Question 19.5: Design of a Semiconductor Design a p-type semiconductor base......
Design of a Semiconductor
Design a p-type semiconductor based on silicon, which provides a constant conductivity of 100 ohm^{-1} ⋅ cm^{-1} over a range of temperatures. Compare the required concentration of acceptor atoms in Si with the concentration of Si atoms.
In order to obtain the desired conductivity, we must dope the silicon with atoms having a valence of +3, adding enough dopant to provide the required number of charge carriers. If we assume that the number of intrinsic carriers is small compared to the dopant concentration, then
σ = N_{a}qμ_{p}
where σ = 100 ohm^{-1} · cm^{-1} and μ_{p} = 480 cm²/(V · s). Note that electron and hole mobilities are properties of the host material (i.e., silicon in this case) and not the dopant species. If we remember that a coulomb can be expressed as ampere-seconds and voltage can be expressed as ampere-ohm, the number of charge carriers required is
N_{a} = \frac{σ}{qμ_{p}}= \frac{100 \ ohm^{-1} \ cm^{-1}}{(1.6 × 10^{-19} \ A · s)[480 \ cm^2/(A · ohm · s)]} = 1.30 × 10^{18} acceptor atoms/cm³
Assume that the lattice constant of Si remains unchanged as a result of doping:
N_{a} = \frac{(1 \ \text{hole/dopant} \ \text{atom})(x \ \text{dopant} \ \text{atom}/\text{Si} \ \text{atom})(8 \ Si \ \text{atoms/unit} \ \text{cell})}{(5.4307 × 10^{-8} \ cm)^3/\text{unit} \ \text{cell}}
x = (1.30 × 10^{18})(5.4307 × 10^{-8})^3/8 = 26 × 10^{-6} dopant atom/Si atom
or 26dopant atoms/10^6 Si atoms
Possible dopants include boron, aluminum, gallium, and indium. High-purity chemicals and clean room conditions are essential for processing since we need 26 dopant atoms per million silicon atoms.