Lecture 8: Phase Diagrams

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