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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+//Calculate Q point in voltage divider
+clear;
+clc;
+//soltion
+//given
+B=100; //dc beta
+Rc=2*10^3;//ohm //resistor connected to collector
+R1=10*10^3;//ohm //voltage divider resistor 1
+R2=1*10^3;//ohm //voltage divider resistor 2
+Re=200;//ohm //resistor connected to emitter
+Vcc=10;//V //Voltage supply across the collector resistor
+Vbe=0.3;//V //base to emitter voltage
+
+I=Vcc/(R1+R2); //current through voltage divider
+Vb=I*R2; //voltage at base
+Ve=Vb-Vbe;
+Ie=Ve/Re;
+Ic=Ie //approaximating Ib is nearly equal to 0
+Vc=Vcc-Ic*Rc;
+Vce=ceil(Vc)-Ve;
+printf("The Q point is (%.1f V, %.0f mA)",Vce,Ic*1000);
+
+Ibc=I/20; //critical value of base current
+Ib=Ic/B; //actual base current
+
+//Since Ib < Ibc, hence assumption is alright