EE 5340 Semiconductor Device Theory Lecture 05 – Spring 2011 Professor Ronald L. Carter

©rlc L05-08Feb20112 Review the Following R. L. Carter’s web page: – EE 5340 web page and syllabus. (Refresh all EE 5340 pages when downloading to assure the latest version.) All links at: – University and College Ethics Policies – Makeup lecture at noon Friday (1/28) in 108 Nedderman Hall. This will be available on the web.

©rlc L05-08Feb20113 First Assignment Send to –On the subject line, put “5340 ” –In the body of message include address: ______________________ Your Name*: _______________________ Last four digits of your Student ID: _____ * Your name as it appears in the UTA Record - no more, no less

©rlc L05-08Feb20114 Second Assignment Submit a signed copy of the document posted at

©rlc L05-08Feb20115 Schedule Changes Due to University Weather Closings Make-up class will be held Friday, February 11 at 12 noon in 108 Nedderman Hall. Additional changes will be announced as necessary. Syllabus and lecture dates postings will be updated in the next 24 hours. Project Assignment will be posted in the next 36 hours.

©rlc L05-08Feb20116 Intrinsic carrier conc. (MB limit) n i 2 = n o p o = N c N v e -Eg/kT N c = 2{2  m* n kT/h 2 } 3/2 N v = 2{2  m* p kT/h 2 } 3/2 E g = 1.17 eV -  T 2 /(T+  )  = 4.73E-4 eV/K  = 636K

©rlc L05-08Feb20117 Classes of semiconductors Intrinsic: n o = p o = n i, since N a &N d << n i, n i 2 = N c N v e -Eg/kT, ~1E-13 dopant level ! 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

©rlc L05-08Feb2011 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 8

©rlc L05-08Feb20119 Equilibrium conc (cont.) For N d > N a (taking the + root) n o = (N d -N a )/2 + {[(N d - N a )/2] 2 +n i 2 } 1/2 For N d >> N a and N d >> n i, can use the binomial expansion, giving n o = N d /2 + N d /2[1 + 2n i 2 /N d 2 + … ] So n o = N d, and p o = n i 2 /N d in the limit of N d >> N a and N d >> n i

©rlc L05-08Feb n-type equilibrium concentrations N ≡ N d - N a, n type  N > 0 For all N, n o = N/2 + {[N/2] 2 +n i 2 } 1/2 In most cases, N >> n i, so n o = N, and p o = n i 2 /n o = n i 2 /N, (Law of Mass Action is al- ways true in equilibrium)

©rlc L05-08Feb 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

©rlc L05-08Feb p-type equilibrium concentrations N ≡ N d - N a, p type  N < 0 For all N, p o = |N|/2 + {[|N|/2] 2 +n i 2 } 1/2 In most cases, |N| >> n i, so p o = |N|, and n o = n i 2 /p o = n i 2 /|N|, (Law of Mass Action is al- ways true in equilibrium)

©rlc L05-08Feb 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

©rlc L05-08Feb 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 )

©rlc L05-08Feb 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 )

©rlc L05-08Feb 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 1st equation minus the 2nd 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

©rlc L05-08Feb Example calculations For N d = 3.2E16/cm 3, n i = 1.4E10/cm 3 n o = N d = 3.2E16/cm 3 p o = n i 2 /N d, (p o is always n i 2 /n o) = (1.4E10/cm 3 ) 2 /3.2E16/cm 3 = 6.125E3/cm 3 (comp to ~1E23 Si) For p o = N a = 4E17/cm 3, n o = n i 2 /N a = (1.4E10/cm 3 ) 2 /4E17/cm 3 = 490/cm 3

©rlc L05-08Feb Sample calculations E fi = (E c + E v )/2 - (kT/2) ln(N c /N v ), so at 300K, kT = 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 = meV ln(280/3), E c - E F = eV =117meV ~3x(E c - E D ) what N d gives E c -E F =E c /3

©rlc L05-08Feb Equilibrium electron conc. and energies

©rlc L05-08Feb Equilibrium hole conc. and energies

©rlc L05-08Feb 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

©rlc L05-08Feb 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.

©rlc L05-08Feb Carrier mobility (cont.) If the rate of a single contribution to the scattering is 1/  i, then the total scattering rate, 1/  coll is

Figure 1.16 (p. 31 M&K) Electron and hole mobilities in silicon at 300 K as functions of the total dopant concentration. The values plotted are the results of curve fitting measurements from several sources. The mobility curves can be generated using Equation with the following values of the parameters [3] (see table on next slide). ©rlc L05-08Feb201124

Figure 1.16 (cont. M&K) ParameterArsenicPhosphorusBoron μ min μ max N ref 9.68 X X X α ©rlc L05-08Feb201125

©rlc L05-08Feb 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

©rlc L05-08Feb 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 )

©rlc L05-08Feb 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)

©rlc L05-08Feb References M&K and 1 Device Electronics for Integrated Circuits, 2 ed., by Muller and Kamins, Wiley, New York, –See Semiconductor Device Fundamen- tals, by Pierret, Addison-Wesley, 1996, for another treatment of the  model. 2 Physics of Semiconductor Devices, by S. M. Sze, Wiley, New York, 1981.

©rlc L05-08Feb References *Fundamentals of Semiconductor Theory and Device Physics, by Shyh Wang, Prentice Hall, **Semiconductor Physics & Devices, by Donald A. Neamen, 2nd ed., Irwin, Chicago. M&K = Device Electronics for Integrated Circuits, 3rd ed., by Richard S. Muller, Theodore I. Kamins, and Mansun Chan, John Wiley and Sons, New York, 2003.