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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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