1. L2 January 171 Semiconductor Device Modeling and Characterization EE5342, Lecture 2-Spring 2002 Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/

2. L2 January 172 K-P E(k) Relationship*

3. L2 January 173 Analogy: a nearly -free electr. model Solutions can be displaced by ka = 2n  Allowed and forbidden energies Infinite well approximation by replacing the free electron mass with an “effective” mass (noting E = p 2 /2m = h 2 k 2 /2m) of

4. L2 January 174 Generalizations and Conclusions The symm. of the crystal struct. gives “allowed” and “forbidden” energies (sim to pass- and stop-band) The curvature at band-edge (where k = (n+1)  ) gives an “effective” mass.

5. L2 January 175 Silicon Covalent Bond (2D Repr) Each Si atom has 4 nearest neighbors Si atom: 4 valence elec and 4+ ion core 8 bond sites / atom All bond sites filled Bonding electrons shared 50/50 _ = Bonding electron

6. L2 January 176 Silicon Band Structure** Indirect Bandgap Curvature (hence m*) is function of direction and band. [100] is x-dir, [111] is cube diagonal E g = 1.17-  T 2 /(T+  )  = 4.73E-4 eV/K,  = 636K

7. L2 January 177 Si Energy Band Structure at 0 K Every valence site is occupied by an electron No electrons allowed in band gap No electrons with enough energy to populate the conduction band

8. L2 January 178 Si Bond Model Above Zero Kelvin Enough therm energy ~kT(k=8.62E-5eV/K) to break some bonds Free electron and broken bond separate One electron for every “hole” (absent electron of broken bond)

9. L2 January 179 Band Model for thermal carriers Thermal energy ~kT generates electron-hole pairs At 300K E g (Si) = 1.124 eV >> kT = 25.86 meV, Nc = 2.8E19/cm3 > Nv = 1.04E19/cm3 >> n i = 1E10/cm3

10. L2 January 1710 Donor: cond. electr. due to phosphorous P atom: 5 valence elec and 5+ ion core 5th valence electr has no avail bond Each extra free el, -q, has one +q ion # P atoms = # free elect, so neutral H atom-like orbits

11. L2 January 1711 Bohr model H atom- like orbits at donor Electron (-q) rev. around proton (+q) Coulomb force, F=q 2 /4  Si  o,q=1.6E-19 Coul,  Si =11.7,  o =8.854E-14 Fd/cm Quantization L = mvr = nh/2  E n = -(Z 2 m*q 4 )/[8(  o  Si ) 2 h 2 n 2 ] ~-40meV r n = [n 2 (  o  Si )h]/[Z  m*q 2 ] ~ 2 nm for Z=1, m*~m o /2, n=1, ground state

12. L2 January 1712 Band Model for donor electrons Ionization energy of donor E i = E c -E d ~ 40 meV Since E c -E d ~ kT, all donors are ionized, so N D ~ n Electron “freeze- out” when kT is too small

13. L2 January 1713 Acceptor: Hole due to boron B atom: 3 valence elec and 3+ ion core 4th bond site has no avail el (=> hole) Each hole adds -(-q) and has one -q ion #B atoms = #holes, so neutral H atom-like orbits

14. L2 January 1714 Hole orbits and acceptor states Similar to free electrons and donor sites, there are hole orbits at acceptor sites The ionization energy of these states is E A - E V ~ 40 meV, so N A ~ p and there is a hole “freeze-out” at low temperatures

15. L2 January 1715 Impurity Levels in Si: E G = 1,124 meV Phosphorous, P: E C - E D = 44 meV Arsenic, As:E C - E D = 49 meV Boron, B: E A - E V = 45 meV Aluminum, Al: E A - E V = 57 meV Gallium, Ga: E A - E V = 65meV Gold, Au: E A - E V = 584 meV E C - E D = 774 meV

16. L2 January 1716 Semiconductor Electronics - concepts thus far Conduction and Valence states due to symmetry of lattice “Free-elec.” dynamics near band edge Band Gap –direct or indirect –effective mass in curvature Thermal carrier generation Chemical carrier gen (donors/accept)

17. L2 January 1717 Counting carriers - 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

18. L2 January 1718 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 *

19. L2 January 1719 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

20. L2 January 1720 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

21. L2 January 1721 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

22. L2 January 1722 Electron Conc. in the MB approx. Assuming the MB approx., the equilibrium electron concentration is

23. L2 January 1723 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 = 1E10/cm3

24. L2 January 1724 Calculating the equilibrium n o The ideal 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

25. L2 January 1725 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

26. L2 January 1726 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) = 0.886 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

27. L2 January 1727 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

28. L2 January 1728 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

29. L2 January 1729 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

30. L2 January 1730 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

31. L2 January 1731 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

32. L2 January 1732 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

33. L2 January 1733 Equilibrium conc (cont.) For N a > N d (taking the + root) p o = (N a -N d )/2 + {[(N a - N d )/2] 2 +n i 2 } 1/2 For N a >> N d and N a >> n i, can use the binomial expansion, giving p o = N a /2 + N a /2[1 + 2n i 2 /N a 2 + … ] So p o = N a in the limit of N a >> N d and N a >> n i

34. L2 January 1734 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

35. L2 January 1735 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.