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// Exa 4.16.21
clc;
clear;
close;
// Given data
V_T = 26;// in mV
V_T = V_T * 10^-3;// in V
Eta = 1;
// I = -90% for Io, so
IbyIo= 0.1;
// I = I_o * ((e^(v/(Eta * V_T)))-1)
V = log(IbyIo) * V_T;// in V
disp(V,"The reverse bias voltage in volts is");
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