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GATE 2014 EC (EC2) – Question 18

Electronic Devices · Energy Bands and Carrier Concentration · 1 mark · Multiple choice

A silicon bar has donor doping $N_D=2.25\times10^{15}$ atoms/cm³. At 300 K, $n_i=1.5\times10^{10}$ cm⁻³. Assuming complete ionization, find equilibrium electron and hole concentrations.

  1. $n_0=1.5\times10^{16},p_0=1.5\times10^5$ cm⁻³
  2. $n_0=1.5\times10^{10},p_0=1.5\times10^{15}$ cm⁻³
  3. $n_0=2.25\times10^{15},p_0=1.5\times10^{10}$ cm⁻³
  4. $n_0=2.25\times10^{15},p_0=10^5$ cm⁻³

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Correct answer: (D) $n_0=2.25\times10^{15},p_0=10^5$ cm⁻³

Explanation

Complete donor ionisation gives n0≈ND=2.25×10¹⁵cm⁻³ because ND is much larger than ni. By mass action p0=ni²/n0=(1.5×10¹⁰)²/(2.25×10¹⁵)=10⁵cm⁻³. Thus **n0=2.25×10¹⁵,p0=10⁵cm⁻³,D**.