ENEE 313, Spr ’09Supplement IIExamples on Doping and Fermi LevelsZeynep Dilli, Oct. 2008, rev. Mar. 2009This is a supplement presenting examples on the relationship between doping, carrier concentrationsand the Fermi level at equilibrium.Required Constants: For silicon at T =300◦K, ni= 1.45 × 10101/cm3. The Boltzmann constantkB= k = 8.61 × 10−5eV/K. Silicon bandgap energy Eg=1.12 eV.1. Consider a silicon crystal at room temperature (300◦K) doped with arsenic atoms so that ND=6 × 10161/cm3. Find the equilibrium electron concentration n0, hole concentration p0, and Fermilevel EFwith respect to the intrinsic Fermi level Eiand conduction band edge EC.This is an n-type material, as it is doped with donor atoms. Therefore:n0≈ ND= 6 × 10161/cm3(1)Then we can use the Law of Mass Action to find the hole concentration:p0=n2in0=2.1 × 10206 × 1016= 3.5 × 1031/cm3(2)To find the Fermi level with respect to the intrinsic Fermi level, we use the expression that linkselectron concentration to Eiand ni:n0= niexp(EF− EikT) ⇒ (EF− Ei) = kT ln(n0ni) (3)At room temperaturekT = 8.61 × 10−5× 300 ≈ 0.026 eV (4)We will be using this quantity often.Then the separation of the Fermi level and intrinsic Fermi level is, from Eqn. 3:(EF− Ei) = 0.026 ln(6 × 10161.45 × 1010) = 0.026 × 15.24 = 0 .396 eV⇒ EF= Ei+ 0.396 eV (5)Drawing the band energy diagram, we can then place the Fermi level in the correct place with respectto the intrinsic Fermi level (middle of bandgap) and also find its separation from the conduction bandedge EC. See Figure 1.1Figure 1: Solution to Problem 1. This is an n-type material doped at 6 × 10161/cm3.22. Consider a silicon crystal at 300◦K, with the Fermi level 0.18 eV above the valence band. Whattype is the material? What are the elec tron and hole concentrations?We start with drawing the band energy diagram for the crystal, as in Figure 2:Figure 2: The energy band diagram for the crystal presented in Problem 2.The separation between EFand EVis given. Then, as can be seen from the figure, we can calculatethe separation between Eiand EF:Ei− EV= Eg/2 = 0.56 eVEi− EV= (Ei− EF) + (EF− EV) ⇒ (Ei− EF) = Eg/2 − (EF− EV)⇒ (Ei− EF) = 0.56 − 0.18 = 0.38 eV (6)The Fermi level is below the intrinsic Fermi level (in other words, it is closer to the valence band thanit is to the conduction band). Therefore this is a p-type material, and we can find the equilibriumhole concentration asp0= niexp(Ei− EFkT) = 1.45 × 1010exp(0.380.026) ≈ 3.23 × 10161/cm3(7)We can then calculate the equilibrium electron concentration by the Law of Mass Action:n0=n2ip0=2.1 × 10203.23 × 1016≈ 6.5 × 1031/cm3(8)3Note that we could also arrive at this result by using the EF-related expression for n0:n0= niexp(EF− EikT) = 1.45 × 1010exp(−0.380.026) = 1.45 × 1010exp(−14.62) ≈ 6.5 × 1031/cm3(9)As an exercise, go back to Problem 1 and show that you could use the expression in Eqn. 7 tocalculate p0in that problem after the calculation of EFin Eqn. 5.43. Consider a silicon crystal at room temperature, doped with both donor and acceptor atoms so thatND= 2 × 10151/cm3and NA= 1 × 10151/cm3. What type of material would this yield? Find thelocation of the Fermi level.Here, ND> NA, so with more donors than acceptors this will be an n-type crystal.n0≈ ND− NA= 1 × 10151/cm3(10)Thenp0=n2in0=2.1 × 10201 × 1015= 2.1 × 1051/cm3(11)and(EF− Ei) = kT ln(n0ni) = 0.026 ln(1 × 10151.45 × 1010) = 0.026 · 11.14 = 0 .29 eV (12)The band energy diagram will be as in Figure 3.Figure 3: The energy band diagram for the crystal presented in Problem