Question 1.2: Objective: Calculate the thermal equilibrium electron and ho...
Objective: Calculate the thermal equilibrium electron and hole concentrations.
(a) Consider silicon at T = 300 K doped with phosphorus at a concentration of N_{d} = 10^{16} cm^{−3} . Recall from Example 1.1 that n_{i} = 1.5 × 10^{10} cm^{−3} .
(b) Consider silicon at T = 300 K doped with boron at a concentration of N_{a} = 5 × 10^{16} cm^{−3}
(a) Since N_{d} \gg n_{i} , the electron concentration is
n_{o} \cong N_{d} = 10^{16} cm^{−3}
and the hole concentration is
p_{o} = \frac{n^{2} _{i}}{N_{d}} = \frac{(1.5 × 10^{10})^{2}}{10^{16}} = 2.25 × 10^{4} cm^{−3}
(b) Since N_{a} \gg n_{i} , the hole concentration is
p_{o} ≅ N_{a} = 5 × 10^{16} cm^{−3}
and the electron concentration is
n_{o} = \frac{n^{2} _{i}}{N_{a}} = \frac{(1.5 × 10^{10})^{2}}{5 × 10^{16}} = 4.5 × 10^{3} cm^{−3}
Comment: We see that in a semiconductor doped with donors, the concentration of electrons is far greater than that of the holes. Conversely, in a semiconductor doped with acceptors, the concentration of holes is far greater than that of the electrons. It is also important to note that the difference in the concentrations between electrons and holes in a particular semiconductor is many orders of magnitude