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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+clear;
+clc;
+//Example 3.10
+Rb=0.24;
+Vcc=12;
+Vbe=0.7;
+Vce=0.1;
+b=75;
+Rc=5;//Ohm
+//for Vt=0 ,transistor is cut off,Ib=Ic=0,Vo=Vcc=12 V,power dissipation is zero
+Vt=12;//(V)
+Ib=(Vt-Vbe)/Rb;
+printf('\nbase current=%0.3f mA\n',Ib)
+Ic=(Vcc-Vce)/Rc;
+printf('\ncollector current=%0.2f A\n',Ic)
+Ib=0.0471;//A
+x=Ic/Ib
+//since Ic/Ib<b transistor is in saturation
+//Vo==Vcc;
+Vo=0.1;
+printf('\noutput voltage=%0.2f V\n',Vo)
+P=Ic*Vce+Ib*Vbe;
+printf('\npower dissipation=%0.3f W\n',P)