Lecture%2015

Lecture%2015 - EEE 352: Lecture 15 Metals, Insulators, and...

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1 EEE 352: Lecture 15 Metals, Insulators, and Semiconductors * Fermi-Dirac distribution *Quantum statistics—getting there! * Broadening of the distribution with temperature * Density of states Enrico Fermi Nobel prize in Physics, 1938 “for discovery of neutron irradiation and slow neutrons” Paul A. M. Dirac Nobel prize in Physics, 1933 “for discovery of new forms of atomic radiation” (positrons and the Dirac equation) Band Structure ρ ( E ) E While this is the Si band structure, it could be any material for the purposes of the present argument. We want to discuss the “number of states per unit energy”… ( E ) E How many of these available states are filled? What is the probability that any one state is filled? Density of states per unit energy per unit volume Energy gap ± We need a QUANTUM MECHANICAL distribution function that gives the probability of electrons OCCUPYING certain energy states AT FINITE temperatures and with the electrons satisfying the Pauli exclusion principle ± This function is known as the FERMI-DIRAC distribution function f ( E ): 1 ] / ) exp[( 1 ) ( + = T k E E E f B F ± f ( E ) IS THE FERMI-DIRAC DISTRIBUTION FUNCTION AND DESCRIBES THE OCCUPATION OF ELECTRON STATES AT FINITE TEMPERATURES ± MORE PRECISELY f ( E ) IS THE PROBABILITY THAT ANY STATE AT ENERGY E WILL BE OCCUPIED BY AN ELECTRON AT AN ARBITRARY TEMPERATURE T AND FOR A GIVEN FERMI ENERGY E F ± THE FUNCTION f ( E ) THEREFORE TAKES VALUES RANGING FROM ZERO TO ONE ± THE PROBABILITY OF THE STATE NOT BEING OCCUPIED AT THE SAME TEMPERATURE MAY THEN EASILY BE WRITTEN AS 1 f ( E ) How do we determine this function?
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2 Fermi-Dirac Distribution We have two constraints on the problem: conservation of particles and energy = i i n N = i i i E n E Sum runs over the individual levels i . Number of particles in each level. Energy of level i . Consider that we have N i particles to distribute into S i quantum states of level i . The number of ways to do this is: () ! ! ! i i i i i N S N S W = Fermi-Dirac Distribution How do we get this probability? Well, the probability of a particle being in box j is: i i j S N P = 1 The probability of a second ball is: 1 1 2 = i i j S N P Finally, for N i balls in box j : + = + = = 1 1 ... 1 1 1 1 1 1 1 1 1 1 2 1 i i i i i i i i i i i i i i i i i i i i jN j j j N S N S N S N S N S S N S N N S S N S N P P P P i Number of balls Number of boxes Fermi-Dirac Distribution Consider an example with 3 buckets and 2 balls (with each bucket containing at most 1 ball): i i i i i i i i i i i i i i i i i i i j W S N S N N S N S N S N S N S S N S N P 1 ! ! ! 1 1 ... 1 1 1 1 1 1 = + = Number of ways the balls can be distributed in the boxes Fermi-Dirac Distribution There are three possible ways in which the balls can be put into the buckets: 3 ) 1 )( 1 2 ( ) 1 2 3 ( )! 2 3 ( ! 2 ! 3 )! ( !
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This note was uploaded on 09/19/2009 for the course EEE 352 taught by Professor Ferry during the Spring '08 term at ASU.

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Lecture%2015 - EEE 352: Lecture 15 Metals, Insulators, and...

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