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A bar of silicon is doped with boron concentration of ...

 

A bar of silicon is doped with boron concentration of $10^{16}\,\text{cm}^{-3}$ and assumed to be fully ionized. It is exposed to light such that electron-hole pairs are generated throughout the volume of the bar at the rate of $10^{20}\,\text{cm}^{-3}\text{s}^{-1}$. If the recombination lifetime is $100\,\mu s$, intrinsic carrier concentration of silicon is $10^{10}\,\text{cm}^{-3}$, and assuming 100% ionization of boron, then the approximate product of steady-state electron and hole concentrations due to this light exposure is

  • (A) $10^{20}\,\text{cm}^{-6}$
  • (B) $2\times10^{20}\,\text{cm}^{-6}$
  • (C) $10^{32}\,\text{cm}^{-6}$
  • (D) $2\times10^{32}\,\text{cm}^{-6}$

Step-by-Step Solution

Step 1 : Identify the Type of Semiconductor

The silicon is doped with boron. Boron is an acceptor impurity, therefore the semiconductor is p-type.

Since boron is fully ionized,

\[ p_0=N_A=10^{16}\,\text{cm}^{-3} \]

This is the equilibrium hole concentration.

Step 2 : Find the Minority Electron Concentration

Use the mass-action law:

\[ n_0p_0=n_i^2 \]

Substitute the given values:

\[ n_0= \frac{(10^{10})^2}{10^{16}} = \frac{10^{20}}{10^{16}} = 10^4\;\text{cm}^{-3} \]

Step 3 : Calculate the Excess Carrier Concentration

Under steady-state illumination,

\[ \Delta n=\Delta p=G\tau \]

Given

\[ G=10^{20}\,\text{cm}^{-3}\text{s}^{-1} \] \[ \tau=100\mu s=10^{-4}s \]

Therefore,

\[ \Delta n = 10^{20}\times10^{-4} = 10^{16} \]

Hence,

\[ \boxed{\Delta n=\Delta p=10^{16}\,\text{cm}^{-3}} \]

Step 4 : Calculate the Total Electron Concentration

The total electron concentration is

\[ n=n_0+\Delta n \]
\[ n=10^4+10^{16} \]

Since

\[ 10^4\ll10^{16}, \]

we can neglect $10^4$.

\[ \boxed{n\approx10^{16}\,\text{cm}^{-3}} \]

Step 5 : Calculate the Total Hole Concentration

The total hole concentration is

\[ p=p_0+\Delta p \]
\[ p = 10^{16}+10^{16} = 2\times10^{16} \]
\[ \boxed{p=2\times10^{16}\,\text{cm}^{-3}} \]

Step 6 : Calculate the Product \(np\)

\[ np = (10^{16}) (2\times10^{16}) \]
\[ np = 2\times10^{32}\,\text{cm}^{-6} \]
Final Answer
\[ \boxed{ np=2\times10^{32}\,\text{cm}^{-6} } \]
Therefore, the correct option is (D).

Fundamental Concepts Used

Concept Formula
Majority carriers (p-type) $p_0=N_A$
Majority carriers (n-type) $n_0=N_D$
Mass-action law $n_0p_0=n_i^2$
Excess carriers under illumination $\Delta n=\Delta p=G\tau$
Total electrons $n=n_0+\Delta n$
Total holes $p=p_0+\Delta p$
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