Electron & hole concentrations after doping silicon.

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SUMMARY

The discussion focuses on calculating electron and hole concentrations in a silicon wafer doped with 2 x 1016 cm-3 Boron and 1016 cm-3 Phosphorous. The resulting hole concentration is determined to be 1 x 1016 cm-3, indicating a p-type semiconductor. The electron concentration is calculated using the intrinsic carrier concentration formula, yielding a value of 2.25 x 104 cm-3. These calculations are essential for understanding the electrical properties of doped silicon at room temperature.

PREREQUISITES
  • Understanding of semiconductor physics
  • Familiarity with doping concepts in silicon
  • Knowledge of intrinsic carrier concentration (ni)
  • Basic skills in algebra for calculations
NEXT STEPS
  • Study the principles of p-type and n-type semiconductors
  • Learn about the Fermi level in semiconductors
  • Explore resistivity calculations in doped silicon
  • Investigate temperature effects on carrier concentrations
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Students and professionals in electrical engineering, materials science, and semiconductor physics who are involved in the study and application of doped silicon materials.

Alex-012
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1. A Silicon wafer is doped with 2*10^16 cm^-3 Boron and 10^16 cm^-3 Phosphorous atoms. Calculate the electron and hole concentrations, the Fermi-level and the resistivity at room temperature

I have no idea how to work out the first part of this question as my lecture notes are lacking in material which wasn't discussed in the lecture and because I am at home for christmas I have no access to my recommended textbooks. If anyone could help me out I would be extremely grateful!
 
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Ok, I think I was overthinking this. It seems easier than I originally thought.
I think the hole concentration is calculated using

Na - Nd = (2-1)x1016

since there are more acceptors than donors so therefore the silicon is p-type with a hole concentration of

1*10^{16} cm^{-3}

Electron concentration is then calculated using:
np = n_{i}^2

which gives

n = (1.5\times10^{10})^2/10^{16} = 2.25\times 10^{4} cm^{-3}

Or am I way off the mark?
 

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