Chegg Solutions for Microelectronic Circuits (Adel S Sedra, Kenneth C Smith) (Z-Library)_parte_420

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Step 1 of 2 3.014E Consider the following expression for the depletion region or junction capacitance, c, at zero reverse bias: Here, the area of the pn junction, the electrical permittivity of silicon, F/cm the charge of electron, q=1.6x10⁻¹⁹ C the acceptor doping concentration or doping concentration of the p side, N, = , the donor doping concentration, = , and the barrier voltage, Substitute these values in the equation for = Simplify further. =3.18 pF Therefore, the junction capacitance at zero reverse bias, is, 3.18 pF Step 2 of 2 Consider the following relation between the depletion capacitance, c, and junction capacitance at zero reverse bias, Here, the value of gradient coefficient, m is, Consider that the reverse bias voltage, = Substitute 3.18 pF for 2V for , 0.814 V for and for m in the equation for 3.18x10⁻¹² = = = 1.8593 =1.71x10⁻¹² Simplify further. Therefore, the junction capacitance, c, is, 1.71 pF