SOLID STATE 17 Packing and Defects
1.
Point Defects – Vacancies, Interstitials and Frenkel Defects
For a crystal containing N atoms, there is an equilibrium number of vacancies, nv, at a temperature T (in Kelvin) which is given by: EV n V N exp k BT
where EV is the energy of vacancy formation (in Joules) and kB is Boltzmann’s constant Diffusion is also thermally controlled; the diffusion coefficient, D, is given by: E D D0 exp D k BT
where ED is the energy of diffusion and Do is a diffusion constant. Point defects in metals can cause increased electrical resistivity at low temperatures dues to extra electron scattering. By cooling quickly from high temperatures T (quenching), nonequilibrium concentrations of vacancies may be obtained. The increased resistivity, R, is proportional to the number of defects, i.e. EV R Cn V CN exp k BT
and hence ln(R) ln(CN)
EV k BT
Impurities are foreign atoms which can substitute for atoms in the structure. intentionally they are called dopant atoms/ions.
If added
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2.
Line Defects and the Burgers’ Vector
Edge dislocation:
Screw dislocation:
For dislocations, we can define a vector B which shows the displacement of an atom due to the dislocation.
In both cases, the atom A would have been at A’ had the dislocation not occurred. Thus the Burgers’ vector, B, is from A’ to A. For the edge dislocation, left, the Burgers’ vector is perpendicular to the dislocation. For the screw dislocation, right, it is parallel to the dislocation. Twins are defects in which part of the crystal is a mirror image of the other:
These can be produced by stacking faults in close packed structures.
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PX3012 3.
Donor and Acceptor Doping of Silicon
Si doped with a group V element is an extrinsic, n-type semiconductor. It is “donor doped”. Si doped with a group III element is an extrinsic, p-type semiconductor. It is “acceptor doped.”
10.3a Classification of Solids Conductor
Insulator
-7
>10 m
ND; NA-ND = NA’ ND-NA = ND’ when ND>NA; Then in the following special cases, we can make simplifications: ND=NA;
n=p=ni
NA’ >> ni;
p ~ NA’, n ~ ni2/NA’
ND’ >> ni;
n ~ ND’, p ~ ni2/ND’
Example Find the electron and hole densities in a semiconductor if ni = 1016 m-3, ND = 1020 m-3 and NA = 1018 m-3.
We can then calculate the conductivity using the equation
= q(en + hp)
where are tabulated values for mobility of electrons (e) and holes (h) in a particular material. 4/5
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PX3012
Concept Question Why should n=p in an intrinsic semiconductor?
Problems Speed of light Boltzmann constant, electron volt
c kB 1 eV
= 2.998 108 ms-1 = 1.381 10-23 J K-1 = 1.602 10-19 J
1.
How many vacancies would you expect in a crystal of 1 million atoms at room temperature (300K) if Ev = 1eV? At 700K? At 1000K?
2.
The table below gives the change in resistivity relative to that measured at 20K for samples of a metal that have been quenched from various temperatures. By plotting a graph of ln(R) against 1/T, find the gradient and hence the energy of vacancy formation in this metal. Temperature, T (K) Resistivity change, %
3.
800 0.46
850 1.07
900 2.18
950 4.14
1000 7.17
The table gives the change in resistivity relative to that measured at 78K for samples of gold that have been quenched from various temperatures. By plotting a graph of ln(R) against 1/T, find the gradient and hence the energy of vacancy formation in gold. Temperature, T (K) Resistivity change, %
920 0.41
970 0.70
1020 1.40
1060 2.30
1220 9.00
4.
What are the equilibrium concentrations of holes and electrons at 300K in: (a) Silicon doped with ND = 3 1014 m-3 donors, ni = 1.5 1016m-3 ; (b) Germanium doped with ND = 3 1014 m-3 donors, ni = 2.5 1019 m-3 ? Comment on your results.
5.
Calculate the equilibrium number densities of holes and electrons for the following cases (assuming full ionisation of donors/acceptors): Silicon, ni = 1.5 1016m-3, doped with 1018 acceptors. Germanium, ni = 2.5 1019m-3, doped with 1022 donors GaAs, ni = 1013m-3, doped with 1018 donors and 1015 acceptors.
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