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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

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

Correct answer: goes down by 0.427 eV

Solution

Boron is an acceptor, so silicon becomes p-type and the Fermi level moves toward the valence band (down). The shift is kT*ln(Na/ni) = 0.025*ln(4e17/1.5e10) = 0.025*17.1 = 0.427 eV. So it goes down by 0.427 eV (option C), not down by 0.13 eV.

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