EE 5340 Semiconductor Device Theory Lecture 04 – Spring 2011 Professor Ronald L. Carter

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EE 5340 Semiconductor Device Theory Lecture 04 – Spring 2011 Professor Ronald L. Carter

©rlc L04-28Jan20112 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 L04-28Jan20113 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 L04-28Jan20114 Second Assignment Submit a signed copy of the document posted at

©rlc L04-28Jan2011 Quantum density of states function 1 dim electron wave #s range for n+1 “atoms” is 2  /L < k < 2  /a where a is “interatomic” distance and L = na is the length of the assembly (k = 2  / ) Shorter s, would “oversample” if n increases by 1, dp is h/L Extn 3D: E = p 2 /2m = h 2 k 2 /2m so a vol of p-space of 4  p 2 dp has h 3 /L x L y L z 5

©rlc L04-28Jan2011 QM density of states (cont.) So density of states, g c (E) is (Vol in p-sp)/(Vol per state*V) = 4  p 2 dp/[(h 3 /L x L y L z )*V] Noting that p 2 = 2mE, this becomes g c (E) = {4  2m n *) 3/2 /h 3 }(E-E c ) 1/2 and E - E c = h 2 k 2 /2m n * Similar for the hole states where E v - E = h 2 k 2 /2m p * 6

©rlc L04-28Jan2011 Fermi-Dirac distribution fctn The probability of an electron having an energy, E, is given by the F-D distr f F (E) = {1+exp[(E-E F )/kT]} -1 Note: f F (E F ) = 1/2 E F is the equilibrium energy of the system The sum of the hole probability and the electron probability is 1 7

©rlc L04-28Jan2011 Fermi-Dirac DF (continued) So the probability of a hole having energy E is 1 - f F (E) At T = 0 K, f F (E) becomes a step function and 0 probability of E > E F At T >> 0 K, there is a finite probability of E >> E F 8

©rlc L04-28Jan2011 Maxwell-Boltzman Approximation f F (E) = {1+exp[(E-E F )/kT]} -1 For E - E F > 3 kT, the exp > 20, so within a 5% error, f F (E) ~ exp[-(E-E F )/kT] This is the MB distribution function MB used when E-E F >75 meV (T=300K) For electrons when E c - E F > 75 meV and for holes when E F - E v > 75 meV 9

©rlc L04-28Jan2011 Electron Conc. in the MB approx. Assuming the MB approx., the equilibrium electron concentration is 10

©rlc L04-28Jan2011 Electron and 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 = 1.45E10/cm3 11

©rlc L04-28Jan2011 Calculating the equilibrium n o The idea is to calculate the equilibrium electron concentration n o for the FD distribution, where f F (E) = {1+exp[(E-E F )/kT]} -1 g c (E) = [4  2m n *) 3/2 (E-E c ) 1/2 ]/h 3 12

©rlc L04-28Jan2011 Equilibrium con- centration for n o Earlier quoted the MB approximation n o = N c exp[-(E c - E F )/kT],(=N c exp  F ) The exact solution is n o = 2N c F 1/2 (  F )/  1/2 Where F 1/2 (  F ) is the Fermi integral of order 1/2, and  F = (E F - E c )/kT Error in n o,  is smaller than for the DF:  = 31%, 12%, 5% for -  F = 0, 1, 2 13

©rlc L04-28Jan2011 Equilibrium con- centration for p o Earlier quoted the MB approximation p o = N v exp[-(E F - E v )/kT],(=N v exp  ’ F ) The exact solution is p o = 2N v F 1/2 (  ’ F )/  1/2 Note: F 1/2 (  ) = 0.678, (  1/2 /2) = Where F 1/2 (  ’ F ) is the Fermi integral of order 1/2, and  ’ F = (E v - E F )/kT Errors are the same as for p o 14

©rlc L04-28Jan2011 Figure 1.10 (a) Fermi-Dirac distribution function describing the probability that an allowed state at energy E is occupied by an electron. (b) The density of allowed states for a semiconductor as a function of energy; note that g(E) is zero in the forbidden gap between E v and E c. (c) The product of the distribution function and the density-of-states function. (p M&K) 15

©rlc L04-28Jan2011 Degenerate and nondegenerate cases Bohr-like doping model assumes no interaction between dopant sites If adjacent dopant atoms are within 2 Bohr radii, then orbits overlap This happens when N d ~ N c (E F ~ E c ), or when N a ~ N v (E F ~ E v ) The degenerate semiconductor is defined by E F ~/> E c or E F ~/< E v 16

©rlc L04-28Jan2011 Figure 1.13 Energy-gap narrowing  E g as a function of electron concentration. [A. Neugroschel, S. C. Pao, and F. A. Lindhold, IEEE Trans. Electr. Devices, ED-29, 894 (May 1982).] taken from p M&K) 17

©rlc L04-28Jan2011 Donor ionization The density of elec trapped at donors is n d = N d /{1+[exp((E d -E F )/kT)/2]} Similar to FD DF except for factor of 2 due to degeneracy (4 for holes) Furthermore n d = N d - N d +, also For a shallow donor, can have E d -E F >> kT AND E c -E F >> kT: Typically E F -E d ~ 2kT 18

©rlc L04-28Jan2011 Donor ionization (continued) Further, if E d - E F > 2kT, then n d ~ 2N d exp[-(E d -E F )/kT],  < 5% If the above is true, E c - E F > 4kT, so n o ~ N c exp[-(E c -E F )/kT],  < 2% Consequently the fraction of un- ionized donors is n d /n o = 2N d exp[(E c -E d )/kT]/N c = 0.4% for N d (P) = 1e16/cm 3 19

©rlc L04-28Jan2011 Figure 1.9 Electron concentration vs. temperature for two n-type doped semiconductors: (a) Silicon doped with 1.15 X arsenic atoms cm - 3 [1], (b) Germanium doped with 7.5 X arsenic atoms cm -3 [2]. (p.12 in M&K) 20

©rlc L04-28Jan 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.