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GATE 2014 EC (EC3) – Question 44

Electronic Devices · P-N Junction and Semiconductor Devices · 2 marks · Multiple choice

An abrupt silicon PN diode has donor concentration $10^{16}$ cm⁻³ and acceptor concentration $5\times10^{18}$ cm⁻³. At 300 K, $n_i=1.5\times10^{10}$ cm⁻³, $kT/q=26$ mV and $\epsilon_{si}=1.04\times10^{-12}$ F/cm. Find built-in potential and equilibrium depletion width.

  1. 0.7 V and $10^{-4}$ cm
  2. 0.86 V and $10^{-4}$ cm
  3. 0.7 V and $3.3\times10^{-5}$ cm
  4. 0.86 V and $3.3\times10^{-5}$ cm

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Correct answer: (D) 0.86 V and $3.3\times10^{-5}$ cm

Explanation

Use Vbi=VTln(NAND/ni²)=0.026ln[(5×10¹⁸×10¹⁶)/(1.5×10¹⁰)²]≈0.8589V. Then W=√[(2εsiVbi/q)(1/NA+1/ND)]≈3.34×10⁻⁵cm. These round to **0.86V and3.3×10⁻⁵cm,D**. Both sides of the junction contribute to the width.