are mostly fi lled with electrons and the energy states above EF are mostly empty. In the degenerate n-type semiconductor, the states between EF and Ec are mostly fi lled with electrons; thus, the electron concentration in the conduction band is very large.
Similarly, in the degenerate p-type semiconductor, the energy states between Ev and EF are mostly empty; thus, the hole concentration in the valence band is very large.
4.4 | STATISTICS OF DONORS AND ACCEPTORS
In the previous chapter, we discussed the Fermi–Dirac distribution function, which gives the probability that a particular energy state will be occupied by an electron.
We need to reconsider this function and apply the probability statistics to the donor and acceptor energy states.
4.4.1 Probability Function
One postulate used in the derivation of the Fermi–Dirac probability function was the Pauli exclusion principle, which states that only one particle is permitted in each quantum state. The Pauli exclusion principle also applies to the donor and acceptor states.
Suppose we have Ni electrons and gi quantum states, where the subscript i in- dicates the ith energy level. There are gi ways of choosing where to put the fi rst particle. Each donor level has two possible spin orientations for the donor electron;
thus, each donor level has two quantum states. The insertion of an electron into one quantum state, however, precludes putting an electron into the second quantum state.
By adding one electron, the vacancy requirement of the atom is satisfi ed, and the addition of a second electron in the donor level is not possible. The distribution func- tion of donor electrons in the donor energy states is then slightly different than the Fermi–Dirac function.
The probability function of electrons occupying the donor state is nd N d
____
1 1 _
2 exp Ed E F
__ kT
(4.50)(a) Conduction band
Filled states (electrons)
Valence band Ev
EF Ec
Electron energy
(b) Conduction band
Empty states (holes)
Valence band EF
Ec
Ev
Electron energy
Figure 4.11 | Simplifi ed energy-band diagrams for degenerately doped (a) n-type and (b) p-type semiconductors.
where nd is the density of electrons occupying the donor level and Ed is the energy of the donor level. The factor _ 12 in this equation is a direct result of the spin factor just mentioned. The 1_ 2 factor is sometimes written as 1g, where g is called a degeneracy factor.
Equation (4.50) can also be written in the form
nd Nd N d (4.51)
where N d is the concentration of ionized donors. In many applications, we will be interested more in the concentration of ionized donors than in the concentration of electrons remaining in the donor states.
If we do the same type of analysis for acceptor atoms, we obtain the expression pa ____ N a
1 1 _ g exp EF E a
__ kT
Na N a (4.52)where Na is the concentration of acceptor atoms, Ea is the acceptor energy level, pa
is the concentration of holes in the acceptor states, and N a is the concentration of ionized acceptors. A hole in an acceptor state corresponds to an acceptor atom that is neutrally charged and still has an “empty” bonding position as we have discussed in Section 4.2.1. The parameter g is, again, a degeneracy factor. The ground state de- generacy factor g is normally taken as 4 for the acceptor level in silicon and gallium arsenide because of the detailed band structure.
4.4.2 Complete Ionization and Freeze-Out
The probability function for electrons in the donor energy state was just given by Equation (4.50). If we assume that (Ed EF) kT, then
nd N d
___
1 _
2 exp Ed E F
__ kT
2Nd exp __ (EdkT E F) (4.53)If (Ed EF) kT, then the Boltzmann approximation is also valid for the electrons in the conduction band so that, from Equation (4.11),
n0 Nc exp (Ec EF)
__ kT
We can determine the relative number of electrons in the donor state compared with the total number of electrons; therefore, we can consider the ratio of electrons in the donor state to the total number of electrons in the conduction band plus donor state. Using the expressions of Equations (4.53) and (4.11), we write
nd
__ nd n0
2Nd exp (Ed EF)
__ kT
________
2Nd exp __ (EdEF)
kT
Nc exp __ (EckT EF) (4.54)4 . 4 Statistics of Donors and Acceptors 133
The Fermi energy cancels out of this expression. Dividing by the numerator term, we obtain
nd
__ nd n0
_____ 1
1 _ Nc
2Nd
exp __ (Ec Ed)
kT
(4.55)The factor (Ec Ed) is just the ionization energy of the donor electrons.
EXAMPLE 4.7
Objective: Determine the fraction of total electrons still in the donor states at T 300 K.
Consider phosphorus doping in silicon, for T 300 K, at a concentration of Nd 1016 cm3.
■ Solution
Using Equation (4.55), we fi nd nd
__ n0 nd
_____ 1 1 2__ .8 1019
2(1016) exp __ 0.045
0.0259
0.0041 0.41%■ Comment
This example shows that there are very few electrons in the donor state compared with the conduction band. Essentially all of the electrons from the donor states are in the conduction band and, since only about 0.4 percent of the donor states contain electrons, the donor states are said to be completely ionized.
■ EXERCISE PROBLEM
Ex 4.7 Repeat Example 4.7 for (a) T 250 K and (b) T 200 K. (c) What can be said about the fraction as the temperature decreases?
[Ans. (a) 7.50
3 10
; (b) 1.75
2 10
; (c) Fraction increases as temperature decreases.]
At room temperature, then, the donor states are essentially completely ionized and, for a typical doping of 1016 cm3, almost all donor impurity atoms have donated an electron to the conduction band.
At room temperature, there is also essentially complete ionization of the accep- tor atoms. This means that each acceptor atom has accepted an electron from the valence band so that pa is zero. At typical acceptor doping concentrations, a hole is created in the valence band for each acceptor atom. This ionization effect and the cre- ation of electrons and holes in the conduction band and valence band, respectively, are shown in Figure 4.12.
The opposite of complete ionization occurs at T 0 K. At absolute zero degrees, all electrons are in their lowest possible energy state; that is, for an n-type semiconductor, each donor state must contain an electron, therefore nd Nd or N d 0. We must have, then, from Equation (4.50) that exp [(Ed EF)kT] 0. Since T 0 K, this will occur for exp () 0, which means that EF Ed. The Fermi energy level must be above the donor energy level at absolute zero. In the case of a p-type semiconductor at absolute zero temperature, the impurity atoms will not contain any electrons, so that the Fermi en- ergy level must be below the acceptor energy state. The distribution of electrons among the various energy states, and hence the Fermi energy, is a function of temperature.
A detailed analysis, not given in this text, shows that at T 0 K, the Fermi en- ergy is halfway between Ec and Ed for the n-type material and halfway between Ea
and Ev for the p-type material. Figure 4.13 shows these effects. No electrons from the donor state are thermally elevated into the conduction band; this effect is called freeze-out. Similarly, when no electrons from the valance band are elevated into the acceptor states, the effect is also called freeze-out.
Between T 0 K, freeze-out, and T 300 K, complete ionization, we have partial ionization of donor or acceptor atoms.
(a)
Ec Ed EFi
Ev
Electron energy
Conduction band
Valence band
ⴙ ⴙ ⴙ ⴙ ⴙ ⴙ ⴙ ⴚ ⴚ ⴚ ⴚ ⴚ ⴚ ⴚ
(b)
Ec
EFi Ea
Electron energy
Conduction band
Ev Valence band
ⴙ ⴙ ⴙ ⴙ ⴙ ⴙ ⴙ ⴚ ⴚ ⴚ ⴚ ⴚ ⴚ ⴚ
Figure 4.12 | Energy-band diagrams showing complete ionization of (a) donor states and (b) acceptor states.
(a)
Ec Ed EF
EFi
Ev
Electron energy
Conduction band
Valence band
(b)
Ec
Ea EF
EFi
Ev
Electron energy
Conduction band
Valence band
Figure 4.13 | Energy-band diagram at T 0 K for (a) n-type and (b) p-type semiconductors.
EXAMPLE 4.8 Objective: Determine the temperature at which 90 percent of acceptor atoms are ionized.
Consider p-type silicon doped with boron at a concentration of Na 1016 cm3.
■ Solution
Find the ratio of holes in the acceptor state to the total number of holes in the valence band plus acceptor state. Taking into account the Boltzmann approximation and assuming the de- generacy factor is g 4, we write
pa
__ p0 pa
_____ 1 1 _ Nv
4Na
exp __ (Ea Ev) kT