SOLUTION FYS4310 problem 200-12
(Defects, Si vacancy charge state, intermediate-easy)
Restate problem:
Assume here the vacancy in Si can exist in 4 different charge states: V+,V*,V-,V= .
(positive, neutral, minus, double minus). In the text book is given a relationship between
the concentration of neutrals and negatively charged vacancies, equation (2-4)
 E i − E V− 
E −E − 
n
n
−
[V ] = [V *] n exp kT  or with N symbols NV − = NV n exp i kT V  [eq1]
i
i
(The index 0 in book indicates equilibrium). Show/argue that this equation is correct.
Solution:
We will solve this the ‘backward’ way, starting with what we are going to show, and
€ to.
derive what showing [eq1] is equivalent
We have from problem 200-9, see the solution for its derivation
[V − ] = n [eq2] where the sub-index i indicates the intrinsic case. Thus we have
[V − ] ni
€
i
n
[V ] = n [V ]
−
showing
€
i
−
i
[eq3] . Now comparing [eq3]and [eq1], showing [eq1] is equal to
 E i − E V− 
 E i − E V− 
n −
n
−
i.e.
V
=
V
*
exp
V
=
V
*
exp
[ ]i [ ] n  kT  [ ]i [ ]  kT  [eq4]
ni
i
since
(The neutral vacancy concentration does not depend upon the Fermi
level), showing [eq4] is equivalent to show that
 E i − E V− 
−
V
=
V
*
exp
[
]

€[eq5]
[ ]i
i
kT


Check the meaning of [eq5]: It relates the concentration of negatively charged vacancies
to the concentration of neutral vacancies – in the intrinsic case. We should now realize
that [eq5] makes perfectly sense; We know that the vacancy concentration is given by an
Arrhenius type expression with the Boltzmann factor exp(-Ea/kT) where Ea is the
activation energy = mean free energy change. That type of expression will apply for each
charge state of the vacancy, but the free energy change for different charge states are
different. Specifically for the negative vacancy we can create the vacancy by first
creating a neutral vacancy and then add an electron to it. The average work needed to do
this is the difference between the Fermi level and E-, the energy of the top occupied level
of the negative vacancy. The Fermi level in the intrinsic case is the intrinsic Fermi level,
Ei. So the task has been accomplished. We can rationalize [eq5]
====
We could then construct an alternative way, by going the opposite way of what we just
did, ie me move forward . Lets start with the expression for the vacancy concentration
given in the lecture notes. Then we have
 ΔS   −ΔE V 
NV = N exp V  exp
 [eq6]
 k   kT 
€
€
€
€
Here N is concentration of lattice sites, ∆S V the difference in entropy of the system after
creation of 1 vacancy, not including the entropy of mixing, and ∆EV is the enthalpy
difference of the system upon creation of a vacancy. It consists mostly of the change in
internal energy. [ Digression, if we compare with the expression in the book, it would
mean that in order to get the expressions equal, we would have to consider the activation
energy in the book as a free energy change (including the entropy difference ), or the N
only on the order of the atom concentration ]
This type of relation [eq6] applies to all the charge states of the vacancy defect. Each
charge state will have its own characteristic activation energy though. Lets write explicit
relations for the negative vacancy V- and the neutral V*. We could then simply write
 ΔS −   −ΔE − 
 ΔS   −ΔE * 
[V − ] = N exp kV  exp kTV  , [eq6a], and [V *] = N exp kV *  exp kTV  [eq6b]
where ∆ E V − is the activation energy for the negative vacancy. Before going further we
should consider the entropy difference. This is going to be a factor on the order of 1. We
can further say that the difference in entropy for different charge states is not going to be
large, so as a first approximation we€can say the exponent with the entropy term is the
€same for all charge states. Lets multiply and divide the right hand side of [eq6a] by the
left and right hand side of [eq6b] respectively. i.e.
 ΔS −   −ΔE V − 
 −ΔE V − 
N exp V  exp
exp


 k   kT 
 kT  [V *] exp −ΔE V − + ΔE V * 
−


≈ [V *]
=
[V ] = [V *]
kT
 −ΔE V * 


 ΔSV *   −ΔE V * 
N exp
exp
 exp


 k   kT 
 kT 
where we have used that the entropy differences are not large and similar.
We realize the change in internal energy for the negative vacancy must me equal to that
€
of the neutral plus the ionization energy, thus
€
€
 −(ΔE V * + (E V − − E F )) + ΔE V * 
 −((E V − − E F + E i − E i )) 
= [V *] exp
 = [V *] exp

kT
kT




where the ionization energy is EV--EF which is the average work done in putting an
electron onto the vacancy
€
 E − EV −   E F − E i 
= [V *] exp i

 exp
 kT   kT 
[eq7]
We have that the electron concentration is given by
 E − Ei 
 E − Ei  n
n = n i exp F
[eq8]
 ⇒ exp F
=
€
 kT 
 kT  n i
putting equation [eq8] into equation [eq7] we have
E −E − 
n
[V − ] = [V *] n exp i kT V  Q.E.D
€
i
€