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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+// Exa 3.10.21
+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");