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+// chapter 13
+// example 13.3
+// page 278
+
+Vcc=15 // V
+Re=2 // kilo ohm
+Rc=1 // kilo ohm
+Rl=1 // kilo ohm
+Vbe=0.7 // V
+
+// dc load line
+
+ // when Ic=0, Vce=Vcc i.e. Vce=15 and when Vce=0, Ic=Vcc/(Rc+Re) i.e. Ic=15/3
+ // so equation of load line becomes Ic=-(1/3)*Vce+15
+
+ clf()
+ x=linspace(0,15,5)
+ y=-(1/3)*x+5
+ plot2d(x,y,style=3,rect=[0,0,16,6])
+ xtitle("dc load line-green ac load line-blue","collector emitter voltage(volts)","collector current(mA)")
+
+ V2=5 // V
+ // since voltage across R2 is V2=5 V and V2=Vbe+Ie*Re we get
+ Ie=(V2-Vbe)/Re
+ Ic=Ie
+ Vce=Vcc-Ic*(Rc+Re)
+
+ printf("the operating point is %.3f V and %.3f mA \n",Vce,Ic)
+
+
+// ac load line
+
+ R_AC=Rc*Rl/(Rc+Rl) // ac load
+ V_ce=Vce+Ic*R_AC // maximum collector emitter voltage
+ I_c=Ic+Vce/R_AC // maximum collector current
+ // the equation of ac load line in terms of V_ce and I_c becomes
+ y=-(I_c/V_ce)*x+I_c
+ plot2d(x,y,style=2,rect=[0,0,10,20])
+
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