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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
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