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Lecture 8 of "Fundamentals of Photovoltaics" course, delivered by Prof. Ken Durose, Nov 7 2014
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Learning objectives
2 Phase diagrams
Be able to read a phase diagramBe able to identify the phases present at a given point Be able to calculate how much of those phases there areBe able to predict the microstructures from the phase diagram
3 Mixing
Understand the thermodynamic basis of mixingUnderstand the origin of the miscibility gapBe able to use the phase diagram to predict miscibility phenomena
1.Solid state reactions
Recap: how Gibbs Free Energy can be used to predict reactions
There will be workedexamples on this bit for you to practice ….
1. Prediction of reactionse.g. solar cell contact – metal M to semiconductor AB
AB
M
AB
M
MB + A
AB
M
or
stable
unstable
M + AB MB + A Will it happen at equilibrium?
keqm = [MB]*[A] /[M]*[AB] If keqm is large, the reaction goes from l to r
)ln( eqmkRTG Find the Gibb’s free energy for the reaction.
If ∆G is large and negative, keqm is large and the reaction goes from l to r
1. Prediction of reactions
initialfinalreaction
initialfinalreaction
initialfinalreaction
reactionreactionreaction
SSS
HHH
GGG
STHG
M + AB MB + A
etc
GGGGG ABMAMBreaction )()(
If ∆G is large and negative, keqm is large and the reaction goes from l to r
In this case the contact would be unstable to reaction with the solar cellmaterials. The contact could become non-Ohmic, the series resistance could increase, and the performance could become unstable.
ISSUES TO ADDRESS...• When we combine two elements... what equilibrium state do we get?• In particular, if we specify... --a composition (e.g., wt% Cu - wt% Ni), and --a temperature (T )
then... How many phases do we get? What is the composition of each phase? How much of each phase do we get?
2. Equilibrium phase diagrams
Phase BPhase A
Nickel atomCopper atom
• Components: The elements or compounds which are present in the mixture (e.g., Al and Cu)
• Phases: The physically and chemically distinct material regions that result (e.g., and ).
Aluminum-CopperAlloy
Components and Phases
(darker phase)
(lighter phase)
Adapted from chapter-opening photograph, Chapter 9, Callister 3e.
Phase Equilibria
CrystalStructure
electroneg r (nm)
Ni FCC 1.9 0.1246
Cu FCC 1.8 0.1278
• Both have the same crystal structure (FCC) and have similar electronegativities and atomic radii (W. Hume – Rothery rules) suggesting high mutual solubility.
Simple solution system (e.g., Ni-Cu solution)
• Ni and Cu are totally miscible in all proportions.
Phase Diagrams• Indicate phases as function of T, Co, and P. • For this course: -binary systems: just 2 components.
-independent variables: T and Co (P = 1 atm is almost always used).
• PhaseDiagramfor Cu-Nisystem
Adapted from Fig. 9.3(a), Callister 7e.(Fig. 9.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH (1991).
• 2 phases: L (liquid)
(FCC solid solution)
• 3 phase fields: LL +
wt% Ni20 40 60 80 10001000
1100
1200
1300
1400
1500
1600T(°C)
L (liquid)
(FCC solid solution)
L + liquidus
solid
us
wt% Ni20 40 60 80 10001000
1100
1200
1300
1400
1500
1600T(°C)
L (liquid)
(FCC solid solution)
L +
liquidus
solid
us
Cu-Niphase
diagram
Phase Diagrams:number and types of phases
• Rule 1: If we know T and Co, then we know: --the number and types of phases present.
• Examples:A(1100°C, 60): 1 phase:
B(1250°C, 35): 2 phases: L +
Adapted from Fig. 9.3(a), Callister 7e.(Fig. 9.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH, 1991).
B (
1250
°C,3
5) A(1100°C,60)
wt% Ni20
1200
1300
T(°C)
L (liquid)
(solid)L +
liquidus
solidus
30 40 50
L +
Cu-Ni system
Phase Diagrams:composition of phases
• Rule 2: If we know T and Co, then we know: --the composition of each phase.
• Examples:TA
A
35Co
32CL
At TA = 1320°C:
Only Liquid (L) CL = Co ( = 35 wt% Ni)
At TB = 1250°C:
Both and L CL = C liquidus ( = 32 wt% Ni here)
C = C solidus ( = 43 wt% Ni here)
At TD = 1190°C:
Only Solid ( ) C = Co ( = 35 wt% Ni)
Co = 35 wt% Ni
Adapted from Fig. 9.3(b), Callister 7e.(Fig. 9.3(b) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH, 1991.)
BTB
DTD
tie line
4C3
• Rule 3: If we know T and Co, then we know: --the amount of each phase (given in wt%).• Examples:
At TA: Only Liquid (L)
W L = 100 wt%, W = 0At TD: Only Solid ( )
W L = 0, W = 100 wt%
Co = 35 wt% Ni
Adapted from Fig. 9.3(b), Callister 7e.(Fig. 9.3(b) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH, 1991.)
Phase Diagrams:weight fractions of phases – ‘lever rule’
wt% Ni20
1200
1300
T(°C)
L (liquid)
(solid)L +
liquidus
solidus
30 40 50
L +
Cu-Ni system
TAA
35Co
32CL
BTB
DTD
tie line
4C3
R S
At TB: Both and L
% 733243
3543wt
= 27 wt%
WL S
R +S
W RR +S
wt% Ni20
1200
1300
30 40 501100
L (liquid)
(solid)
L +
L +
T(°C)
A
35Co
L: 35wt%Ni
Cu-Nisystem
• Phase diagram: Cu-Ni system.
• System is: --binary i.e., 2 components: Cu and Ni. --isomorphous i.e., complete solubility of one component in another; phase field extends from 0 to 100 wt% Ni.
Adapted from Fig. 9.4, Callister 7e.
• Consider Co = 35 wt%Ni.
e.g.: Cooling in a Cu-Ni Binary
4635
4332
: 43 wt% Ni
L: 32 wt% Ni
L: 24 wt% Ni
: 36 wt% Ni
B: 46 wt% NiL: 35 wt% Ni
C
D
E
24 36
Equilibrium cooling
• The compositions that freeze are a function of the temperature
• At equilibrium, the ‘first to freeze’ composition must adjust on further cooling by solid state diffusion
• We now examine diffusion processes – how fast does diffusion happen in the solid state?
Diffusion
Fick’s law – for steady state diffusion
J = flux – amount of material per unit area per unit time
C = conc
dxdC
DJ
C
x
Co
C
x
Co/4Co/2
3Co/4
x2
t = 0
t = 0
t = t
Dtx 2
Diffusion cont….• Diffusion coefficient D (cm2s-1)
• Non-steady state diffusion
Fick’s second law
• Diffusion is thermally activated
RTQ
DD dexp0
2
2
xC
DtC
Non – equilibrium
cooling α
L
α + L
• C changes as we solidify.• Cu-Ni case:
• Fast rate of cooling: Cored structure
• Slow rate of cooling: Equilibrium structure
First to solidify has C = 46 wt% Ni.
Last to solidify has C = 35 wt% Ni.
Cored vs Equilibrium Phases
First to solidify: 46 wt% Ni
Uniform C:
35 wt% Ni
Last to solidify: < 35 wt% Ni
Mechanical Properties: Cu-Ni System• Effect of solid solution strengthening on:
--Tensile strength (TS) --Ductility (%EL,%AR)
--Peak as a function of Co --Min. as a function of Co
Adapted from Fig. 9.6(a), Callister 7e. Adapted from Fig. 9.6(b), Callister 7e.
Ten
sile
Str
eng
th (
MP
a)
Composition, wt% NiCu Ni0 20 40 60 80 100
200
300
400
TS for pure Ni
TS for pure Cu
Elo
nga
tion
(%
EL
)Composition, wt% Ni
Cu Ni0 20 40 60 80 10020
30
40
50
60
%EL for pure Ni
%EL for pure Cu
: Min. melting TE
2 componentshas a special compositionwith a min. melting T.
Adapted from Fig. 9.7, Callister 7e.
Binary-Eutectic Systems
• Eutectic transition
L(CE) (CE) + (CE)
• 3 single phase regions (L, ) • Limited solubility: : mostly Cu : mostly Ag • TE : No liquid below TE
• CE
composition
Ex.: Cu-Ag system Cu-Agsystem
L (liquid)
L + L+
Co wt% Ag in Cu/Ag alloy20 40 60 80 1000
200
1200T(°C)
400
600
800
1000
CE
TE 8.0 71.9 91.2779°C
L+L+
+
200
T(°C)
18.3
C, wt% Sn20 60 80 1000
300
100
L (liquid)
183°C 61.9 97.8
• For a 40 wt% Sn-60 wt% Pb alloy at 150°C, find... --the phases present: Pb-Sn
system
e.g. Pb-Sn Eutectic System (1)
+ --compositions of phases:
CO = 40 wt% Sn
--the relative amount of each phase:
150
40Co
11C
99C
SR
C = 11 wt% SnC = 99 wt% Sn
W=C - CO
C - C
=99 - 4099 - 11
= 5988
= 67 wt%
SR+S
=
W =CO - C
C - C=R
R+S
=2988
= 33 wt%=40 - 1199 - 11
Adapted from Fig. 9.8, Callister 7e.
• 2 wt% Sn < Co < 18.3 wt% Sn• Result:
Initially liquid + then alonefinally two phases
poly-crystal fine -phase inclusions
Adapted from Fig. 9.12, Callister 7e.
Microstructures in Eutectic Systems: II
Pb-Snsystem
L +
200
T(°C)
Co , wt% Sn10
18.3
200Co
300
100
L
30
+
400
(sol. limit at TE)
TE
2(sol. limit at Troom)
L
L: Co wt% Sn
: Co wt% Sn
Lamellar Eutectic Structure
Adapted from Figs. 9.14 & 9.15, Callister 7e.
Intermetallic Compounds
Mg2Pb
Note: intermetallic compound forms a line - not an area - because stoichiometry (i.e. composition) is exact.
Adapted from Fig. 9.20, Callister 7e.
Eutectoid & Peritectic
Cu-Zn Phase diagram
Adapted from Fig. 9.21, Callister 7e.
Eutectoid transition +
Peritectic transition + L
• Phase diagrams are useful tools to determine:--the number and types of phases,--the wt% of each phase,--and the composition of each phase
for a given T and composition of the system.
• Alloying to produce a solid solution usually--increases the tensile strength (TS)--decreases the ductility.
• Binary eutectics and binary eutectoids allow for a range of microstructures.
Summary
3)Thermodynamics of mixing• Why do some things mix and others
not?• Why does heating promote mixing?• Is an alloy of two semiconductors a
random solid solution, or is it just a mixture of two phases?
Solid solution Mixture of phases ?OR
Free energy of mixing
• Enthalpy term – bond energies• Entropy term – stats of mixing
behaviour of mixtures
consider the case: Pure A + Pure B mixture AxB1-x
mixmixmix STHG
Enthalpy term: Compare the energy in bonds (VAB etc) before and after
mixingHeat of mixing Binding energy
in mixture compared to average in components
Binding energy change on mixing
ΔHmix
Exothermic VAB > 1/2(VAA + VBB) Stronger -ve
Ideal VAB = 1/2(VAA + VBB) Same 0
endothermic VAB < 1/2(VAA + VBB) Weaker +ve
ΔHmix derivedΔHmix = (H for mixture AB) – (H for pure A + H for pure B) • ΔHmix = zNxA xB {1/2(VAA + VBB) – VAB}
NAA etc = number of bonds of type AA, AB etcNA, NB, number of atoms of A, B. NA + NB = NVAA, VBB, VAB, = binding energies of pairs A-A, xA etc = NA/N = mole fraction of AxB = NB/N = mole fraction of B. xA = (1-xB )z = co-ordination number
For reference only
ΔSmix derived
ΔS = (Sfinal – Sinitial)
Sfinal from the statistical definition S = kBlnw
ΔSmix = kBln
Using Stirling’s approximation lnn! = nlnn-n gives• ΔSmix = -NkB[ xAlnxA + xBlnxB]
)!()!(
!
BA NxNxN
For reference only
ΔGmix - the formula
• ΔGmix = zNxA xB {1/2(VAA + VBB) – VAB}
+ NkBT[ xAlnxA + xBlnxB]
Binding energy term {...} may be +ve or –ve
Entropy term [...] is negative
mixmixmix STHG
For reference only
changing ΔHmix e.g. #3
iv) ΔHmix large and positive
• Enthalpy dominates
• miscibility gap exists
• In gap, solid segregates into xA’ and xA’’
0 0.51
En
erg
y
ΔSΔHΔG
x’ x’’
Miscibilitygap
x
T
Binary isomorphous
phase diagram
Mole fraction x
T
L
L + solid
solid
Phase diagram with
miscibility gap
Mole fraction x
T
L
L + solid
solid
x’ x’’
02
2
xG
locus
or ‘spinode’
A
B
Example phase diagramCdTe-CdS
L+
+
200
T(°C)
C, wt% Sn20 60 80 1000
300
100
L (liquid)
L+
183°C
• For a 40 wt% Sn-60 wt% Pb alloy at 200°C, find... --the phases present: Pb-Sn
system
Adapted from Fig. 9.8, Callister 7e.
Pb-Sn Eutectic System (2) + L
--compositions of phases:CO = 40 wt% Sn
--the relative amount of each phase:
W =CL - CO
CL - C=
46 - 40
46 - 17
= 6
29= 21 wt%
WL =CO - C
CL - C=
23
29= 79 wt%
40Co
46CL
17C
220SR
C = 17 wt% SnCL = 46 wt% Sn
Worked example of earlier concept