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+clear;
+clc;
+//Example 5.13
+Kn1=0.2;
+Kn2=0.1;
+Kn3=0.1;
+Kn4=0.1;
+Vtn1=1;
+Vtn2=1;
+Vtn3=1;
+Vtn4=1;
+V2=-5;
+Vgs3=(sqrt(Kn4/Kn3)*(-V2-Vtn4)+Vtn3)/(1+sqrt(Kn4/Kn3));
+printf('\nVgs3=%.2f V\n',Vgs3)
+Iq=Kn3*(Vgs3-Vtn3)^2;
+printf('\nbias current=%.3f mA\n',Iq)
+Vgs1=sqrt(Iq/Kn1)+Vtn1;
+printf('\ngate to source voltage on M1=%.2f V\n',Vgs1)
+Vds2=-V2-Vgs1;
+printf('\ndrain to source voltage on M2=%.2f V\n',Vds2)
+Vgs2=Vgs3;
+Vdssat=Vgs2-Vtn2
+//since Vds2>Vdssat M2 is biased in saturation region