Concepts Of Physics MCQ Edition [Volume 2]PhysicsSemiconductors and Semiconductor Devices
An ideal p - n junction diode has a current-voltage characteristic described by i = i₀ ! (e^ eV/kT - 1 ) , in which the drift current i₀ is 10 A . The temperature T is taken as 300 K . Evaluate the voltage at which the dynamic resistance becomes 0.2 .
Options
- A0.25 V
- B1.25 V
- C0.50 V
- D0.12 V
Correct answer
A. 0.25 V
Step-by-step solution
The dynamic resistance r_d of a diode is given by the reciprocal of the derivative of current with respect to voltage: r_d = dV di Given the current-voltage characteristic: i = i₀ (e^ eV/kT - 1 ) Differentiating with respect to V : di dV = i₀ ( e kT ) e^ eV/kT Therefore, the dynamic resistance is: r_d = kT e i₀ e^ eV/kT At T = 300 K , the thermal voltage kT e is approximately: kT e = 1.38 10⁻²³ 300 1.6 10⁻¹⁹ 0.026 V Substituting the given values r_d = 0.2 and i₀ = 10 A = 10⁻⁵ A : 0.2 = 0.026 10⁻⁵ e^ V/0.026 e^ V/0.