GATE 2014 EC (EC4) – Question 44
A silicon sample has $N_D=10^{15}$ cm⁻³ donors and $n_i=1.5\times10^{10}$ cm⁻³. It is additionally doped with $N_A=10^{18}$ cm⁻³ acceptors. Find the approximate electron density (cm⁻³) at 300 K.
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Correct answer: 224.9 to 225.1
Explanation
The acceptor density10¹⁸cm⁻³dominates the original10¹⁵donors, so the official approximation takes hole density p≈10¹⁸cm⁻³. Mass action then gives n≈ni²/p=(1.5×10¹⁰)²/10¹⁸=**225cm⁻³**. Retaining donor compensation gives p≈9.99×10¹⁷and n≈225.225cm⁻³, slightly outside the narrow official224.9–225.1range; the key neglects the smaller donor concentration.