Question 3 of 6: Doping Germanium for a Target Conductivity

Question 3: Doping Germanium for a Target Conductivity (15 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Target conductivity σ = 2000 (Ω·cm)^-1; Ge electron mobility μ_n = 3800 cm²/V·s; Ge hole mobility μ_p = 1820 cm²/V·s; germanium is diamond-cubic with lattice parameter a = 5.658 Å (8 atoms/cell, standard crystallographic data).

Find. The doping level, in at%, of (i) phosphorous and (ii) gallium needed to reach σ = 2000 (Ω·cm)^-1, and which semiconductor type each dopant produces.

Approach. Phosphorus is a Group-V donor (produces n-type Ge) and gallium is a Group-III acceptor (produces p-type Ge); at room temperature such shallow dopants are essentially fully ionized, so σ = Nqμ with N the required dopant density. Dividing N by Ge's own atomic density (from its diamond-cubic unit cell) converts it to at%.

1. Germanium's own atomic density. Diamond-cubic structure carries 8 atoms per unit cell: $$V_{cell}=a^3=(5.658\times10^{-8})^3=1.811\times10^{-22}\ \text{cm}^3,\qquad N_{Ge}=\dfrac{8}{V_{cell}}=4.417\times10^{22}\ \text{cm}^{-3}$$
2. Phosphorus (donor) doping — produces n-type Ge. Assuming full ionization, σ = N_Dqμ_n: $$N_D=\dfrac{\sigma}{q\mu_n}=\dfrac{2000}{(1.6\times10^{-19})(3800)}=3.289\times10^{18}\ \text{cm}^{-3}$$ $$at\%\ P=\dfrac{N_D}{N_{Ge}}\times100=\boxed{7.45\times10^{-3}\ \text{at\%}}$$
3. Gallium (acceptor) doping — produces p-type Ge. Assuming full ionization, σ = N_Aqμ_p: $$N_A=\dfrac{\sigma}{q\mu_p}=\dfrac{2000}{(1.6\times10^{-19})(1820)}=6.868\times10^{18}\ \text{cm}^{-3}$$ $$at\%\ Ga=\dfrac{N_A}{N_{Ge}}\times100=\boxed{1.56\times10^{-2}\ \text{at\%}}$$

Gallium requires roughly twice the at% of phosphorus for the same target conductivity, directly reflecting its lower hole mobility (1820 vs 3800 cm²/V·s) — a smaller mobility must be compensated by a larger carrier density to reach the same σ.