In the growth of superlattice ABAB , the growth is

1
Part 7 Solid Phase Amorphization, Crystallization,
and Epitaxy
?In the growth of superlattice (A/B/A/B ), the
growth is asymmetrical. If A wets B, B will
not wet A. In selecting a pair of materials
for superlattice or 3-dimensional structure
growth, we must consider their surface
energies as well as the interfacial energy
between them.
B
A
The interfacial misfit and thermal strain can
be calculated. However, the surface energies
are hard to measure and are generally not
available. Knowing the surface energy
difference between two materials will enable
us to find ways to bring them close in order
to grow the superlattice with flat interfaces.
2
In the following, an indirect method of
measuring surface energy by the kinetics of
nucleation will be described. Crystallization
of amorphous CoSi2 and amorphous Si thin films
will be the examples. ?The Metastable State
Amorphous phase are in metastable state.
Metastable state is a state relative to stable
state. First, define the stable state. A
stable state has the lowest free energy at a
given set of thermodynamic variables and is
defined by equilibrium phase diagram (e.g. p-T
phase diagram). The phase boundaries indicate
?p ?T
solid
?T
liquid
p
TP
?p
gas
T
Isotherm
liquid
supersaturation
p
LG
gas
V
3
reversible transition between the phases. It
is possible to cross these phase boundaries
without a transition to obtain metastable
phases. For example, to have an undercooled
liquid without solidification, one could follow a
proper kinetic path of cooling (?T) to
suppressed the nucleation of the stable phase.
Barrier exists in the process of nucleation
of new phases. In the isotherm curve of the
pressure-volume phase diagram, the broken line
which represents the supersaturation (p/p0 gt
1) of the gas phase. The supersaturated gas is
in a metastable state. Metastable phases can
also be formed by cooling under a high
pressure, e.g. to form diamond. The nucleation
and growth of metastable phases do occur for
the case of diamond, but, with different
configuration.
4
The processes lead to metastable phase
formation involve changes in temperature
and/or pressure. The process could be very
fast, like explosion (p, T) and quenching (T).
In the case of thin film reactions, the processes
is conducted at constant temperature and
constant pressure. What do we vary to produce
the metastable (amorphous) phase? The answer
is composition! The Gibb free energy is
For a binary system in reaction at constant p
and T, we have
The reaction must always occur in the
direction toward equilibrium, i.e. to lower
its free energy. The reaction seeks a minimum
in the free energy vs composition curve G(c1).
In a typical binary system which forms
intermetallic compounds, there can be several
minima in the G(c1)
5
curve. The reaction between A and B can
form either amorphous AB or A?B phase. The
composition of the product phase is determined
by the tangents extending from the free
energy curves to the chemical potential of the
element A and B.
GoB
G
?-AB
GoA
A?B
?C?eq
c1
The transformation from the elements to the
amorphous alloy or to the intermetallic
compound can not be achieved by a continuous
change in composition (since the process
requires the free energy to increase). Therefore,
the transformation occurs by a jump from one
composition to another, which means that
nucleation is required.
6
Because nucleation barrier for different
phases can be different. ? phase with lower
barrier forms first, if the amorphous alloy
can nucleate preferably ? solid phase
amorphization.
Kinetic Processes to form Metastable Phases
Variables Slow Process
Fast Process Temperature slow
cooling to form Quenching
supercooled liquid
Pressure slow build-up of the
Explosion
pressure in a mechanical
mixing
supersaturated gas, Concentration slow
reaction by solid Ion Implantation
state amorphization,
slow dissolution of
sugar in water to
form supersaturated
solution
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?Solid Phase Amorphization Amorphous solid is
a phase with atoms arranged without long range
order. The first nearest neighbor covalent bonds
are virtually unchanged, however, the bond
angles are changed so that the long range
correlation of atomic positions are lost.
Amorphous could be formed by processes that
involve rapid changes in temperature,
pressure, composition, or a combination of them
(a high rate of energy change). Here, we
will focus on the formation of amorphous
alloys by a slow reaction of bilayer thin
film and consider the rate of energy change
in the process. Examples Ni-Zr, Al-Pt, Rh-Si,
Ti-Si, etc.
S.L. Cheng, et. al., Micron 33, 543 (2002)
8
Bilayer State A/B
(1) kinetic barrier ?H1 (2) ?G1 (3)
?GTransformation (4) kinetic barrier ?H0 (5)
?G0 (6) linetic barrier ?H2
Amorphous State - ?AB
Energy
(4)
(1)
(6)
(3)
(5)
(2)
Stable state A?B
Reaction coordination
The kinetic barriers ?H0 and ?H1 are
activation energies for nucleating the
critical nuclei of the stable phase and the
amorphous phase at the interface, respectively.
The ratio of their nucleation rates at the
interface is
The next question is finding the ?H. Following
the classical nucleation theory, assume the
free energy
9
changes from the old phase to the new phase
for an atom is ?G and the interfacial energy
for the newly form phase is ?. When a new
cluster of n atoms forms from the old phase,
the total energy change is where ? is the
geometrical shape factor.
?GT
The nucleation will occur only when ?G is
negative, i.e. the atoms have lower free
energy in the new phase. The question is to
find the critical energy for the phase to
nucleate and define the energy to be the
nucleation barrier (?H) for the phase
transformation.
??n2/3
?H
n
nc
n?G
Therefore, ?H is ?GT(nc).
10
A geometrical related factor!
Therefore, ?H0 A0(?0)3/(?G0)2 and ?H1
A1(?1)3/(?G1)2.
Since ?G0 is very close to ?G1 ? the factor
involves both terms is close to 1. Therefore,
the reaction is controlled by ? terms. To
have the transformation stop at the metastable
state, the barrier for the transformation of
the metastable state to the stable state (i.e.
?H2) should be higher than ?H1. Otherwise, the
transformation could reach the final stable
state once it started. The amorphous phase
nucleates because its barrier is lower
11
than that of the crystalline phase. This has
been proposed as a key condition for the
occurrence of solid state amorphization. If
the difference between ?G0 and ?G1 is about
10, a 10 decrease in surface energy will make
the nucleation of the amorphous phase
overwhelming. In further considering the
metastability of the amorphous phase, the
nucleation of the crystalline phase has to be
considered. The incubation time of the
crystalline phase has been defined as the
metastable period of the amorphous phase (i.e.
the nucleation and growth of the crystalline
phase from the amorphous phase have to be
considered.).
12
?-AB has grown to a thickness of X. The
thickness W is required for A?B to form a
continue layer. It means physically that the
grains of the compound have nucleated and have
grown to joint each other. When A?B is
continuous, the amorphous layer can not
coexist with it. Since, the red line in the
following figure (G curve) is higher than
the yellow line.
B
A?B
W
?-AB
X
A
GoB
G
?-AB
That is the decomposition of ?-AB into A
and A?B is more energetically favorable.
Therefore, the metastablilty of the amorphous
phase is defined as the period that the
crystalline phase has gained the thickness W.
GoA
A?B
?C?eq
c1
13
In the competing growth between two phases, we
shall consider the role of growth rate. Then
the change of the free energy is rate
dependent. Express the free energy change in a
given time t0 as
Further define the rate of energy change,
F driving force of the reaction v (dx/dt)the
growth rate
The meaning of the above two equations is if
the G- curve of the amorphous phase does not
lie much above the tangent line between A and
A?B (i.e. the driving force for the two phases
are comparable), the initial phase is
determined by the nucleation and growth rate of
the phase. Therefore, the amorphous phase is
favored.
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As ?-AB gets thicker, the growth become
diffusion- controlled and must slow down. Then
the advantage disappears. The initial rapid
growth ? not enough time for the crystalline
phase to nucleate ? ?-AB gets thicker ? the
growth becomes diffusion controlled ? growth
rate slows down ? to the point where the time
required for the amorphous layer to advance a
distance W (W/v) is comparable to the time
needed for forming the continuous crystalline
layer ? the metastability of ?-AB is over. The
time required for the crystalline phase to form
a continuous layer is the sum of its
incubation time to nucleate plus the grain
growth time to joint each other. gt the
condition when the amorphous phase begins to
disappear is
Define a critical thickness of the amorphous
phase beyond which it is unstable.
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In the above discussion, the metastability of
slow amorphization at a constant temperature
and pressure thin film reaction has been
considered.
?Amorphous-to-Crystalline Transformation and
Avramis Equation The transformation from
an amorphous state to crystalline state is an
important subjects frequently deal with for
alloyed and semiconductor thin films. In
crystallization, the product can have a
different composition from the amorphous phase
? phase separation into a crystalline and an
amorphous phase, or into two crystalline
phases ? long range diffusion is required.
If the amorphous phase has the same composition
as the crystalline phase ? no long range
diffusion is required.
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Example co-deposited amorphous films of
CoSi2, crystallize to form CoSi2 grains. See
pictures in the textbook. In the initial
stage, there is a small number of grains
nucleated randomly. Random in space, but not
random in time. They are typically assumed to
have a steady state nucleation rate. The size
and the number of grains increase with time.
Impingement of some grains will occur and form
a grain boundary. ? polycrystalline
microstructure. Tailoring the nucleation and
growth process ? control the evolution of
microstructure ? control extrinsic film
properties!
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The classical analysis of the phase
transformation by Mehl, Johnson, and Avrami
was used to describe the crystallization
process by nucleation and growth. Define XT
the fraction of volume transformed X Xext
fraction of the extended volume the following
equation is used to describe the overall behavior
To understand the physical meaning of the
equation V total volume involved in the
transformation Vu untransformed volume VT
transformed volume
Xu fraction of volume untransformed
Assume the transformation occurs by nucleation
and growth and the nucleation is random in
time and in space. The nucleation rate is I
( of nuclei/unit time-volume).
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In the interval between t and t d?, a number
of newly transformed regions nucleated in Vu,
Further assume that each of the transformed
regions has the same isotropic growth rate of
G, the volume of a transformed region which
originated at t is
? in the period d?, the differential change of
the transformed volume is
- sign represents volume reduction
? the total transformed volume at time t is
In the initial stage of the transformation, VT
ltlt Vu, Vu ? V,
Correct only at the initial stage where the
transformed region do not interfere with each
other!
19
To correct the interference, introducing the
concept of the extended volume Vext in the
period between t and t d? to be
Comparing this equation to
The difference is that the dVext includes the
random nucleation and isotropic growth in both
the transformed and untransformed regions.
Phantom nuclei
The nuclei in the transformed region are
phantom nuclei and their growth does not
affect the untransformed region. The growth
impingement also makes no contribution to
increase the fraction of the transformed
region. since
Growth Impingement
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Only the fraction of untransformed region
counts!
constant
Vext 0 when VT 0, C 0
If I and G are constants
Assume I and G follow the Boltzmanns distribution
Therefore,
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1
Experimentally one could measure XT as a
function of time and temperature, and we
can choose a constant value of XT, e.g.
0.5, then we obtain
T3 gtT2gtT1
Fraction recrystallized
T2
T3
T1
Log time
0
t1
t2
t3
lnt
constant
Slope (?Hn3?HG)/4kB
If we could measure one of them (?Hn or
?HG) separately ? know the other. If ?Hn is
known ? interfacial energy between amorphous
and crystalline ?AC using the relation
1/T
We could measure ?GAC by calorimetry
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The t4 dependence is only good for the above
assumption. In general, the growth could be
isotropic or anisotropic, it can be linear
with time or it can vary with the square root
of time (when it is diffusion limited). So, the
general form of Avramis equation is
Value of n in Kinetic law
(a) Polymorphic changes, discontinuous
precipitation, interface controlled growth,
etc.
Increase nucleation rate
n gt 4 Constant nucleation rate
n 4 Decreasing nucleation
rate 4 gt n gt 3 Zero
nucleation rate
n 3 Grain edge nucleation after saturation
n 2 Grain boundary nucleation after
saturation n 1
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(b) Diffusion Controlled Growth
All shape growing from small dimensions,
Increase nucleation rate
n gt 2.5 Constant nucleation
rate n
2.5 Decreasing nucleation rate
2.5 gt n gt 1.5 Zero
nucleation rate
n 1.5 Growth of particles of
appreciable initial volume 1.5 gt n gt
1 Needles and plates of finite long dimensions,
small in comparison with their separation
n 1 Thickening of long cylinders
n
1 Thickening of large plates
n 0.5 Precipitation on
dislocations (very early stage) n
2/3
From J.W. Christian, The Theory of
Transformations in Metals and alloys Pergamon
Press, 1975.
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?Measurement of Crystallization of Amorphous Thin
Film To measure XT Direct method
sectioning technique to measure the
transformed region directly Indirect method
such as integrating X-ray diffraction
intensity, resistivity change, and
differential scanning calorimetry
all rely on assuming that the
measured quantity is proportional to the volume
of phase change (?) Example
crystallization of CoSi2 from ?-CoSi2, the
resistivity of ?-CoSi2 (??) is larger than
c-CoSi2 (?c), the XT is expresses in terms of
the resistivity of the film at time t (?t),
the linear relation has been found to exist
for XT in the range of 0 to 0.4 by comparing
with direct measurement.
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0
T1
T3
Take XT below 0.4
n 4
XT
ln ln1/1-XT(t)
1
Annealing time
ln (t/?)
?HT (?Hn ?HG)/4 0.9 eV was obtained for
CoSi2. ?HG has been reported to be 1.09 eV for
the CoSi2 growth ? ?Hn 0.3 eV The heat of
crystallization of ?-CoSi2 is ?HAC 0.05
eV/atom ? ?AC (0.3 ? 0.05 ? 0.05)1/3 0.09
eV. Critical nucleus (Nc) 8 atoms For Si
?HT 3.4 eV, ?HG 2.7 eV, ?Hn 5.5 eV, ?HAC
0.1 eV, ?AC 0.4 eV, Nc 64 atoms.
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?Unique Features of Phase Changes in Thin Films
(1) the growth becomes 2-D when the crystal
reach a size greater than the film
thickness (2) the nucleation favors
heterogeneous sites on the film surface
and the interface with substrate rather than
homogeneous sites within the film (3) the
substrate may influence the transition in the
film, especially when the film react
with the substrate (4) thin film samples can
be observed in-situ during phase changes
?Hyperbolic Grain Boundary When two grains
meet each other, a grain boundary is formed.
If the two grains are equal in size, i.e., if
they nucleate at the same time and grow at the
same rate ? the boundary formed is straight.
The boundary will be
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curved if the two grains are unequal in size.
If we assume that (a) the nucleation is
random and continuous, (b) the growth is
constant and isotropic, and (c) the grain
boundary do not migrate. ? all the grain
boundaries formed are hyperbolic (consider 2D
only, in 3D the boundaries is hyperboloids).
Assume two grains separate by a distance 2c,
the time interval between the nucleation events
at A and B be ?t, and the growth rate is u ?
the hyperbola is
y
x
a u?t/2, b (c2-a2)/2
all parameters c, u, ?t can be measured
experimentally. For a pair of grains, the ?t
can be calculated by knowing c and u.
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all parameters c, u, ?t can be measured
experimentally. For a pair of grains, the ?t
can be calculated by knowing c and u. The
triple point in the amorphous-to- crystalline
interfaces must maintain a fixed geometry in
order to satisfy the equilibrium condition (?
2?icos?).
?i
?
?
?
The formation of grain boundary starts with an
angle smaller than ? and then become greater
than ?. The macroscopic nonequilibrium
situation can be due to a dominant isotropic
growth.
?Grain Growth The driving force for the grain
growth is the reduction of the total grain
boundary energy. In some cases, the grain
boundary move against the trend due to other
kinds of
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driving force, such as chemical diffusion
induced grain boundary migration (DIGM). Here
we only focus on the driving force for the
reduction of grain boundary area. The velocity
of a diffusion atom
F driving force
The pressure difference on a curved grain
boundary is
? grain boundary energy per unit area r radius
?p is the force per unit area, define M as the
mobility per unit area ? Take v dr/dt
gt describe the shrinking of the grain For
growth, assume the growing grain has an average
radius of R and assume that R r constant.
120o
Shrinking of neighboring grain growth of the
grain
Stable
Shrink
grow
30
B Temp. dependent and contains M and ? n 0.5
Experimentally, n has been found to be about
0.5 to 0.4. The deviation is explained as the
delay created when a new set of grain begins
to be consumed. Estimation of the work
required by grain growth p?V 2??/r ?
2.7?10-23 cm3 take ? 500 ergs/cm2 r 100
nm ? p?V 1.5?10-3 eV/atom. 2 to 3 order of
magnitude smaller than typical chemical energy
per atom is 0.1 1 eV/atom. ? the
driving force to move the amorphous-crystalline
interface is 2 to 3 orders of magnitude greater
than that to move the hyperbolic grain
boundaries. When the mobility of the interface
and grain boundaries are comparable, the one
with weak driving force (grain boundary) does
not move.
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The mechanism for the mobility of grain
boundary is unclear. It is known that when a
pair of grains can be related by the
coincidence lattice sites their grain boundary
has a high mobility. Impurities have a very large
effect on grain boundary mobility, except for
some interstitial solutes. In annealing
thin films, some exceptionally large grains
are formed due to the secondary grain growth
under a driving force of reduction in surface
energy (grains with low index surface).
Consider a film (with film thickness h) after
normal grain growth, a section of film with
surface area A composed of a lot of small
grains is established. If we compare the
surface energy for the grains with orientation
that minimize surface energy (?min) with the
average surface
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energy of the normal grains ( ), one can
define
a measure of the surface energy anisotropy
If area A undergoes secondary grain growth,
the surface energy per unit volume Fs is
reduced by
assume
Driving force for surface energy driven secondary
grain growth
Secondary grain growth
Further approximation the secondary grains
are cylindrical with radius rs, the normal
grain are hexagonal cylinders with uniform
diameter dn (rn dn/2) ? total driving force
for the secondary grain growth is
h
? geometrical factor ?gb average grain
boundary energy
If rn h, ? 1
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For a highly textured thin film, the driving
force of secondary grain growth is removed (??
0). The grain boundary of the textured thin
film is typically tilt-type. In this
situation, the grain growth of a large grain can
still take place by coalescence (for small
tilt angle only).
?Solid Phase Epitaxy (without a medium) Solid
phase epitaxy (SPE) means to convert a
non-epitaxial film to a epitaxial film on a
single crystal substrate. Example
Ion implant
From the viewpoints of device application,
it is of interest when Si on SiO2 (SOI, silicon
on insulator) can be achieved by combining
SPE and lateral growth of Si over SiO2
surfaces. Two methods can be used to measure
the
?
heating
?
SPE
34
SPE growth (1) Rutherford (Ion)
backscattering spectrometry (RBS) can be
used to measure the SPE growth. An ion beam
is directed along a major channeling direction
of the single crystal wafer, so
channeling occurs in the substrate but
not in the amorphous Si film. ? ?-Si
backscatters the ion beam more than the Si
substrate ? separating the two regions.
(2) Time resolved reflectivity (TRR)
measurement the interference between the
light reflected from the air/?-Si interface and
?-Si/c-Si interface ? thickness of the
?-Si. If ? 632.2 nm The
growth rate (v) can be measured in-situ
as different T.
reflectivity
?/2 326 nm
Thickness of ?
35
For Si v0 3?10-8 cm/sec ?HG 2.7 eV
A measure for ?HG ? ?Hn 4?HT - 3?HG for Si,
?HT 3.4 eV ? ?Hn 5.5 eV Since ?HAC
0.1 eV/atom (DSC) ? ?AC 0.4 eV/atom, much
larger than the surface energy of CoSi2 ?
Since the mismatch between Si and CoSi2 are small
? Epitaxial growth is possible the growth of
CoSi2 on Si is OK, while the growth of Si on
top of CoSi2 tends to have pinhole formation
(due to large surface energy of Si)! The
knowledge of the surface energy is obtained
indirectly.
?Solid Phase Epitaxy (with a medium) One can
achieve SPE with a medium for mass transport.
For example, one can interpose a layer of PdSi2
or Au between ?-Si and c-Si to achieve SPE.
36
The advantage of the medium is that it can
reduce the growth temperature. The medium
chosen should contain an element which react
easily with the growth materials, e.g. one of
the noble and near-noble metals for Si. SPE
without a medium interfacial reaction process
SPE with a medium interfacial reaction and
diffusion controlled process. For the
case of ?-Si/Au/c-Si, one can imagine that the
Si atoms in the ?-Si dissolve into Au and
deposit on the c-Si to achieve SPE growth.
In the case of ?-Si/Pd2Si/c-Si, one can again
imagine the Pd2Si serves as a transport medium
for Si to diffuse from the ?-Si to c-Si.
?-Si
Au
c-Si
Csi
Concentration gradient
37
Alternative thinking to achieve the same
goal. One can imagine that Pd2Si decomposes
into Pd and Si at the Pd2Si/c-Si interface.
The decomposed Si will grow epitaxially on the
c-Si and Pd diffuses across the Pd2Si to the
?-Si/ Pd2Si interface and react with ?-Si to
form new layers of Pd2Si. In one case, the
diffusing species is Si and in another case Pd
is the diffusing species. Using experiments to
determined which one is the case (will be
discussed in next part).
?-Si
Pd2Si
c-Si
CPd
Concentration gradient
A(?)
A(c)
A?B
Consider a general case JA the flux of
direct transport of A atoms from the
A(?) to A(c).
JA
JB
JA
marker
x1
x2
xm
38
JB the flux of B atoms due to decomposition
of A?B at the A?B/A(?) interface
JA the release of A atoms due to the
decomposition of A?B at the A?B/A(?)
interface
The relation between JA and JB is governed by
the stochiometry of the compound A?B. The
fluxes with in the compound are
DA, DB diffusivity of A and B atoms in A?B phase
At position x1, the growth of the crystalline
phase A(c) is
the concentration of A atoms in A(c)
the volume of A atom in A(c)
Assume the whole system is in steady state ?
layer thickness of A?B is invariant ?
39
At the marker position xm, one can regard its
motion as the sum of dx1/dt and the relative
motion of marker with respect to x1,
The marker motion with respect to x1 is due to
the flux of JB only!
?B is the molecular volume of an A?B molecule in
the compound
Assume that the atomic volume of A in all
phases are equal.

40
From the above two equations if xm - x1 0 ?
JB 0 and JA -(1/?B)(dx1/dt), i.e. the
epitaxial growth is due to JA along if the
marker migrates ? some decomposition at the
interface. To estimate the amount of
decomposition, one can evaluate the flux ratio
By measuring xm, one can determine the ratio
of the fluxes, and in turn the dominant
diffusion species. In the case of Pd2Si, the
diffusion species is Pd and the interfacial
decomposition dominated the reaction.