Question

(a) The room-temperature electrical conductivity of a semiconductor specimen is 13 (Ω-m)-1. The hole concentration is known to be 4.0E+20 m-3. The electron and hole mobilities for this material are 0.38 and 0.18 m2/V-s, respectively. Compute the electron concentration. (Use scientific notation.) m-3

(b) On the basis of the result in part (a), is the specimen intrinsic, n-type extrinsic, or p-type extrinsic? the tolerance is +/-2%

Answer 1

Answer: a) 0.24E+20 m-3. b) p-type extrinsic.

Explanation:

The current density in a semiconductor is composed by two types of charge carriers: electrons and holes.

This parameter, is proportional to the Electric field within the semiconductor, being the proportionality constant, the electrical conductivity of the material, that takes into account the charge carrier concentrations, and the mobility for each type.

The expression for electrical conductivity is as follows:

σ = q . ne . µe + q . np . µp

Replacing by the given values, and the value of q (charge of an electron), we can get the only unknown that remains, ne , as follows:

ne =( σ – (q . np . µp)) / q µe = (13 (Ω.m)-1 – (1.6E-19) coul(4.0E+20) m-3.0.18) m2/V-s /( (1.6E-19).0.38) coul.m2/V-s

ne = 0.24E+20.

As ne is smaller than np, this means that the semiconductor behaves like a p-type extrinsic one.