Transcript EE 5340©
EE 5340 Semiconductor Device Theory Lecture 2 - Fall 2003 Professor Ronald L. Carter [email protected] http://www.uta.edu/ronc ©L02 Aug 28 1 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 • From Poynting’s theorem (em waves), momentum density = energy density/c • Postulate a Photon “momentum” p = h/l = hk, h = h/2p wavenumber, k = 2p /l ©L02 Aug 28 2 Wave-particle duality • Compton showed Dp = hkinitial - hkfinal, so an photon (wave) is particle-like ©L02 Aug 28 3 Wave-particle duality • DeBroglie hypothesized a particle could be wave-like, l = h/p ©L02 Aug 28 4 Wave-particle duality • Davisson and Germer demonstrated wave-like interference phenomena for electrons to complete the duality model ©L02 Aug 28 5 Newtonian Mechanics • Kinetic energy, KE = mv2/2 = p2/2m Conservation of Energy Theorem • Momentum, p = mv Conservation of Momentum Thm • Newton’s second Law F = ma = m dv/dt = m d2x/dt2 ©L02 Aug 28 6 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)| ©L02 Aug 28 7 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 2y x 8p2m E V ( x ) y x 0 x 2 h2 ©L02 Aug 28 8 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. * y x y x dx 1 ©L02 Aug 28 9 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 2 npx y x sin , n = 1,2,3,... a a 2 2 2 2 h n h k h hk En ,p 2 2 l 2p 8ma 8mp ©L02 Aug 28 10 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. ©L02 Aug 28 11 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 ©L02 Aug 28 12 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 ©L02 Aug 28 13 K-P Potential Function* ©L02 Aug 28 14 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 ©L02 Aug 28 15 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 ©L02 Aug 28 16 K-P Solutions* ©L02 Aug 28 17 K-P E(k) Relationship* ©L02 Aug 28 18 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 2 2 1 h E * m 2 2 4p k ©L02 Aug 28 19 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. ©L02 Aug 28 20 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 ©L02 Aug 28 21 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 ©L02 Aug 28 22 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 ©L02 Aug 28 23 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) ©L02 Aug 28 24 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. ©L02 Aug 28 25