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//chapter-2,Example2_11,pg 486
R1=10*10^3
R4=10*10^3
Idss=1*10^-3//drain saturation current
Vp=2.2//peak voltage
Vo=10//output voltage
V2=2//input-1
V1=-2//input-2
R5=((R1*R4)/Vo)*((-2*Idss/(Vp^2)))*V1*V2
printf("R5=%.2f ohm",R5)
//R5 should satisfy the condition R5=((1+R1*(-2*Idss*Vp)/R2)*R3*R6) and with Vp negative it is obiviously possible
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