FE Electrical and Computer Properties of Electrical Materials Practice Problems

The current FE Electrical and Computer specification assigns 4-6 of the 110 exam questions to Properties of Electrical Materials. Form A contains 4 original problems in this area. The sample below is published in full, with the same worked-solution format used throughout the book.

Silicon at 300 K is doped with donors at $3 \times 1 0^{16} cm^{- 3}$ and acceptors at $1 \times 1 0^{16} cm^{- 3}$ . The intrinsic carrier concentration of silicon at this temperature is $1.5 \times 1 0^{10} cm^{- 3}$ . The equilibrium hole concentration is most nearly:

(A) $1.1 \times 1 0^{4} cm^{- 3}$
(B) $2.3 \times 1 0^{4} cm^{- 3}$
(C) $5.6 \times 1 0^{3} cm^{- 3}$
(D) $7.5 \times 1 0^{3} cm^{- 3}$

Show worked solution

Answer: (A)

Compensation leaves a net donor density of $2 \times 1 0^{16} cm^{- 3}$ , and the mass-action law puts the equilibrium hole concentration at about $1.1 \times 1 0^{4} cm^{- 3}$ .

Compensated material carries a majority density equal to the net dopant excess, since each acceptor consumes one donor electron.

n = N_{D} - N_{A} = 3 \times 1 0^{16} - 1 \times 1 0^{16} = 2 \times 1 0^{16} cm^{- 3}

Mass action then fixes the minority population, which in heavily doped silicon sits many orders below the intrinsic level.

p = \frac{n _{i}^{2}}{n} = \frac{( 1.5 \times 1 0 ^{10} ) ^{2}}{2 \times 1 0 ^{16}} = \frac{2.25 \times 1 0 ^{20}}{2 \times 1 0 ^{16}}

p = 1.1 \times 1 0^{4} cm^{- 3}

FE Reference Handbook — Properties of Electrical Materials: Intrinsic and Extrinsic Carrier Concentration (Mass Action Law)

Why the other choices appear

(B)Takes the acceptor concentration $1 \times 1 0^{16} cm^{- 3}$ as the majority density: $2.25 \times 1 0^{20} / (1 \times 1 0^{16}) = 2.25 \times 1 0^{4} cm^{- 3}$ .
(C)Adds the acceptor density to the donor density instead of subtracting: $2.25 \times 1 0^{20} / (4 \times 1 0^{16}) = 5.6 \times 1 0^{3} cm^{- 3}$ .
(D)Ignores compensation and divides by the donor density alone: $2.25 \times 1 0^{20} / (3 \times 1 0^{16}) = 7.5 \times 1 0^{3} cm^{- 3}$ .