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EE 5340 Semiconductor Device Theory Lecture 2 - Fall 2003. Professor Ronald L. Carter [email protected] http://www.uta.edu/ronc. Photon: A particle -like wave. E = hf, the quantum of energy for light. (PE effect & black body rad.) f = c/ l , c = 3E8m/sec, l = wavelength - PowerPoint PPT Presentation
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©L02 Aug 28 1
EE 5340Semiconductor Device TheoryLecture 2 - Fall 2003
Professor Ronald L. [email protected]
http://www.uta.edu/ronc
©L02 Aug 28 2
Photon: A particle-like wave• E = hf, the quantum of energy for light.
(PE effect & black body rad.)• f = c/, c = 3E8m/sec, = wavelength• From Poynting’s theorem (em waves),
momentum density = energy density/c• Postulate a Photon “momentum”
p = h/= hk, h = h/2 wavenumber, k =2/
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Wave-particle duality
• Compton showed p = hkinitial - hkfinal, so an photon (wave) is particle-like
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Wave-particle duality
• DeBroglie hypothesized a particle could be wave-like, = h/p
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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 = p2/2mConservation of Energy Theorem
• Momentum, p = mvConservation of Momentum Thm
• Newton’s second LawF = 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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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.