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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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+// Exa 2.3
+clc;
+clear;
+close;
+// Given data
+Vt= 0.7;// in V
+I_D= 100;// in µA
+I_D=I_D*10^-6;// in A
+// When
+V_GS= 1.2;// in V
+V_DS= V_GS;// in V
+// At this condition, device is in saturation region, so I_D= unCox*W/(2*L)*(V_GS-VT)^2
+unCoxWby2L= I_D/(V_GS-Vt)^2;
+// For
+V_DS= 3;// in V
+V_GS= 1.5;// in V
+// In this condition, device is in saturation region, so
+I_D= unCoxWby2L*(V_GS-Vt)^2;// in A
+disp(I_D*10^6,"For V_DS= 3V and V_GS= 1.5 V, The value of I_D in µA is : ")
+// For
+V_GS= 3.2;// in V
+r_DS= 1/(2*unCoxWby2L*(V_GS-Vt));// in Ω
+disp(r_DS,"For V_GS = 3.2 V, Drain to source resistance in Ω is : ")