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Silicon is doped with boron to a concentration of 4 × 1017 atoms/cm3. Assuming 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?
  • a)
    Goes down by 0.13eV
  • b)
    Goes up by 0.13eV
  • c)
    Goes down by 0.427eV
  • d)
    Goes up by 0.427eV
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
Silicon is doped with boron to a concentration of 4 × 1017 atoms...
E2 – E1 = kTln (NA/ni)
NA = 4 × 1017
ni = 1.5 × 1010
E2 – E1 = 25 × 10-3 e ln 4 × 1017/1.5×1010 = 0.427eV
Hence Fermi level goes down by 0.427 eV as silicon is doped with boron.
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Most Upvoted Answer
Silicon is doped with boron to a concentration of 4 × 1017 atoms...
× 10^16 atoms/cm^3. This type of doping is known as p-type doping. When boron is added to silicon, it creates holes in the crystal lattice, which can be thought of as positively charged vacancies. These holes can attract and combine with free electrons in the silicon, effectively reducing the number of available electrons and creating a net positive charge in the material. This makes p-type silicon a good candidate for building p-n junctions, which are essential components in many electronic devices, including solar cells.
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Silicon is doped with boron to a concentration of 4 × 1017 atoms/cm3. Assuming the intrinsic carrier concentration of silicon to be 1.5×1010/cm3 and the value of kT/q to be 25 mV at 300 KCompared to undoped silicon, the Fermi level of doped silicon?a)Goes down by 0.13eVb)Goes up by 0.13eVc)Goes down by 0.427eVd)Goes up by 0.427eVCorrect answer is option 'C'. Can you explain this answer?
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