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+clear;
+clc;
+//Example 17.9
+bf=25;
+b=bf;
+br=0.1;
+Vcc=5;
+R1=4;
+Vbc=0.7;
+Vy=0.1;
+Vx=0.1;
+R2=1.6;
+Vbe=0.8;
+Rc=4;
+Vce=0.1;
+vB2=Vx+Vce;
+printf('\nvB2=%.2f V\n',vB2)
+vB1=Vx+Vbe;
+printf('\nbase voltage=%.2f V\n',vB1)
+i1=(Vcc-vB1)/R1;
+printf('\ncurrent i1=%.2f mA\n',i1)
+vB1=Vbe+Vbe+Vbc;
+printf('\nvB1=%.2f V\n',vB1)
+vC2=Vbe+Vce;
+printf('\ncollector voltage=%.2f V\n',vC2)
+i1=(Vcc-vB1)/R1;
+printf('\ncurrent i1 =%.2fmA\n',i1)
+iB2=(1+2*br)*i1;
+printf('\niB2=%.2f mA\n',iB2)
+i2=(Vcc-vC2)/R2;
+printf('\ni2 =%.2fmA\n',i2)
+iE2=i2+iB2;
+printf('\niE2=%.2f mA\n',iE2)
+Rb=1;
+i4=Vbe/Rb;
+printf('\ncurrent in the pull down resistor=%.2f mA\n',i4)
+iBo=iE2-i4;
+printf('\nbase drive to the output transistor=%.2f mA\n',iBo)
+i1=(Vcc-Vce)/Rc;
+printf('\ni1=%.2f mA\n',i1)