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 = S(E)f(E)dE band 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) * 1

1 2mdse S ( E ) 2 2 E EC 2 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) * 1

1 2mdsh S ( E ) 2 2 EV E 2 2 Energy Band Diagram Eelectron E(x)

conduction band EC S(E) EV valence band Ehole

x 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 = S(E)f(E)dE band 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) (Can derive this statistical mechanics.) f(E) = 1/{1+exp[(E-EF)/kT]} 1

T=0 oK T1>0 T2>T1 f(E) 0.5 0 EF E

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} 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 T=0 oK T1>0

T2>T1 S(E) 0.5 0 E EV EF EC n(E)=S(E)f(E)

E Putting the pieces together: for holes, p(E) fp(E) T=0 oK

1 T1>0 T2>T1 0.5 S(E) 0 EV EF EC

p(E)=S(E)f(E) hole energy E Finding no and po Ec (max) n0

S ( E ) f ( E )dE Ec * dse 2 1 2m

2 2 N C exp[ ( EC E F ) / kT ] 3/ 2

E EC e E EF / kT dE Ec * dse

m kT ...where N C 2 2 2

3/ 2 the effective density of states in the conduction band Ev * dsh

2 1 2m p0 S ( E ) f p ( E )dE 2 2 Ev (min) NV exp[ ( E F EV ) / kT ]

3 / 2 Ev EV E e EF E / kT dE *

dsh m kT ...where NV 2 2 2 3/ 2

Energy Band Diagram intrinisic semiconductor: no=po=ni E(x) conduction band EC n(E) p(E) EF=Ei

EV valence band x where Ei is the intrinsic Fermi level Energy Band Diagram n-type semiconductor: no>po n0 N C exp[ ( EC E F ) / kT ]

E(x) conduction band EC n(E) p(E) EF

EV valence band x Energy Band Diagram p-type semiconductor: po>no p0 NV exp[ ( EF EV ) / kT ]

E(x) conduction band EC n(E) p(E) EF

EV valence band x A very useful relationship n0 p0 N C exp[ ( EC E F ) / kT ] NV exp[ ( E F EV ) / kT ] N C NV e ( Ec Ev ) / kT N C NV e

E g / kT 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) E / kT n0 p0 N C NV e g ni2

ni N C NV e E g / 2 kT The intrinsic carrier density n0 p0 N C N V e ni N C NV e Eg / kT

2 i n E g / 2 kT is sensitive to the energy bandgap, temperature, and m* *

dse m kT N C 2 2 2 3/ 2

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

Consider a semiconductor doped with NA ionized acceptors (-q) and ND ionized donors (+q), do not assume that ni is small high temperature expression. 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 + 4nni2]1/2 we use the + solution since no should be increased by ni

n = N - N in the limit that n <

EC ED1 + + + + Eg(ND) 1. The doping concentration is so high that EF moves within a few kT of the band edge (EC or EV). Boltzman approximation not valid.

2. 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. Eg0 This effectively decreases the

bandgap by for ND > 1018 cm-3 in Si EV Eg = Eg0 Eg(ND). Degenerate Semiconductors As the doping conc. increases more, EF rises above EC

available impurity band states filled impurity band states Eg EC (intrinsic) EF EC (degenerate) ~ ED

apparent band gap narrowing: Eg* (is optically measured) Eg* is the apparent band gap: - an electron must gain energy Eg* = EF-EV 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- Eg*)/kT] = NV exp[-Ego/kT]exp[Eg*)/kT] nopo = NDNVexp[-Ego/kT] exp[Eg*)/kT]

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