1. Semiconductor Device Modeling and Characterization – EE5342 Lecture 5 – Spring 2011 Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/

2. ©rlc L05-28Jan20112 First Assignment e-mail to listserv@listserv.uta.edu –In the body of the message include subscribe EE5342 This will subscribe you to the EE5342 list. Will receive all EE5342 messages If you have any questions, send to ronc@uta.edu, with EE5342 in subject line.

3. ©rlc L05-28Jan20113 Second Assignment Submit a signed copy of the document that is posted at www.uta.edu/ee/COE%20Ethics%20Statement%20Fall%2007.pdf

4. ©rlc L05-28Jan20114 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

5. ©rlc L05-28Jan20115 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

6. ©rlc L05-28Jan20116 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

7. ©rlc L05-28Jan20117 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

8. ©rlc L05-28Jan20118 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

9. ©rlc L05-28Jan20119 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 )

10. ©rlc L05-28Jan201110 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- typeE 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 )

11. ©rlc L05-28Jan201111 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

12. ©rlc L05-28Jan201112 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

13. ©rlc L05-28Jan201113 Equilibrium electron conc. and energies

14. ©rlc L05-28Jan201114 Equilibrium hole conc. and energies

15. ©rlc L05-28Jan201115 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

16. ©rlc L05-28Jan201116 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.

17. ©rlc L05-28Jan201117 Carrier mobility (cont.) If the rate of a single contribution to the scattering is 1/  i, then the total scattering rate, 1/  coll is

18. ©rlc L05-28Jan201118 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

19. ©rlc L05-28Jan201119 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 )

20. ©rlc L05-28Jan201120 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)

21. ©rlc L05-28Jan201121 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. 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. 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.