio of 0.5:1, net deposition must result. The process is unlikely to be highly selective. However, an inlet mixture of 3:1 HCl:SiH2Cl2 will give an inlet Si:Cl ratio very near that of the equilibrium mixture. The small energy difference between polycrystalline silicon growing on oxide and the more stable epitaxial silicon can be the difference between net deposition and net etching. Some Reference DataIt is a fair approximation to associate a specific enthalpy change with each bond in a molecule -- only fair because electrons are not in fact that localized, and the strength of interaction, as we noted, depends on the geometry of a molecule as well as the atoms that compose it. We provide below a table of some useful bond energies in KJ/mole, with which one can make a rough estimate of the enthalpy of formation of species when the measured data is not available. |
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bonds with hydrogen |
H-H 435 H-C 414 H-N 389 H-O 464 H-Si 318 H-Cl 430 |
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bonds with silicon |
Si-O 451 Si-F 564 Si-Si 221 Si-Cl 380 |
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bonds with carbon |
C-C 347 C=C 610 C-N 305 C=N 614 C-F 484 C-O 359 C=O 735 C-Si 301 C-Cl 339 |
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bonds with nitrogen, oxygen, fluorine |
N-O 221 N-F 272 O-O 195 O=O 250 O-F 188 |
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