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L3 January 221 Semiconductor Device Modeling and Characterization EE5342, Lecture 3-Spring 2002 Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/

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L3 January 222 Classes of semiconductors Intrinsic: n o = p o = n i, since N a &N d << n i =[N c N v exp(E g /kT)] 1/2,(not easy to get) n-type: n o > p o, since N d > N a p-type: n o < p o, since N d < N a Compensated: n o =p o =n i, w/ N a - = N d + > 0 Note: n-type and p-type are usually partially compensated since there are usually some opposite-type dopants

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L3 January 223 Equilibrium concentrations Charge neutrality requires q(p o + N d + ) + (-q)(n o + N a - ) = 0 Assuming complete ionization, so N d + = N d and N a - = N a Gives two equations to be solved simultaneously 1. Mass action, n o p o = n i 2, and 2. Neutralityp o + N d = n o + N a

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L3 January 224 For N d > N a >Let N = N d -N a, and (taking the + root) n o = (N)/2 + {[N/2] 2 +n i 2 } 1/2 For N d+ = N d >> n i >> N a we have >n o = N d, and >p o = n i 2 /N d Equilibrium conc n-type

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L3 January 225 For N a > N d >Let N = N d -N a, and (taking the + root) p o = (-N)/2 + {[-N/2] 2 +n i 2 } 1/2 For N a- = N a >> n i >> N d we have >p o = N a, and >n o = n i 2 /N a Equilibrium conc p-type

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L3 January 226 Electron Conc. in the MB approx. Assuming the MB approx., the equilibrium electron concentration is

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L3 January 227 Hole Conc in MB approx Similarly, the equilibrium hole concentration is p o = N v exp[-(E F -E v )/kT] So that n o p o = N c N v exp[-E g /kT] n i 2 = n o p o, N c,v = 2{2 m* n,p kT/h 2 } 3/2 N c = 2.8E19/cm3, N v = 1.04E19/cm3 and n i = 1E10/cm3

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L3 January 228 Position of the Fermi Level E fi is the Fermi level when n o = p o E f shown is a Fermi level for n o > p o E f < E fi when n o < p o E fi < (E c + E v )/2, which is the mid- band

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L3 January 229 E F relative to E c and E v Inverting n o = N c exp[-(E c -E F )/kT] gives E c - E F = kT ln(N c /n o ) For n-type material: E c - E F =kTln(N c /N d )=kTln[(N c P o )/n i 2 ] Inverting p o = N v exp[-(E F -E v )/kT] givesE F - E v = kT ln(N v /p o ) For p-type material: E F - E v = kT ln(N v /N a )

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L3 January 2210 E F relative to E fi Letting n i = n o gives E f = E fi n i = N c exp[-(E c -E fi )/kT], so E c - E fi = kT ln(N c /n i ). Thus E F - E fi = kT ln(n o /n i ) and for n-type E F - E fi = kT ln(N d /n i ) Likewise E fi - E F = kT ln(p o /n i ) and for p-type E fi - E F = kT ln(N a /n i )

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L3 January 2211 Locating E fi in the bandgap Since E c - E fi = kT ln(N c /n i ), and E fi - E v = kT ln(N v /n i ) The sum of the two equations gives E fi = (E c + E v )/2 - (kT/2) ln(N c /N v ) Since N c = 2.8E19cm -3 > 1.04E19cm -3 = N v, the intrinsic Fermi level lies below the middle of the band gap

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L3 January 2212 Sample calculations E fi = (E c + E v )/2 - (kT/2) ln(N c /N v ), so at 300K, kT = 25.86 meV and N c /N v = 2.8/1.04, E fi is 12.8 meV or 1.1% below mid-band For N d = 3E17cm -3, given that E c - E F = kT ln(N c /N d ), we have E c - E F = 25.86 meV ln(280/3), E c - E F = 0.117 eV =117meV ~3x(E c - E D ) what N d gives E c -E F =E c /3

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L3 January 2213 Equilibrium electron conc. and energies

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L3 January 2214 Equilibrium hole conc. and energies

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L3 January 2215 Carrier Mobility In an electric field, E x, the velocity (since a x = F x /m* = qE x /m*) is v x = a x t = (qE x /m*)t, and the displ x = (qE x /m*)t 2 /2 If every coll, a collision occurs which “resets” the velocity to = 0, then = qE x coll /m* = E x

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L3 January 2216 Carrier mobility (cont.) The response function is the mobility. The mean time between collisions, coll, may has several important causal events: Thermal vibrations, donor- or acceptor-like traps and lattice imperfections to name a few. Hence thermal = q thermal /m*, etc.

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L3 January 2217 Carrier mobility (cont.) If the rate of a single contribution to the scattering is 1/ i, then the total scattering rate, 1/ coll is

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L3 January 2218 Drift Current The drift current density (amp/cm 2 ) is given by the point form of Ohm Law J = (nq n +pq p )(E x i+ E y j+ E z k), so J = ( n + p )E = E, where = nq n +pq p defines the conductivity The net current is

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L3 January 2219 Drift current resistance Given: a semiconductor resistor with length, l, and cross-section, A. What is the resistance? As stated previously, the conductivity, = nq n + pq p So the resistivity, = 1/ = 1/(nq n + pq p )

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L3 January 2220 Drift current resistance (cont.) Consequently, since R = l/A R = (nq n + pq p ) -1 (l/A) For n >> p, (an n-type extrinsic s/c) R = l/(nq n A) For p >> n, (a p-type extrinsic s/c) R = l/(pq p A)

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L3 January 2221 Drift current resistance (cont.) Note: for an extrinsic semiconductor and multiple scattering mechanisms, since R = l/(nq n A) or l/(pq p A), and ( n or p total ) -1 = i -1, then R total = R i (series Rs) The individual scattering mechanisms are: Lattice, ionized impurity, etc.

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L3 January 2222 Exp. mobility model function for Si 1 ParameterAsPB min 52.268.544.9 max 14171414470.5 N ref 9.68e169.20e162.23e17 0.6800.7110.719

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L3 January 2223 Exp. mobility model for P, As and B in Si

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L3 January 2224 Carrier mobility functions (cont.) The parameter max models 1/ lattice the thermal collision rate The parameters min, N ref and model 1/ impur the impurity collision rate The function is approximately of the ideal theoretical form: 1/ total = 1/ thermal + 1/ impurity

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L3 January 2225 Carrier mobility functions (ex.) Let N d = 1.78E17/cm3 of phosphorous, so min = 68.5, max = 1414, N ref = 9.20e16 and = 0.711. Thus n = 586 cm2/V-s Let N a = 5.62E17/cm3 of boron, so min = 44.9, max = 470.5, N ref = 9.68e16 and = 0.680. Thus n = 189 cm2/V-s

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L3 January 2226 Lattice mobility The lattice is the lattice scattering mobility due to thermal vibrations Simple theory gives lattice ~ T -3/2 Experimentally n,lattice ~ T -n where n = 2.42 for electrons and 2.2 for holes Consequently, the model equation is lattice (T) = lattice (300)(T/300) -n

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L3 January 2227 Ionized impurity mobility function The impur is the scattering mobility due to ionized impurities Simple theory gives impur ~ T 3/2 /N impur Consequently, the model equation is impur (T) = impur (300)(T/300) 3/2

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L3 January 2228 Net silicon (ex- trinsic) resistivity Since = -1 = (nq n + pq p ) -1 The net conductivity can be obtained by using the model equation for the mobilities as functions of doping concentrations. The model function gives agreement with the measured (N impur )

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L3 January 2229 Net silicon extr resistivity (cont.)

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L3 January 2230 Net silicon extr resistivity (cont.) Since = (nq n + pq p ) -1, and n > p, ( = q /m*) we have p > n Note that since 1.6(high conc.) < p / n < 3(low conc.), so 1.6(high conc.) < n / p < 3(low conc.)

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L3 January 2231 Net silicon (com- pensated) res. For an n-type (n >> p) compensated semiconductor, = (nq n ) -1 But now n = N = N d - N a, and the mobility must be considered to be determined by the total ionized impurity scattering N d + N a = N I Consequently, a good estimate is = (nq n ) -1 = [Nq n (N I )] -1

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L3 January 2232 References 1 Device Electronics for Integrated Circuits, 2 ed., by Muller and Kamins, Wiley, New York, 1986. 2 Physics of Semiconductor Devices, by S. M. Sze, Wiley, New York, 1981.

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