Discussion
An intrinsic semiconductor has conduction electron concentration, n = 1012 cm–3. The mobility of both electrons and holes are identical = 4 × 104 cm2 V–1 s–1. If a voltage of 100 V is applied on two parallel end faces of the cube (edge length 1 cm) through Ohmic contacts, the current through the cube would be (in mA)
(Given: charge of electron = 1.6 × 10–19 C)
Answer: B
The current is 1280 mA.
In an intrinsic semiconductor the electron and hole densities are equal, so both carriers contribute to the conductivity:
\[ \begin{aligned} \sigma &= ne(\mu_{n} + \mu_{p}) \\ &= (10^{12})(1.6\times10^{-19}) \\ &\quad \times (8\times10^{4}) \\ &= 1.28\times10^{-2} \text{ S/cm} \end{aligned} \]
With 100 V across a 1 cm cube the field is \(E = 100\) V/cm, so the current density is \(J = \sigma E = 1.28\) A/cm2. Across the 1 cm2 face, \(I = 1.28\) A = 1280 mA.
640 mA counts only the electrons (the hole current is equal to it), and 6400 and 12800 are five and ten times too large through a power-of-ten slip in the carrier density.
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