Question 2 of 9intermediate🔧 ApplyNumerical3 marks

Suppose a pure Si crystal has 5×10285 \times 10^{28} atoms m−3^{-3}. It is doped by 1 ppm concentration of pentavalent As. Calculate the number of electrons and holes. Given that ni=1.5×1016n_i = 1.5 \times 10^{16} m−3^{-3}.

Correct Answer

Number of electrons = 5×10225 \times 10^{22} m−3^{-3}

Number of holes = 4.5×1094.5 \times 10^{9} m−3^{-3}

Exercise: Example 14.2 | Q: Example 14.2 | (Page Page 10)
For More Understanding

Explanation

This is an n-type semiconductor problem. When Si is doped with pentavalent As, each As atom donates one free electron. The concentration of dopant atoms (1 ppm of 5×10285 \times 10^{28}) gives ND=5×1022N_D = 5 \times 10^{22} m−3^{-3}. Since thermally generated electrons (ni∼1016n_i \sim 10^{16}) are negligibly small compared to doping contribution, ne≈NDn_e \approx N_D. The hole concentration is then found using mass action law nenh=ni2n_e n_h = n_i^2.

Solution Steps

  1. Step 1: Calculate dopant concentration (NDN_D). 1 ppm means 1 part per million = 10−610^{-6}. So, ND=5×1028×10−6=5×1022N_D = 5 \times 10^{28} \times 10^{-6} = 5 \times 10^{22} m−3^{-3}.

  2. Step 2: Determine electron concentration (nen_e). Since thermally generated electrons are negligibly small compared to those from doping, ne≈ND=5×1022n_e \approx N_D = 5 \times 10^{22} m−3^{-3}.

  3. Step 3: Calculate hole concentration (nhn_h) using mass action law. nenh=ni2n_e n_h = n_i^2. Therefore, nh=ni2ne=(1.5×1016)25×1022=2.25×10325×1022=4.5×109n_h = \frac{n_i^2}{n_e} = \frac{(1.5 \times 10^{16})^2}{5 \times 10^{22}} = \frac{2.25 \times 10^{32}}{5 \times 10^{22}} = 4.5 \times 10^{9} m−3^{-3}.

PreviousNext