
Schottky
defects use
the second simple possibility to maintain
charge equilibrium in ionic crystals with two atoms in the base; they consist
of the two possible types of vacancies
which automatically carry opposite charge. 


Look at the illustrations in the link for visualizations
of Schottky defects as a well as other defects in ionic crystals. 


You may call the single vacancy in metals a Schottky defect, if you
like (some do), but that somehow misses the point. 


As pointed out in the context of
Frenkel defects, the vacancy in ionic
crystal carries a net
charge the same way a hole  a missing electron  carries a charge. 

We have postulated, that charge
neutrality must be maintained, but we have not proved it. Moreover, even for
equal numbers of oppositely charged vacancies (or Frenkel defects for that
matter), charge neutrality can only be maintained on
average; on a scale comparable to the average distance between the
point defects, we must have electrical fields which only (on average) cancel
each other for larger volumes. 


The total formation energy therefore must contain some electrostatic energy part because
we do have electrical fields around single point defects. The same is true for
the interstitials in Frenkel defects or in the general case of mixed defects,
and the consideration we are going to make for vacancies applies in an
analogous way to interstitials, too. 


Moreover, the electrical field of one vacancy
will be felt by other charged point defects, which means that there is also
some electrostatic interaction between vacancies, or vacancies and other
charged point defects. This interaction is stronger and has a much larger range
than the elastic interactions caused by the lattice deformation around a
defect. 

Let's look at a relatively simple
example. Here we are only going through the physical argumentation, the
details of the calculations
are contained in the link. 

Calling the formation energy of the
anion vacancy (= missing anion = positively charged vacancy) H
^{+}, the formation energy of the cation vacancy H
^{–}, and the binding energy between close pairs H
^{B}, we obtain for DG, the change in free enthalpy, upon
introducing n^{+}, n^{–}, and
n^{B} anion vacancies, cation vacancies, and vacancy
pairs, resp., is 

DG
= 
ó
õ 

ó
õ
V 

ó
õ 
æ
è 
H ^{+} ·
n^{+}(r) + H ^{–}
· n^{–}(r) +
[H^{+} + H ^{–}
– H^{B}] ·
n^{B} (r) + 1/2
r (r) V (r)
– T S_{n} 
ö
ø 
dx dy dz 




With r(r) =
local charge concentration, V(r) = electrical
potential, S_{n} = entropy of mixing;
r is the space vector. The usual sum is replaced by an
integral because the electrical potential is a smooth function and not strongly
localized. 

The number of point defects is now
dependent on r. The (noncompensated) electrical charge
stems from the charged vacancies, the net electrical charge density at any
point r is thus given by 


r (r) 
= 
e · 
æ
è 
n^{+}
(r_{ }) – n^{–}
(r) 
ö
ø 




With e being the elementary charge. 

The electrostatic potential follows
from the charge density via the
Poisson
equation, we have 


DV (r)

= – 
1_{ }
ee_{0} 
· r (r) 




With e = dielectric constant
of the material and e_{0} = vacuum
constant 

For equilibrium condition,
DG must be minimal, i.e. we
have to solve the variation problem 





for variations of the n's.
Using the conventional approximations one obtains solutions being rather
obvious on hindsight: 


n^{+}(r) 
= 
N · exp 
H ^{+} – e · V
(r)
kT 

n^{–}(r) 
= 
N · exp – 
H ^{–} + e · V
(r)
kT 

n^{B} 
= 
N · z · exp 
H ^{+} + H ^{–}
– H^{B}
kT 




For uncharged vacancies (where the – eV term would
not apply), this is the old result, except that now the divacancy is
included. 

We still must find the electrostatic
potential as a function of space. It may be obtained by expressing the charge
density now with the formulas for the charged vacancy densities given above,
and then solving
Poissons
equation. We obtain 


DV(
r ) = – _{ } 
e · N
ee_{0} 
· 
æ
ç
è 
exp – 
H ^{–} + e
· V (r)
k T 
– exp –

H ^{+} – e
· V (r)
k T 
ö
÷
ø 



This is a differential equation for
the electrostatic potential; the problem now is a purely mathematical one:
Solving a tricky differential equation. 


It is now useful to introduce a
"normalized" potential v by shifting the zero point in a
convenient manner, and by utilizing a useful abbreviation. We define 


v 
:= 
e · V(r) – 0,5(H
^{+} + H^{ –})
kT 
c^{–2} := 
2 · N ·
e^{2}
ee_{0} ·
kT

· exp – 
H ^{+} + H
^{–}
kT 




This gives us a simple looking differential equation for our
new quantities 


Dv (r) 
= 
c_{ }^{2} · sinh {v
(r)}^{ } 



c^{–1} has the dimension of a length, it
is nothing (as it will turn out) but the (hopefully) well known
Debye length for our
case. 

The "simple" differential
equation obtained above, however, still cannot be solved easily. We must resort
to the usual approximations and linearize, i.e. use the approximate relation
sinh v » v for small v´s.



We also need some boundary conditions as always woth
differential equations. They must come from the physics of the problem. 


The first guess is always to assume an infinite crystal with
v = 0 for x = ± ¥.
The solutions for an infinite crystal are trivial, however, we therefore assume
a crystal infinite in y and zdirection, but with
a surface in xdirection at x = 0; from there the
crystal extends to infinity. 


Now we have a onedimensional problem with
r = x. One general solution is (please
appreciate that I didn't state "obvious solution") 


v(x) = 
v_{0}^{2} –

æ
ç
è 
v_{0}'
c 
ö
÷
ø 
2 
+ 
æ
ç
è 
v_{0}+ 
v_{0}'
c

ö
÷
ø 
2 
· exp [2c
· (x – x_{0})] 
æ
ç
è 
v_{0} + 
v_{0}'
c

ö
÷
ø 
· exp [c
· (x – x_{0})] 






with v_{0} = v(x = 0) and
v_{0}' = integration constant which needs to be determined. 

Cool, but we can do better yet: 


If we do not want infinities, the divergent term exp
[2c(x – x_{0})]
must disappear for x approaching infinity, this means 


lim
x®¥ 
v(x) = 0 
Þ^{ }
v_{0} = – 
v_{0}'
c_{ } 
Þ
v(x) = 2v_{0}
· exp – c · x 



c^{–1} obviously determines at which
distance from a charged surface, or more generally, from any charge, the (normalized) potential (and
therefore also the real potential) decreases to 1/e  this is akin to
the definition of the Debye
length. 


For a charged surface and x >>
c (i.e. the bulk of the crystal) we obtain
v(x >> c) = 0, and therefore



V(r >> c) 
= 
H ^{+} + H ^{–}
2e 



If we substitute
V(r) into the equations for
the equilibrium concentrations above, we obtain the final equations for the vacany concentrations in the
case of Schottky defects 


n^{+} 
= 
n^{–} = N · exp – 
H^{+} + H ^{–}
2kT 




i.e. both concentrations are identical, and
charge neutrality is maintained! 


The energy costs of not doing it would be very large! That is
exactly what we expected all along, except
that now we proved it. 

More important,
however, we now can calculate what happens if there are electrical fields that
do not have their origin in the vacancies
themselves, but may originate from fixed charges on e.g. surfaces
and interfaces (including grain boundaries or precipitates), or from the
outside world. 


In this case the concentrations of the defects may be quite
different from the equilibrium concentrations in a neighborhood "Debye
Length" (= c) from the fixed
charges. 


We also see that we have (average) electrical neutrality in
the bulk and a statistical distribution of vacancies there, but this is not necessarily true in regions within one Debye
length c next to an external or
internal surface. 

Charged surfaces
thus may change point defect concentrations within about one Debye length. And
charged point defects, if they are mobile, may carry an electrical current or
redistribute (and then changing potentials), if surface or interface charges
change. 

This is the basic
principle of using ionic conductors (and, to some extent, semiconductors) for
sensor applications! 

The interaction between point
defects, the electrical potential and the Debye length may be demonstrated
nicely by plotting the relevant curves for different sets of parameters; this
can be done with the following JAVA module. 



We see that the
Debye length as expressed in
the formulas is strongly dependent on temperature. Only for high temperatures
do we have enough charged vacancies so that their redistribution can screen a
external potential on short distances. For NaCl we have, as an example,
the following values 


T
[K] 
c^{ –1} [cm] 
1100 
1.45 · 10^{
–7} 
900 
4.55 · 10^{
–7} 
700 
2.83 · 10^{
–6} 
500 
8.21 · 10^{
–5} 
300 
2.20 · 10^{
–1} 



The large values, however, are unrealistic in
real crystals, because grain boundaries, other charged defects, and especially
impurities must also be considered in this region; they will always decrease
the Debye length. 

