Q5Electrical Materials

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Q.5. A P-type material having majority carrier density 10¹⁶/cm³ is further doped with 10¹⁴/cm³ Boron; then find its final electron and hole concentration. Assume intrinsic carrier concentration is 10¹⁰/cm³.

Answer 1

Boron is a trivalent acceptor, so the additional 10¹⁴/cm³ boron doping adds to the existing acceptor population of the P-type sample. The final hole (majority) concentration is:

p \approx 1 0^{16} + 1 0^{14} = 1.01 \times 1 0^{16} cm^{- 3}

The electron (minority) concentration follows from the mass-action law np = ni²:

n = \frac{n _{i}^{2}}{p} = \frac{( 1 0 ^{10} ) ^{2}}{1.01 \times 1 0 ^{16}} \approx 9.9 \times 1 0^{3} cm^{- 3}

So the final concentrations are p ≈ 1.01×10¹⁶ cm⁻³ and n ≈ 9.9×10³ cm⁻³ — the extra acceptor doping slightly raises the hole majority and correspondingly suppresses the electron minority population.

Q5Electrical Materials

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