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EE 5340Semiconductor Device TheoryLecture 2 - Fall 2003

Professor Ronald L. Carterronc@uta.edu

http://www.uta.edu/ronc

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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.