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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/

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

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

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

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

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

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

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

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 )

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 )

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

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

L3 January 2213 Equilibrium electron conc. and energies

L3 January 2214 Equilibrium hole conc. and energies

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

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.

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

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

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 )

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)

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.

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

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

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

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

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

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

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 )

L3 January 2229 Net silicon extr resistivity (cont.)

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.)

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

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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