Calculation of ternary, quaternary, and higher-order phase diagrams from binary diagrams and binary thermodynamic data

1993 ◽  
Vol 14 (6) ◽  
pp. 710-717 ◽  
Author(s):  
M. Hoch
1989 ◽  
Vol 4 (5) ◽  
pp. 1209-1217 ◽  
Author(s):  
K. Maex ◽  
G. Ghosh ◽  
L. Delaey ◽  
V. Probst ◽  
P. Lippens ◽  
...  

The thermodynamic equilibrium of structures consisting of a thin film silicide (TiSi2 or CoSi2) on doped Si (with As or B) is investigated. Isothermal sections of the ternary phase diagrams for Ti–Si–B, Co–Si–B, Ti–Si–As, and Co–Si–As have been evaluated, indicating the stability of high B concentrations in Si underneath a CoSi2 layer, the instability of high As concentrations in Si underneath a CoSi2 layer, and of B and As concentrations underneath a TiSi2 layer. The obtained thermodynamic predictions agree very well with experimental results (i) on the redistribution of dopants during silicide formation, (ii) on the diffusion of dopants from an ion implanted silicide, and (iii) on the stability of highly doped regions underneath the silicide, both for the case of TiSi2 and CoSi2. It is shown that even though the inaccuracy of reported thermodynamic data is substantial, thermodynamic calculations provide a useful guidance and are consistent with the experimental results.


1975 ◽  
Vol 12 (1) ◽  
pp. 345-355 ◽  
Author(s):  
Robert B. Griffiths

2000 ◽  
Vol 55 (7) ◽  
pp. 627-637 ◽  
Author(s):  
P. Schmidt ◽  
H. Oppermann

Abstract Pseudoternary System Bi2O3/Bi2Se3/Bi2Te3, Phase Diagram, Thermodynamic Data The phase diagram of the pseudoternary system Bi2O3/Bi2Se3/Bi2Te3 is found to include a quaternary solid solution Bi2O2 (TexSe1-x) and ternary, intermetallic mixed crystals Bi2(TexSe1-x)3. Using thermodynamic modeling of the solid solutions it is possible to calcu­ late complex heterogeneous equilibria between all phases of this phase diagram. As a result we can thermodynamically describe the observed phase relations:Bi2(TexSe1-x)3 ⊿H°m(298) = 0; ⊿S°m(298) = R[xlnx + (1-x)ln(1-x)]Bi2O2(TexSe1-x) ⊿H°m(298) = Ω · x(1-x); O⊿S°m(298) = R/4 [xlnx + (1-x)ln(1-x)]Ω = 0,6 kcal/mol


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