Examples on Doping and Fermi Levels Zeynep Dilli, Oct

ENEE 313, Spr ’09 Supplement II Examples on Doping and Fermi Levels Zeynep Dilli, Oct. 2008, rev. Mar. 2009

This is a supplement presenting examples on the relationship between doping, carrier concentrations and the Fermi level at equilibrium. ◦ 10 3 Required Constants: For silicon at T =300 K, ni = 1.45 × 10 1/cm . The Boltzmann constant −5 kB = k = 8.61 × 10 eV/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 = 16 3 6 × 10 1/cm . Find the equilibrium electron concentration n0, hole concentration p0, and Fermi level EF with respect to the intrinsic Fermi level Ei and conduction band edge EC .

This is an n-type material, as it is doped with donor atoms. Therefore:

16 3 n0 ≈ ND = 6 × 10 1/cm (1)

Then we can use the Law of Mass Action to ﬁnd the hole concentration:

2 20 ni 2.1 × 10 3 3 p0 = = 16 = 3.5 × 10 1/cm (2) n0 6 × 10

To ﬁnd the Fermi level with respect to the intrinsic Fermi level, we use the expression that links electron concentration to Ei and ni:

EF − Ei n0 n0 = ni exp( ) ⇒ (EF − Ei) = kT ln( ) (3) kT ni

At room temperature kT = 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:

6 × 1016 (E − E ) = 0.026 ln( ) = 0.026 × 15.24 = 0.396 eV F i 1.45 × 1010 ⇒ EF = Ei + 0.396 eV (5)

Drawing the band energy diagram, we can then place the Fermi level in the correct place with respect to the intrinsic Fermi level (middle of bandgap) and also ﬁnd its separation from the conduction band edge EC . See Figure 1.

1 Figure 1: Solution to Problem 1. This is an n-type material doped at 6 × 1016 1/cm3.

2 2. Consider a silicon crystal at 300 ◦K, with the Fermi level 0.18 eV above the valence band. What type is the material? What are the electron 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 EF and EV is given. Then, as can be seen from the ﬁgure, we can calculate the separation between Ei and EF :

Ei − EV = Eg/2 = 0.56 eV

Ei − 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 than it is to the conduction band). Therefore this is a p-type material, and we can ﬁnd the equilibrium hole concentration as

E − E 0.38 p = n exp( i F ) = 1.45 × 1010 exp( ) ≈ 3.23 × 1016 1/cm3 (7) 0 i kT 0.026 We can then calculate the equilibrium electron concentration by the Law of Mass Action:

2 20 ni 2.1 × 10 3 3 n0 = = 16 ≈ 6.5 × 10 1/cm (8) p0 3.23 × 10

3 Note that we could also arrive at this result by using the EF -related expression for n0: E − E −0.38 n = n exp( F i ) = 1.45 × 1010 exp( ) = 1.45 × 1010 exp(−14.62) ≈ 6.5 × 103 1/cm3 (9) 0 i kT 0.026

As an exercise, go back to Problem 1 and show that you could use the expression in Eqn. 7 to calculate p0 in that problem after the calculation of EF in Eqn. 5.

4 3. Consider a silicon crystal at room temperature, doped with both donor and acceptor atoms so that 15 3 15 3 ND = 2 × 10 1/cm and NA = 1 × 10 1/cm . What type of material would this yield? Find the location of the Fermi level.

Here, ND > NA, so with more donors than acceptors this will be an n-type crystal.

15 3 n0 ≈ ND − NA = 1 × 10 1/cm (10)

Then 2 20 ni 2.1 × 10 5 3 p0 = = 15 = 2.1 × 10 1/cm (11) n0 1 × 10 and 15 n0 1 × 10 (EF − Ei) = kT ln( ) = 0.026 ln( 10 ) = 0.026 · 11.14 = 0.29 eV (12) ni 1.45 × 10 The band energy diagram will be as in Figure 3.

Figure 3: The energy band diagram for the crystal presented in Problem 3.