Verified Solution • 2025–26 Session moderate

QPages 419-421: Q21: HC Verma: Concepts of Physics - Volume 2 Semiconductors and Semiconductor Devices Pages 419-421: Q21 Solution: The current_voltage characteristic of an ideal p-n junction…

The current_voltage characteristic of an ideal p-n junction diode is given by i=i0eeV/kT−1)i = i_0 e^{eV/kT} - 1) where, the drift current i0i_0 equals 10 �A. Take the temperature T to be 300 K. a) Find the voltage V0V_0 for which eeV/kT=100e^{eV/kT} = 100. One can neglect the term 1 for voltages greater than this value. b) Find an expression for the dynamic resistance of the diode as a function of V for V>V0V > V_0. c) Find the voltage for which the dynamic resistance is 0.2 Ω0.2 \ \Omega. Use Planck constant h=4.14×10−15 eV-sh = 4.14 \times 10^{-15} \text{ eV-s}, Boltzmann constant k=8.62×10−5 eV/Kk = 8.62 \times 10^{-5} \text{ eV/K}.)

Step-by-Step Verified Solution

**Answer:** (a) V_ = 0.12 V (b) R_d = (kT/e)/i (c) V = 0.25 V. **Step-by-Step Explanation:** Given: i_ = 10 _A, T = 300 K. (a) For e^(eV/kT) = 100: eV/kT = ln100 = 4.605. V_ = 4.605 _ kT/e = 4.605 _ 0.02585 = 0.119 V � 0.12 V. (b) Dynamic resistance: R_d = dV/di = (kT/e)/(i + i_) � (kT/e)/i for i _ i_. (c) R_d = 0.2 �: i = (kT/e)/0.2 = 0.02585/0.2 = 0.129 A. V = (kT/e)ln(i/i_) = 0.02585 _ ln(12900) = 0.02585 _ 9.46 = 0.245 V � 0.25 V .

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