Question

In an n-type semiconductor, electron-hole combination is a continuous process at room temperature. Yet the electron concentration is always greater than the hole concentration in it. Explain.

Hint:

Doping determines the majority charge carrier. In an \( n \)-type semiconductor, added donor atoms supply extra electrons, ensuring electrons outnumber thermally generated holes.

Answer 1

In an \( n \)-type semiconductor, a pentavalent impurity (such as phosphorus or arsenic) is added to a pure (intrinsic) semiconductor like silicon. Each dopant atom donates one extra electron, which increases the number of free electrons in the conduction band. At room temperature, thermal energy continuously generates electron-hole pairs, leading to recombination of electrons and holes. However, the concentration of electrons remains much higher than that of holes because: \begin{itemize} \item The majority carriers in an \( n \)-type semiconductor are electrons (due to doping). \item The minority carriers (holes) are generated thermally and are much fewer in number. \item Although recombination happens, the large number of donor electrons maintains a higher equilibrium concentration of electrons. \end{itemize} Hence, even with continuous recombination at room temperature, the electron concentration remains significantly greater than the hole concentration in an \( n \)-type semiconductor.