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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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+//chapter12
+//example12.3
+//page240
+
+Vbb=2 // V
+Vcc=9 // V
+Rc=2 // kilo ohm
+Rb=100 // kilo ohm
+gain_beta=50
+
+// by Kirchoff voltage law on base side, we get Ib*Rb+Vbe=Vbb so
+Ib=Vbb/Rb // Vbe is negligible
+Ic=gain_beta*Ib
+
+// by Kirchoff voltage law on collector side, we get Ic*Rc+Vce=Vcc so
+Vce=Vcc-Ic*Rc
+
+// now for Rb=50 kilo ohm
+Rb2=50 // kilo ohm
+
+// since Rb is halved, Ib is doubled so
+Ib2=2*Ib
+Ic2=Ib2*gain_beta
+Vce2=Vcc-Ic2*Rc
+
+printf("for Rb = 100 kilo ohm, collector current = %.3f mA \nand collector emitter voltage = %.3f V \n \n",Ic,Vce)
+printf("for Rb = 50 kilo ohm, collector current = %.3f mA \nand collector emitter voltage = %.3f V \n",Ic2,Vce2)