Suppose a pure \(\mathit{Si}\) crystal has \(5\times 10^{28} \; \text{atoms m}^{-3}.\) It is doped by \(1\)ppm concentration of pentavalent (As). Calculate the number of electron and holes. Given that \(n_i=1.5\times 10^{16}\text{m}^{-3}\).
Answer
\(n_e=5\times 10^{22}\ \text{m}^{-3}\) and \(n_h=4.5 \times10^9\ \text{m}^{-3}\)
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Solution
Given: Number of atoms/meter in pure silicon atom = \(5\times 10^{28}.\)
And, it is doped by \(1\;\text{ppm}\) concentration of pentavalent.
We know that; \(1\ \text{ppm}=\dfrac{1}{10^6}\)
Number of pentavalent \(\text{As}\) atoms doped in given \(\text{Si}\) crystal is;
\(\dfrac{5\times 10^{28}}{10^6}=5\times 10^{22}\ \text{m}^{-3}\).
As one pentavalent atom donates one free electron to the crystal structure. The number of free electrons in given silicon crystal is, \(n_e=5\times 10^{22}\ \text{m}^{-3}\).
Number of holes is,
\(n_h=\dfrac{n_i^2}{n_e}\)
\(n_h=\dfrac{(1.5\times 10^{16})^2}{5\times 10^{22}}\)
\(n_h=4.5 \times10^9\ \text{m}^{-3}\).
And, it is doped by \(1\;\text{ppm}\) concentration of pentavalent.
We know that; \(1\ \text{ppm}=\dfrac{1}{10^6}\)
Number of pentavalent \(\text{As}\) atoms doped in given \(\text{Si}\) crystal is;
\(\dfrac{5\times 10^{28}}{10^6}=5\times 10^{22}\ \text{m}^{-3}\).
As one pentavalent atom donates one free electron to the crystal structure. The number of free electrons in given silicon crystal is, \(n_e=5\times 10^{22}\ \text{m}^{-3}\).
Number of holes is,
\(n_h=\dfrac{n_i^2}{n_e}\)
\(n_h=\dfrac{(1.5\times 10^{16})^2}{5\times 10^{22}}\)
\(n_h=4.5 \times10^9\ \text{m}^{-3}\).
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