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EE 5340Semiconductor Device TheoryLecture 2 - Fall 2009
Professor Ronald L. [email protected]
http://www.uta.edu/ronc
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Quantum Concepts
• Bohr Atom
• Light Quanta (particle-like waves)
• Wave-like properties of particles
• Wave-Particle Duality
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Wave-particle duality
• Davisson and Germer demonstrated wave-like interference phenomena for electrons to complete the duality model
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Newtonian Mechanics
• Kinetic energy, KE = mv2/2 = p2/2m Conservation of Energy Theorem
• Momentum, p = mvConservation of
Momentum Thm• Newton’s second Law
F = ma = m dv/dt = m d2x/dt2
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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”, (x,t)
• Prob. density = |(x,t)• (x,t)|
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Schrodinger Equation
• Separation of variables gives(x,t) = (x)• (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
2
2
280
x
x
mE V x x
h2 ( )
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Solutions for the Schrodinger Equation• Solutions of the form of (x) =
A exp(jKx) + B exp (-jKx) K = [82m(E-V)/h2]1/2
• Subj. to boundary conds. and norm.(x) is finite, single-valued, conts.d(x)/dx is finite, s-v, and conts.
1dxxx
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Infinite Potential Well• V = 0, 0 < x < a• V --> inf. for x < 0 and x > a• Assume E is finite, so
(x) = 0 outside of well
2,
88E
1,2,3,...=n ,sin2
2
22
2
22
nhkh
pmkh
manh
axn
ax
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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 , andevanescent solutions for E <
Vo
• A reflection coefficient can be def.
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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
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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 (x+L) = (x) exp(jkL), Bloch’s Thm
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K-P Potential Function*
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K-P Static Wavefunctions• Inside the ions, 0 < x < a
(x) = A exp(jx) + B exp (-jx) = [82mE/h]1/2
• Between ions region, a < x < (a + b) = L (x) = C exp(x) + D exp (-x) = [82m(Vo-E)/h2]1/2
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K-P Impulse Solution• Limiting case of Vo-> inf. and b -> 0,
while 2b = 2P/a is finite• In this way 2b2 = 2Pb/a < 1, giving
sinh(b) ~ b and cosh(b) ~ 1• The solution is expressed by
P sin(a)/(a) + cos(a) = cos(ka)• Allowed valued of LHS bounded by +1• k = free electron wave # = 2/
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K-P Solutions*
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K-P E(k) Relationship*
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Analogy: a nearly-free electr. model• Solutions can be displaced by ka = 2n• Allowed and forbidden energies• Infinite well approximation by replacing
the free electron mass with an “effective” mass (noting E = p2/2m = h2k2/2m) of
1
2
2
2
2
4
k
Ehm
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Generalizationsand 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)) gives an “effective” mass.
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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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Silicon BandStructure**• Indirect Bandgap• Curvature (hence
m*) is function of direction and band. [100] is x-dir, [111] is cube diagonal
• Eg = 1.17-T2/(T+) = 4.73E-4 eV/K = 636K
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Si Energy BandStructure 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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Si Bond ModelAbove 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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Band Model forthermal carriers• Thermal energy
~kT generates electron-hole pairs
• At 300K Eg(Si) = 1.124 eV
>> kT = 25.86 meV,Nc = 2.8E19/cm3
> Nv = 1.04E19/cm3>> ni = 1.45E10/cm3
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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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Bohr model H atom-like orbits at donor• Electron (-q) rev. around proton (+q)
• Coulomb force, F=q2/4Sio,q=1.6E-19 Coul, Si=11.7, o=8.854E-14 Fd/cm
• Quantization L = mvr = nh/2• En= -(Z2m*q4)/[8(oSi)2h2n2] ~-40meV
• rn= [n2(oSi)h2]/[Zm*q2] ~ 2 nm
for Z=1, m*~mo/2, n=1, ground state
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Band Model fordonor 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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Acceptor: Holedue 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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Hole orbits andacceptor 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
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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 meVEC - ED = 774 meV
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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.