1. Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/
Semiconductor Device Modeling and Characterization EE5342, Lecture 2-Spring 2004
Professor Ronald L. Carter
L2 January 22

2. Web Pages Bring the following to the first class
R. L. Carter’s web page
EE 5342 web page and syllabus
University and College Ethics Policies
www2.uta.edu/discipline/
L2 January 22

3. First Assignment e-mail to listserv@listserv.uta.edu
In the body of the message include subscribe EE5342
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If you have any questions, send to with EE5342 in subject line.
L2 January 22

4. Quantum Mechanics
Schrodinger’s wave equation developed to maintain consistence with wave-particle duality and other “quantum” effects
Position, mass, etc. of a particle replaced by a “wave function”, Y(x,t)
Prob. density = |Y(x,t)• Y*(x,t)|
L2 January 22

5. Schrodinger Equation Separation of variables gives Y(x,t) = y(x)• f(t)
The time-independent part of the Schrodinger equation for a single particle with Total E = E and PE = V. The Kinetic Energy, KE = E - V
L2 January 22

6. Solutions for the Schrodinger Equation
Solutions of the form of y(x) = A exp(jKx) + B exp (-jKx) K = [8p2m(E-V)/h2]1/2
Subj. to boundary conds. and norm. y(x) is finite, single-valued, conts. dy(x)/dx is finite, s-v, and conts.
L2 January 22

7. Infinite Potential Well
V = 0, 0 < x < a
V --> inf. for x < 0 and x > a
Assume E is finite, so y(x) = 0 outside of well
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8. Step Potential V = 0, x < 0 (region 1) V = Vo, x > 0 (region 2)
Region 1 has free particle solutions
Region 2 has free particle soln. for E > Vo , and evanescent solutions for E < Vo
A reflection coefficient can be def.
L2 January 22

9. Finite Potential Barrier
Region 1: x < 0, V = 0
Region 1: 0 < x < a, V = Vo
Region 3: x > a, V = 0
Regions 1 and 3 are free particle solutions
Region 2 is evanescent for E < Vo
Reflection and Transmission coeffs. For all E
L2 January 22

10. Kronig-Penney Model
A simple one-dimensional model of a crystalline solid
V = 0, 0 < x < a, the ionic region
V = Vo, a < x < (a + b) = L, between ions
V(x+nL) = V(x), n = 0, +1, +2, +3, …, representing the symmetry of the assemblage of ions and requiring that y(x+L) = y(x) exp(jkL), Bloch’s Thm
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11. K-P Potential Function*
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12. K-P Static Wavefunctions
Inside the ions, 0 < x < a y(x) = A exp(jbx) + B exp (-jbx) b = [8p2mE/h]1/2
Between ions region, a < x < (a + b) = L y(x) = C exp(ax) + D exp (-ax) a = [8p2m(Vo-E)/h2]1/2
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13. K-P Impulse Solution
Limiting case of Vo-> inf. and b -> 0, while a2b = 2P/a is finite
In this way a2b2 = 2Pb/a < 1, giving sinh(ab) ~ ab and cosh(ab) ~ 1
The solution is expressed by P sin(ba)/(ba) + cos(ba) = cos(ka)
Allowed valued of LHS bounded by +1
k = free electron wave # = 2p/l
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14. K-P Solutions*
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15. K-P E(k) Relationship*
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16. Analogy: a nearly -free electr. model
Solutions can be displaced by ka = 2np
Allowed and forbidden energies
Infinite well approximation by replacing the free electron mass with an “effective” mass (noting E = p2/2m = h2k2/2m) of
L2 January 22

17. Generalizations and Conclusions
The symm. of the crystal struct. gives “allowed” and “forbidden” energies (sim to pass- and stop-band)
The curvature at band-edge (where k = (n+1)p) gives an “effective” mass.
L2 January 22

18. Silicon Covalent Bond (2D Repr)
Each Si atom has 4 nearest neighbors
Si atom: 4 valence elec and 4+ ion core
8 bond sites / atom
All bond sites filled
Bonding electrons shared 50/50
_ = Bonding electron
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19. Silicon Band Structure**
Indirect Bandgap
Curvature (hence m*) is function of direction and band. [100] is x-dir, [111] is cube diagonal
Eg = 1.17-aT2/(T+b) a = 4.73E-4 eV/K b = 636K
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20. Si Energy Band Structure at 0 K
Every valence site is occupied by an electron
No electrons allowed in band gap
No electrons with enough energy to populate the conduction band
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21. Si Bond Model Above Zero Kelvin
Enough therm energy ~kT(k=8.62E-5eV/K) to break some bonds
Free electron and broken bond separate
One electron for every “hole” (absent electron of broken bond)
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22. Band Model for thermal carriers
Thermal energy ~kT generates electron-hole pairs
At 300K
Eg(Si) = eV
>> kT = meV,
Nc = 2.8E19/cm3
> Nv = 1.04E19/cm3
>> ni = 1.45E10/cm3
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23. Donor: cond. electr. due to phosphorous
P atom: 5 valence elec and 5+ ion core
5th valence electr has no avail bond
Each extra free el, -q, has one +q ion
# P atoms = # free elect, so neutral
H atom-like orbits
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24. Bohr model H atom- like orbits at donor
Electron (-q) rev. around proton (+q)
Coulomb force, F=q2/4peSieo,q=1.6E-19 Coul, eSi=11.7, eo=8.854E-14 Fd/cm
Quantization L = mvr = nh/2p
En= -(Z2m*q4)/[8(eoeSi)2h2n2] ~-40meV
rn= [n2(eoeSi)h2]/[Zpm*q2] ~ 2 nm
for Z=1, m*~mo/2, n=1, ground state
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25. Band Model for donor electrons
Ionization energy of donor Ei = Ec-Ed ~ 40 meV
Since Ec-Ed ~ kT, all donors are ionized, so ND ~ n
Electron “freeze-out” when kT is too small
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26. Acceptor: Hole due to boron
B atom: 3 valence elec and 3+ ion core
4th bond site has no avail el (=> hole)
Each hole, adds --q, has one -q ion
#B atoms = #holes, so neutral
H atom-like orbits
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27. Hole orbits and acceptor states
Similar to free electrons and donor sites, there are hole orbits at acceptor sites
The ionization energy of these states is EA - EV ~ 40 meV, so NA ~ p and there is a hole “freeze-out” at low temperatures
L2 January 22

28. Impurity Levels in Si: EG = 1,124 meV
Phosphorous, P: EC - ED = 44 meV
Arsenic, As: EC - ED = 49 meV
Boron, B: EA - EV = 45 meV
Aluminum, Al: EA - EV = 57 meV
Gallium, Ga: EA - EV = 65meV
Gold, Au: EA - EV = 584 meV EC - ED = 774 meV
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29. Quantum density of states function
1 dim electron wave #s range for n+1 “atoms” is 2p/L < k < 2p/a where a is “interatomic” distance and L = na is the length of the assembly (k = 2p/l)
Shorter ls, would “oversample”
if n increases by 1, dp is h/L
Extn 3D: E = p2/2m = h2k2/2m so a vol of p-space of 4pp2dp has h3/LxLyLz
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30. QM density of states (cont.)
So density of states, gc(E) is (Vol in p-sp)/(Vol per state*V) = 4pp2dp/[(h3/LxLyLz)*V]
Noting that p2 = 2mE, this becomes gc(E) = {4p(2mn*)3/2/h3}(E-Ec)1/2 and E - Ec = h2k2/2mn*
Similar for the hole states where Ev - E = h2k2/2mp*
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31. Fermi-Dirac distribution fctn
The probability of an electron having an energy, E, is given by the F-D distr fF(E) = {1+exp[(E-EF)/kT]}-1
Note: fF (EF) = 1/2
EF is the equilibrium energy of the system
The sum of the hole probability and the electron probability is 1
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32. Fermi-Dirac DF (continued)
So the probability of a hole having energy E is 1 - fF(E)
At T = 0 K, fF (E) becomes a step function and 0 probability of E > EF
At T >> 0 K, there is a finite probability of E >> EF
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33. Maxwell-Boltzman Approximation
fF(E) = {1+exp[(E-EF)/kT]}-1
For E - EF > 3 kT, the exp > 20, so within a 5% error, fF(E) ~ exp[-(E-EF)/kT]
This is the MB distribution function
MB used when E-EF>75 meV (T=300K)
For electrons when Ec - EF > 75 meV and for holes when EF - Ev > 75 meV
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34. Electron Conc. in the MB approx.
Assuming the MB approx., the equilibrium electron concentration is
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35. References
*Fundamentals of Semiconductor Theory and Device Physics, by Shyh Wang, Prentice Hall, 1989.
**Semiconductor Physics & Devices, by Donald A. Neamen, 2nd ed., Irwin, Chicago.
M&K = Device Electronics for Integrated Circuits, 3rd ed., by Richard S. Muller, Theodore I. Kamins, and Mansun Chan, John Wiley and Sons, New York, 2003.
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