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+clear;
+clc;
+//Example 6.15
+Vtn1=1.2;
+Vtn2=1.2;
+Kn1=0.8;
+Kn2=0.8;
+//x=R1+R2+R3=300
+x=300;
+Rs=10;
+Idq=0.4;
+Vdsq1=2.5;
+Vdsq2=2.5;
+Vs1=Idq*Rs-5;
+printf('\ndc voltage at source of M1=%.2f V\n',Vs1)
+Vgs=sqrt(Idq/Kn1)+Vtn1;
+printf('\ngate to source voltage=%.2f V\n',Vgs)
+R3=(Vgs+Vs1)*x/5;
+printf('\nR3=%.2f KOhm\n',R3)
+Vs2=Vdsq2+Vs1;
+printf('\nvoltage at source of M2=%.2f V\n',Vs2)
+//y=R2+R3
+y=(Vgs+Vs2)*x/5;
+printf('\nR2+R3=%.2f KOhm\n',y)
+R2=150;
+R1=x-y;
+printf('\nR1=%.2f KOhm\n',R1)
+R3=y-R2;
+printf('\nR3=%.2f KOhm\n',R3)
+Vd2=Vdsq2+Vs2;
+printf('\nvoltage at drain of M2 =%.2fV\n',Vd2)
+Rd=(5-Vd2)/Idq;
+printf('\ndrain resistance=%.2f KOhm\n',Rd)