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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+//Calculate max and min value of emitter current
+clear;
+clc;
+//soltion
+//given
+
+//At B=50
+
+B=50; //dc beta
+Rc=75;//ohm //resistor connected to collector
+Re=100;//ohm //resistor connected to emitter
+Rb=10*10^3;//ohm //resistor connected to base
+Vcc=6;//V //Voltage supply across the collector resistor
+Vbe=0.3;//V //for germanium
+Ib=(Vcc-Vbe)/(Rb+(1+B)*Re); //Base current
+Ie=(1+B)*Ib;
+Vce=Vcc-(Rc+Re)*Ie
+printf("Minimum emitter current %.2f mA\n",Ie*10^3);
+printf("The collector to emitter volatge is %.2f V\n",Vce);
+
+//At B=300
+
+B1=300; //dc beta
+Ib1=(Vcc-Vbe)/(Rb+(1+B1)*Re); //Base current
+Ie1=(1+B1)*Ib1;
+Vce1=Vcc-(Rc+Re)*Ie1
+//Here Vce1= -1.4874 V but can never have negative voltage because Ie1 is wrong as it can't be more than saturation value therefore
+Ie1=Vcc/(Rc+Re);
+
+//And Vce=0 V
+
+Vce1=0;//V
+printf("Maximum emitter current %.2f mA\n",Ie1*10^3);
+printf("The collector to emitter volatge(saturation) is %.0f V\n",Vce1);
+