Homogeneous Semiconductors Dopants Use Density of States and Distribution Function to: Find the Number of Holes and Electrons.

Energy Levels in Hydrogen Atom

Energy Levels for Electrons in a Doped Semiconductor

Assumptions for Calculation

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

GOAL: The density of electrons (no) 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” no = bandS(E)f(E)dE

S(E) = (1/2p2) (2mdse*/2)3/2(E - EC)1/2 For quasi-free electrons in the conduction band: 1. We must use the effective mass (averaged over all directions) 2. the potential energy Ep is the edge of the conduction band (EC) S(E) = (1/2p2) (2mdse*/2)3/2(E - EC)1/2 For holes in the valence band: 1. We still use the effective mass (averaged over all directions) 2. the potential energy Ep is the edge of the valence band (EV) S(E) = (1/2p2) (2mdsh*/2)3/2(EV - E)1/2

Energy Band Diagram E(x) Eelectron S(E) Ehole x conduction band valence band EC EV x E(x) Eelectron S(E) Ehole note: increasing electron energy is ‘up’, but increasing hole energy is ‘down’.

Reminder of our GOAL: The density of electrons (no) 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” no = bandS(E)f(E)dE

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

f(E) = 1/{1+exp[(E-EF)/kT]} EF E T=0 oK T1>0 T2>T1 0.5 All energy levels are filled with e-’s below the Fermi Energy at 0 oK

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: fp(E) = 1 – f(E) = 1 - 1/{1+exp[(E-EF)/kT]} ={1+exp[(E-EF)/kT]}/{1+exp[(E-EF)/kT]} - 1/{1+exp[(E-EF)/kT]} = {exp[(E-EF)/kT]}/{1+exp[(E-EF)/kT]} =1/{exp[(EF - E)/kT] + 1} for the last line, multiply by 1 = exp(-(E-Ef)/kT)/exp(-(E-Ef)/kT)

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

Putting the pieces together: for electrons, n(E) f(E) 1 EF E T=0 oK T1>0 T2>T1 0.5 EV EC S(E) E n(E)=S(E)f(E)

Putting the pieces together: for holes, p(E) fp(E) T=0 oK 1 T1>0 T2>T1 0.5 S(E) E EV EF EC p(E)=S(E)f(E) hole energy

Finding no and po the effective density of states in the conduction band

Energy Band Diagram intrinisic semiconductor: no=po=ni E(x) conduction band EC n(E) EF=Ei p(E) EV valence band x where Ei is the intrinsic Fermi level

Energy Band Diagram n-type semiconductor: no>po E(x) conduction band EC EF n(E) p(E) EV valence band x

Energy Band Diagram p-type semiconductor: po>no E(x) conduction band EC n(E) p(E) EF EV valence band x

A very useful relationship …which is independent of the Fermi Energy Recall that ni = no= po for an intrinsic semiconductor, so nopo = ni2 for all non-degenerate semiconductors. (that is as long as EF is not within a few kT of the band edge)

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

The intrinsic Fermi Energy (Ei) For an intrinsic semiconductor, no=po and EF=Ei which gives Ei = (EC + EV)/2 + (kT/2)ln(NV/NC) so the intrinsic Fermi level is approximately in the middle of the bandgap.

no = 0.5(ND-NA)  0.5[(ND-NA)2 + 4ni2]1/2 Space-charge Neutrality Consider a semiconductor doped with NA ionized acceptors (-q) and ND ionized donors (+q). positive charges = negative charges po + ND = no + NA using ni2 = nopo ni2/no + ND = no+ NA ni2 + no(ND-NA) - no2 = 0 no = 0.5(ND-NA)  0.5[(ND-NA)2 + 4ni2]1/2 we use the ‘+’ solution since no should be increased by ni no = ND - NA in the limit that ni<<ND-NA

Similarly, po= NA - ND if NA-ND>>ni T (K) no/(ND-NA) po/(NA-ND) 1 po= no= ni intrinsic material freeze-out insufficient energy to ionize the dopant atoms carrier conc. vs. temperature 300 100 600

Degenerate Semiconductors …the doping concentration is so high that EF moves within a few kT of the band edge (EC or EV) EC EV + ED1 Eg0 Eg(ND) for ND > 1018 cm-3 in Si impurity band High donor concentrations cause the allowed donor wavefunctions to overlap, creating a band at EDn First only the high states overlap, but eventually even the lowest state overlaps. This effectively decreases the bandgap by DEg = Eg0 – Eg(ND).

Degenerate Semiconductors As the doping conc. increases more, EF rises above EC EV EC (intrinsic) available impurity band states EF DEg EC (degenerate) ~ ED filled impurity band states apparent band gap narrowing: DEg* (is optically measured) - Eg* is the apparent band gap: an electron must gain energy Eg* = EF-EV

Electron Concentration in degenerately doped n-type semiconductors The donors are fully ionized: no = ND The holes still follow the Boltz. approx. since EF-EV>>>kT po = NV exp[-(EF-EV)/kT] = NV exp[-(Eg*)/kT] = NV exp[-(Ego- DEg*)/kT] = NV exp[-Ego/kT]exp[DEg*)/kT] nopo = NDNVexp[-Ego/kT] exp[DEg*)/kT] = (ND/NC) NCNVexp[-Ego/kT] exp[DEg*)/kT] = (ND/NC)ni2 exp[DEg*)/kT]

Summary non-degenerate: nopo= ni2 degenerate n-type: nopo= ni2 (ND/NC) exp[DEg*)/kT] degenerate p-type: nopo= ni2 (NA/NV) exp[DEg*)/kT]