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+clear;
+clc;
+//Example 3.9
+b=100;
+Vbe=0.7;
+V1=-5;
+V2=12;
+Rb=10;
+Re=5;
+Rc=5;
+Rl=5;
+//Q point values:: using KVL eq around B-E loop
+Ib=-(V1+Vbe)/(Rb+(1+b)*Re);
+printf('\nbase current=%f mA\n',Ib)
+Ic=b*Ib;
+printf('\ncollector current=%f mA\n',Ic)
+Ie=(1+b)*Ib;
+printf('\nemitter current=%f mA\n',Ie)
+//at collector node we can write Ic=(V2-Vo)/Rc-Vo/Rl
+Vo=(V2/Rc-Ic)*Rc*Rl/(Rc+Rl);
+printf('\noutput voltage=%0.3f V\n',Vo)
+Vce=Vo-Ie*Re-V1;
+printf('\ncollector emitter voltage=%f V\n',Vce)
+//load line::
+Rth=Rl*Rc/(Rl+Rc);
+printf('\nThevenin rquivalent resistance=%f KOhm\n',Rth)
+Vth=(Rl/(Rl+Rc))*V2;
+printf('\nThevenin equivalent voltage=%f V\n',Vth)
+//fig.3.36(c) KVL law
+//Vce=6-V1-Ic*Rth-Ie*Re;
+Vce=[0,2,4.7,3.5,4,6,8,10]
+Ic=(11-Vce)/7.5;
+xset('window',1)
+plot2d(Vce,Ic,style=3,rect=[0,0,12,2])
+title("load line")
+xlabel("Vce")
+ylabel("Ic")