4. Density of States (Appendix D) Energy Distribution Functions (Section 2.9) Carrier Concentrations (Sections 2.10-12) ECE G201

5. GOAL: The density of electrons (n o ) can be found precisely if we know 1. the number of allowed energy states in a small energy range, dE: S(E)dE “the density of states” 2. the probability that a given energy state will be occupied by an electron: f(E) “the distribution function” n o =  band S(E)f(E)dE

6. For quasi-free electrons in the conduction band: 1. We must use the effective mass (averaged over all directions) 2. the potential energy E p is the edge of the conduction band (E C ) S(E) = (1/2   ) (2m dse * /  2 ) 3/2 (E - E C ) 1/2 For holes in the valence band: 1. We still use the effective mass (averaged over all directions) 2. the potential energy E p is the edge of the valence band (E V ) S(E) = (1/2   ) (2m dsh * /  2 ) 3/2 (E V - E) 1/2

7. Energy Band Diagram conduction band valence band ECEC EVEV x E(x) S(E) E electron E hole note: increasing electron energy is ‘up’, but increasing hole energy is ‘down’.

8. Reminder of our GOAL: The density of electrons (n o ) can be found precisely if we know 1. the number of allowed energy states in a small energy range, dE: S(E)dE “the density of states” 2. the probability that a given energy state will be occupied by an electron: f(E) “the distribution function” n o =  band S(E)f(E)dE

9. Fermi-Dirac Distribution The probability that an electron occupies an energy level, E, is f(E) = 1/{1+exp[(E-E F )/kT]} –where T is the temperature (Kelvin) –k is the Boltzmann constant (k=8.62x10 -5 eV/K) –E F is the Fermi Energy (in eV)

10. f(E) = 1/{1+exp[(E-E F )/kT]} All energy levels are filled with e - ’s below the Fermi Energy at 0 o K f(E) 1 0 EFEF E T=0 o K T 1 >0 T 2 >T 1 0.5

11. Fermi-Dirac Distribution for holes Remember, a hole is an energy state that is NOT occupied by an electron. Therefore, the probability that a state is occupied by a hole is the probability that a state is NOT occupied by an electron: f p (E) = 1 – f(E) = 1 - 1/{1+exp[(E-E F )/kT]} ={1+exp[(E-E F )/kT]}/{1+exp[(E-E F )/kT]} - 1/{1+exp[(E-E F )/kT]} = {exp[(E-E F )/kT]}/{1+exp[(E-E F )/kT]} =1/{exp[(E F - E)/kT] + 1}

12. The Boltzmann Approximation If (E-E F )>kT such that exp[(E-E F )/kT] >> 1 then, f(E) = {1+exp[(E-E F )/kT]} -1  {exp[(E-E F )/kT]} -1  exp[-(E-E F )/kT] …the Boltzmann approx. similarly, f p (E) is small when exp[(E F - E)/kT]>>1: f p (E) = {1+exp[(E F - E)/kT]} -1  {exp[(E F - E)/kT]} -1  exp[-(E F - E)/kT] If the Boltz. approx. is valid, we say the semiconductor is non-degenerate.

13. Putting the pieces together: for electrons, n(E) f(E) 1 0 EFEF E T=0 o K T 1 >0 T 2 >T 1 0.5 EVEV ECEC S(E) E n(E)=S(E)f(E)

14. Putting the pieces together: for holes, p(E) f p (E) 1 0 EFEF E T=0 o K T 1 >0 T 2 >T 1 0.5 EVEV ECEC S(E) p(E)=S(E)f(E) hole energy

15. Finding n o and p o the effective density of states in the conduction band

16. Energy Band Diagram intrinisic semiconductor: n o =p o =n i conduction band valence band ECEC EVEV x E(x) n(E) p(E) E F =E i where E i is the intrinsic Fermi level

17. Energy Band Diagram n-type semiconductor: n o >p o conduction band valence band ECEC EVEV x E(x) n(E) p(E) EFEF

18. Energy Band Diagram p-type semiconductor: p o >n o conduction band valence band ECEC EVEV x E(x) n(E) p(E) EFEF

19. A very useful relationship …which is independent of the Fermi Energy Recall that n i = n o = p o for an intrinsic semiconductor, so n o p o = n i 2 for all non-degenerate semiconductors. (that is as long as E F is not within a few kT of the band edge)

20. The intrinsic carrier density is sensitive to the energy bandgap, temperature, and m *

21. The intrinsic Fermi Energy (E i ) For an intrinsic semiconductor, n o =p o and E F =E i which gives E i = (E C + E V )/2 + ( kT / 2 )ln(N V /N C ) so the intrinsic Fermi level is approximately in the middle of the bandgap.

22. Space-charge Neutrality Consider a semiconductor doped with N A ionized acceptors (-q) and N D ionized donors (+q). positive charges = negative charges p o + N D = n o + N A using n i 2 = n o p o n i 2 /n o + N D = n o + N A n i 2 + n o (N D -N A ) - n o 2 = 0 n o = 0.5(N D -N A )  0.5[(N D -N A ) 2 + 4n i 2 ] 1/2 we use the ‘+’ solution since n o should be increased by n i n o = N D - N A in the limit that n i << N D -N A

23. Similarly, p o = N A - N D if N A -N D >> n i T (K) n o /(N D -N A ) p o /(N A -N D ) 1 p o = n o = n i intrinsic material freeze-out insufficient energy to ionize the dopant atoms carrier conc. vs. temperature 300 100600

24. Degenerate Semiconductors …the doping concentration is so high that E F moves within a few kT of the band edge (E C or E V ) High donor concentrations cause the allowed donor wavefunctions to overlap, creating a band at E Dn First only the high states overlap, but eventually even the lowest state overlaps. This effectively decreases the bandgap by  E g = E g0 – E g (N D ). ECEC EVEV ++++ E D1 E g0 E g (N D ) for N D > 10 18 cm -3 in Si impurity band

25. Degenerate Semiconductors As the doping conc. increases more, E F rises above E C EVEV E C (intrinsic) available impurity band states EFEF EgEg E C (degenerate) ~ E D filled impurity band states apparent band gap narrowing:  E g * (is optically measured) - E g * is the apparent band gap: an electron must gain energy E g * = E F -E V

26. Electron Concentration in degenerately doped n-type semiconductors The donors are fully ionized: n o = N D The holes still follow the Boltz. approx. since E F -E V >>>kT p o = N V exp[-(E F -E V )/kT] = N V exp[-(E g * )/kT] = N V exp[-(E go -  E g * )/kT] = N V exp[-E go /kT]exp[  E g * )/kT] n o p o = N D N V exp[-E go /kT] exp[  E g * )/kT] = (N D /N C ) N C N V exp[-E go /kT] exp[  E g * )/kT] = (N D /N C )n i 2 exp[  E g * )/kT]

27. Summary non-degenerate: n o p o = n i 2 degenerate n-type: n o p o = n i 2 (N D /N C ) exp[  E g * )/kT] degenerate p-type: n o p o = n i 2 (N A /N V ) exp[  E g * )/kT]