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authorpriyanka2015-06-24 15:03:17 +0530
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+clear;
+clc;
+//Example 16.15
+Vdd=3;
+Kn=60;
+Vtnd=0.5;
+//W/L=x
+xd=2;
+Vtnl=-1;
+xl=0.5;
+R=2;//(MOhm)
+Vgsl=0;
+//solution with Depletion load
+iD=Kn*xl*(Vgsl-Vtnl)^2/2;
+printf('\nfrain currents in M1 and M3 =%.fmicroA\n',iD)
+P=iD*Vdd;
+printf('\npower dissipation in the circuit=%.f microW\n',P)
+//iD=Kn/2*x*(2*Vgsd-Vtnd)Vdsd-Vdsd^2
+Q=poly(0,'Q');
+p=poly([0.25 -5 1],'Q','c');
+roots(p)
+disp("Q=50.5 (mV)")
+//solution with Resistor load
+//(Vdd-Q)/R=Kn/2*xd*(2*Vgsd-Vtnd)Q-Q^2
+Q=poly(0,'Q');
+p=poly([3 -599 120],'Q','c');
+roots(p)
+Q=0.005;
+disp("Q=5 (mV)")
+iD=(Vdd-Q)/R;
+printf('\ndrain current =%.2f microA\n',iD)
+P=iD*Vdd;
+printf('\npower dissipation in the circuit =%.2f microW\n',P)