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GATE 2014 EC (EC2) – Question 46

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

A reverse-biased silicon diode has $N_A=9\times10^{16}$ cm⁻³, $N_D=10^{16}$ cm⁻³ and total depletion width 3 µm. Silicon permittivity is $1.04\times10^{-12}$ F/cm. Find p-side depletion width and maximum field, respectively.

  1. 2.7 µm and $2.3\times10^5$ V/cm
  2. 0.3 µm and $4.15\times10^5$ V/cm
  3. 0.3 µm and $0.42\times10^5$ V/cm
  4. 2.1 µm and $0.42\times10^5$ V/cm

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Correct answer: (B) 0.3 µm and $4.15\times10^5$ V/cm

Explanation

Charge neutrality gives $N_Ax_p=N_Dx_n$. Since $N_A=9N_D$ and $x_p+x_n=3$ µm, $x_p=0.3$ µm and $x_n=2.7$ µm. Then $E_{max}=qN_Dx_n/\epsilon_{si}=1.6\times10^{-19}\times10^{16}\times2.7\times10^{-4}/(1.04\times10^{-12})=4.15\times10^5$ V/cm. Therefore **B**.