Question

Assume T= 300K 2. (4 points) A silicon junction diode has v = 0.7V at I= 1 mA a) Find the voltage atI 5 mA b) Find the current į at voltage v 0.8V.

Answer 1

Given T = 300 k V = 0.7 V and I = 1 mA For a silicon junction diode, I_D = I_s [e^q.v/etakT - 1] For silicon eta = 2 a) I m A = I_s [e^1.6 times 10^-11 times 0.7/2 times 1.31 times 10^-20 times 300 - 1] implies I m A = I_s [e^13.526 - 1] implies I_s = I m A/749055.76 implies I_s = 1.335 n A I_s = saturation current But given, I_D changes to 5 mA now implies 5 mA = 1.335 n A = [e^1.6 times 10^-19 times V/2 times 1.38 times 10^-23 times 300 - 1] implies 374519.352 = e^19.32V implies Applying 'log_e' on both sides implies ln (3745319.352) = 19.32 V implies 15.136 = 19.32 V Therefore V = 0.783 Volt Therefore Voltage at I = 5 mA is 0.783 volts \r\n b) Now the voltage changes to V = 0.8 V implies I_0 = I_s [e^qV/etakT - 1] implies I_D = 1.335 n A times {5172863.723} implies I_D = 6.9057 m A Therefore Current 'i' at voltage V = 0.8 volt is i = 6.9 mA