Fundamentals of Materials Science and Engineering: An Integrated Approach, 3e

GTBL042-12 GTBL042-Callister-v3 October 2, 2007 13:41

2nd Revised Pages

12.13 Factors That Affect Carrier Mobility • 487

Since this material is extrinsic andp-type (i.e.,pn), the electrical con-

ductivity is a function of both hole concentration and hole mobility according to

Equation 12.17. In addition, it will be assumed that at room temperature all of the

acceptor dopant atoms have accepted electrons to form holes (i.e., that we are in

the “extrinsic region” of Figure 12.17), which is to say that the number of holes is

approximately equal to the number of acceptor impuritiesNa.

This problem is complicated by the fact thatμhis dependent on impurity

content per Figure 12.18. Consequently, one approach to solving this problem is

trial and error: assume an impurity concentration, and then compute the con-

ductivity using this value and the corresponding hole mobility from its curve of

Figure 12.18. Then on the basis of this result, repeat the process assuming another

impurity concentration.

For example, let us select anNavalue (i.e., apvalue) of 10^22 m−^3. At this

concentration the hole mobility is approximately 0.04 m^2 /V-s (Figure 12.18); these

values yield a conductivity of

σ=p|e|μh=(10^22 m−^3 )(1. 6 × 10 −^19 C)(0.04 m^2 /V-s)

=64 (-m)−^1

which is a little on the high side. Decreasing the impurity content an order of

magnitude to 10^21 m−^3 results in only a slight increase ofμhto about 0.045 m^2 /V-s

(Figure 12.18); thus, the resulting conductivity is

σ=(10^21 m−^3 )(1. 6 × 10 −^19 C)(0.045 m^2 /V-s)

= 7 .2(-m)−^1

With some fine-tuning of these numbers, a conductivity of 50 (-m)−^1 is achieved

whenNa=p∼= 8 × 1021 m−^3 ; at thisNavalue,μhremains approximately 0.04

m^2 /V-s.

It next becomes necessary to calculate the concentration of acceptor impurity

in atom percent. This computation first requires the determination of the number

of silicon atoms per cubic meter,NSi, using Equation 5.2, which is as follows:

NSi=

NAρSi

ASi

=

(6. 02 × 1023 atoms/mol)(2.33 g/cm^3 )(10^6 cm^3 /m^3 )

28 .09 g/mol

= 5 × 1028 m−^3

The concentration of acceptor impurities in atom percent (Ca′) is just the ratio

ofNaandNa+NSimultiplied by 100 as

Ca′=

Na

Na+NSi

× 100

=

8 × 1021 m−^3

(8× 1021 m−^3 )+(5× 1028 m−^3 )

× 100 = 1. 60 × 10 −^5

Thus, a silicon material having a room-temperaturep-type electrical conduc-

tivity of 50 (-m)−^1 must contain 1.60× 10 −^5 at% boron, aluminum, gallium, or

indium.