Question

Consider a long rectangular bar of direct bandgap $p$-type semiconductor. The equilibrium hole density is $10^{17}\,\text{cm}^{-3}$ and the intrinsic carrier concentration is $10^{10}\,\text{cm}^{-3}$. Electron and hole diffusion lengths are $2\,\mu$m and $1\,\mu$m, respectively. The left side of the bar ($x=0$) is uniformly illuminated with a laser having photon energy greater than the bandgap of the semiconductor. Excess electron-hole pairs are generated ONLY at $x=0$ because of the laser. The steady state electron density at $x=0$ is $10^{14}\,\text{cm}^{-3}$ due to laser illumination. Under these conditions and ignoring electric field, the closest approximation (among the given options) of the steady state electron density at $x=2\,\mu\text{m}$ is ______________.

• A. $0.37 \times 10^{14}\,\text{cm}^{-3}$
• B. $0.63 \times 10^{13}\,\text{cm}^{-3}$
• C. $3.7 \times 10^{14}\,\text{cm}^{-3}$
• D. $10^{3}\,\text{cm}^{-3}$

Hint:

Injected minority carriers decay exponentially with distance from the point of generation, with the decay length equal to the diffusion length.

Answer 1

The Correct Option is A

Since the semiconductor is $p$-type and excess electrons are injected only at $x=0$, the steady-state excess electron concentration decays exponentially with distance as: \[ \Delta n(x) = \Delta n(0)\, e^{-x/L_n}, \] where $L_n = 2\,\mu\text{m}$ is the electron diffusion length.
Step 1: Substitute values.
\[ \Delta n(0) = 10^{14}\,\text{cm}^{-3}, \qquad x=2\,\mu\text{m}, \qquad L_n=2\,\mu\text{m}. \] \[ \Delta n(2) = 10^{14} e^{-2/2} = 10^{14}e^{-1}. \] Step 2: Evaluate the exponential.
\[ e^{-1} \approx 0.37. \] Thus, \[ \Delta n(2) \approx 0.37 \times 10^{14}\,\text{cm}^{-3}. \] Step 3: Add equilibrium electron concentration if needed.
The equilibrium electron density in $p$-type semiconductor is: \[ n_0 = \frac{n_i^2}{p_0} = \frac{10^{20}}{10^{17}} = 10^{3}\,\text{cm}^{-3}, \] which is negligible compared to $10^{14}$. Hence, the steady state electron density $\approx \Delta n(2)$.
Therefore the closest value is: \[ 0.37 \times 10^{14}\,\text{cm}^{-3}. \] Final Answer: $0.37 \times 10^{14}\,\text{cm}^{-3}$