Question

Due to illumination by light, the electron and hole concentrations in a heavily doped N-type semiconductor increase by \(\Delta n\) and \(\Delta p\), respectively. If \(n_i\) is the intrinsic concentration, then

• A. \( \Delta n<\Delta p \)
• B. \( \Delta n>\Delta p \)
• C. \( \Delta n = \Delta p \)
• D. \( \Delta n \times \Delta p = n_i^2 \)

Hint:

The generation of carriers by light (photogeneration) always creates electron-hole pairs. This means the excess electron concentration is always equal to the excess hole concentration (\(\Delta n = \Delta p\)), regardless of the doping type or level.

Answer 1

The Correct Option is C

Step 1: Understand the physical process. Illumination of a semiconductor with light (photons) having energy greater than the band gap creates electron-hole pairs. For each photon absorbed, one electron is excited from the valence band to the conduction band, leaving one hole behind in the valence band.
Step 2: Relate the process to concentration changes. This process of photogeneration creates excess electrons and excess holes in equal numbers. The rate of generation is the same for both carriers. Therefore, the increase in the electron concentration (\(\Delta n\)) must be equal to the increase in the hole concentration (\(\Delta p\)). \[ \Delta n = \Delta p \]
Step 3: Consider the effect of doping. The problem states the semiconductor is heavily doped N-type. This means that in thermal equilibrium (before illumination), the electron concentration \(n_0\) is much larger than the hole concentration \(p_0\) (\(n_0 \gg n_i \gg p_0\)). However, the increase due to light is independent of the initial concentrations. The generation of one electron is always accompanied by the generation of one hole. The total concentrations under illumination become \(n = n_0 + \Delta n\) and \(p = p_0 + \Delta p\). While \(n\) will still be much larger than \(p\), the increments \(\Delta n\) and \(\Delta p\) are equal.
Step 4: Evaluate the other options. Option (D), \( \Delta n \times \Delta p = n_i^2 \), is incorrect. The mass-action law \(np=n_i^2\) holds for thermal equilibrium only. Under illumination (a non-equilibrium condition), \(np>n_i^2\).