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Review of Basic Semiconductor Physics Copyright © by John Wiley & Sons 2003 Semiconductor Physics - 1 Current Flow and Conductivity • Charge in volume Ax = Q = q n A x = q n A vt • Current density J = (Q/t)A-1 =qnv • Metals - gold, platinum, silver, copper, etc. • n = 1023 cm-3 ; = 107 mhos-cm • Insulators - silicon dioxide, silicon nitride, aluminum oxide • n < 103 cm-3 ; < 10-10 mhos-cm • Semiconductors - silicon, gallium arsenide, diamond, etc. • 108 < n <1019 cm-3 ; 10-10 << 104 mhos-cm Copyright © by John Wiley & Sons 2003 Semiconductor Physics - 2 Thermal Ionization • Si atoms have thermal vibrations about equilibrium point. • Small percentage of Si atoms have large enough vibrational energy to break covalent bond and liberate an electron. Copyright © by John Wiley & Sons 2003 Semiconductor Physics - 3 Electrons and Holes - • T3 > T2 > T1 A t=T 1 • Density of free electrons = n : Density of free holes = p • p = n = ni(T) = intrinsic carrier density. • ni2(T) = C exp(-qEg/(kT = 1020 cm-6 at 300 K B - )) generation of B + A t=T • T = temp in K • k = 1.4x10-23 joules/ K • Eg = energy gap = 1.1 eV in silicon •q= 1.6x10-19 coulombs Copyright © by John Wiley & Sons 2003 recombination of B - apparent movement of "Hole" A t=T Semiconductor Physics - 4 3 2 Doped Semiconductors • Extrinsic (doped) semiconductors:p = po ≠ n = no ≠ ni • Carrier density estimates: • Law of mass action nopo = ni2(T) • Charge neutrality Na + no = Nd + po • P-type silicon with Na >> ni: po ≈ Na , no ≈ ni2/ Na Copyright © by John Wiley & Sons 2003 • N-type silicon with Nd >> ni: no ≈ Nd , po ≈ ni2/ Nd Semiconductor Physics - 5 Nonequilibrium and Recombination • Thermal Equilibrium - Carrier generation = Carrier recombination • n = no and p = po • Nonequilibrium - n > no and p > po • n = no + n and p = no + n ; n = excess carrier density • Excess holes and excess electrons created in equal numbers by breaking of covalent bonds • Generation mechanisms -light (photoelectric effect), injection, impact ionization • Recombination - removal of excess holes and electrons • Mechanisms - free electron captured by empty covalent bond (hole) or trapped by impurity or crystal imperfection • Rate equation: d(n)/dt = - n • Solution n = n (0) e -t Copyright © by John Wiley & Sons 2003 Semiconductor Physics - 6 Carrier Lifetimes • = excess carrier lifetime • Usually assumed to be constant. Changes in two important situations. • increases with temperature T • decreases at large excess carrier densities ; = o/[1 + (n/nb)2 ] • Control of carrier lifetime values. • Switching time-on state loss tradeoff mandates good lifetime control. • Control via use of impurities such as gold - lifetime killers. • Control via electron irradiation - more uniform and better control. Copyright © by John Wiley & Sons 2003 Semiconductor Physics - 7 Drift Current Flow Diffusion p n J p Jn + - x x • Jdrift = q µn n E + q p µp E • µn = 1500 cm2/V-sec for silicon at room temp. and Nd < 1015 cm-3 • µp = 500 cm2/V-sec for silicon at room temp. and Na < 1015 cm-3 Copyright © by John Wiley & Sons 2003 • Jdiff = Jn + Jp = q Dndn/dx - q Dp dp/dx • Dn/n = Dp/p = kT/q ; Einstein relation • D = diffusion constant, = carrier mobility • Total current density J = Jdrift + Jdiff Semiconductor Physics - 8 PN Junction metallurgical junction P N A N N A ND NA ND NA x - N D Step (abrupt) junction Copyright © by John Wiley & Sons 2003 x - N D Linearly graded junction Semiconductor Physics - 9 Formation of Space Charge Layer metallurgical junction • Diffusing electrons and holes leave the region near metallurgical junction depleted of free carriers (depletion region). x ionized acceptors ionized donors + P + N Electric field opposing diffusion + - • Exposed ionized impurities form space charge layer. • Electric field due to space charge opposes diffusion. Diffusing electrons - - + + + + + Diffusing holes + space charge layer width = W Copyright © by John Wiley & Sons 2003 Semiconductor Physics - 10 Quantitative Description of Space Charge Region • Assume step junction. d2 = - dx 2 = - qNa ; x < 0 = qNd ; x > 0 E(x ) = d = - E(x ) dx qNa(x +x p ) ; - xp < x < 0 qNd (x - x n) E(x ) = ; 0< x < x n xn c = - E(x )dx - xp qNax p 2 + qNd x n2 c = 2 Copyright © by John Wiley & Sons 2003 Semiconductor Physics - 11 Contact (Built-in, Junction) Potential • In thermal equilibrium Jn = q µn n d dn + q Dn =0 dx dx n(xn) (xn) Dn dn • Separate variables and integrate ; d = µn n (xp) n(xp) kT NaNd • (xn) - (xp) = c = q ln 2 ; c = contact potential ni • Example • • • Room temperature kT/q = 0.025 eV Na = Nd = 1016 cm-3 ; ni2 = 1020 cm-6 Fc = 0.72 eV Copyright © by John Wiley & Sons 2003 Semiconductor Physics - 12 Reverse-Biased Step Junction • St art ing equat ions • W( V) = x n( V) + x p ( V) qNax p 2 + qNd x n2 • V + c = 2 • Charge neut ralit y qNax p = qNd x n • Solv e equat ions simult aneously • W( V) = Wo • Wo = 1 + V/ c 2 c ( Na+ Nd ) qNaNd 2 c • Emax = Wo 1 + V/ c Copyright © by John Wiley & Sons 2003 Semiconductor Physics - 13 Forward-Biased PN Junction • Forward bias favors diffusion over drift. • Excess minority carrier injection into both p and n drift regions. • Minority carrier diffusion lengths. • Ln = [Dnn]0.5 • Lp = [Dpp]0.5 Copyright © by John Wiley & Sons 2003 Semiconductor Physics - 14 Ideal PN Junction I-V Characteristics • Excess carriers in drift regions recombined and thus more must be constantly injected if the distributions np(x) and pn(x) are to be maintained. • Constant injection of electrons and holes results in a current density J given by Qn Qp Lp Ln 2 J = + = q ni + n p N N an d p J= Js qV exp( ) kT - 1 - 1 Lp Ln 2 ; Js = q ni + Nd p Nan i J J qV exp( ) kT v - Js v forward bias Copyright © by John Wiley & Sons 2003 reverse bias combined characteristic v Semiconductor Physics - 15 Reverse Saturation Current V • Carrier density gradient immediately adjacent to depletion region causes reverse saturation current to flow via diffusion. + + + + + P N • Js independent of reverse voltage V because carrier density gradient unaffected by applied voltage. Wo W(V) n po p no n p(x) • Js extremely temperature sensitivity because of dependence on ni2(T.) + p (x) n x Electric field, Copyright © by John Wiley & Sons 2003 J s Semiconductor Physics - 16 Impact Ionization • E ≥ EBD ; free electron can acquire sufficient from the field between lattice collisions (tc ≈ 10-12 sec) to break covalent bond. Si - - Si • Energy = 0.5mv2 = q Eg ; v = q EBDtc - • Solving for EBD gives EBD = Si 2 Eg m q tc2 Electric field E - • Numerical evaluation • m = 10-27 grams, Eg = 1.1 eV, tc = 10-12 sec. • EBD = • (2) (1.1) (1027) = 3x105 V/cm -19 -24 (1.6x10 ) (10 ) Experimental estimates are 2-3.5x105 V/cm Copyright © by John Wiley & Sons 2003 Semiconductor Physics - 17