Silicon is doped with boron to a concentration of 4 × 1017 atoms/cm3. Assume the intrinsic carrier concentration of silicon to be 1.5 × 1010/cm3 and the value of kT/q to be 25 mV at 300 K. Compared to undoped silicon, the Fermi level of doped silicon

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UPSC ESE (Prelims) Electronics and Telecommunication Engineering 19 Feb 2023 Official Paper
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  1. goes down by 0.13 eV
  2. goes up by 0.13 eV
  3. goes down by 0.427 eV
  4. goes up by 0.427 eV

Answer (Detailed Solution Below)

Option 3 : goes down by 0.427 eV
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The correct option is 3

Concept:

\({\phi _{fp}} = {E_i} - {E_f}\)

\( = \frac{{KT}}{q}\ln \left( {\frac{{{N_A}}}{{{n_i}}}} \right)\)

\( = 25 \times {10^{ - 3}} \times \ln \left( {\frac{{4 \times {{10}^{17}}}}{{1.5 \times {{10}^{10}}}}} \right)\)

= 0.427 eV

As the impurity is acceptor type the Fermi level goes down by 0.427 eV than the intrinsic level.

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