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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+//chapter 11
+//example 11.7
+//page 328
+clear all;
+clc ;
+//given
+Rl=40;//load resistance in kohm
+Vcc=14;//supply voltage V
+Ve1=5;Vce1=3;Vce2=3;
+Vc1=Ve1+Vce1;
+Vb2=Vc1;
+Ve2=Vb2-0.7;
+VR5=Vcc-Ve2-Vce2;
+R5=Rl/10;
+printf("\nR5=%d kohm,use standard value 3.9 kohm ",R5);R5=3.9;
+Ic2=1000*VR5/R5;
+R6=1000*Ve2/Ic2;
+printf("\nR6=%.1f kohm,use standard value 8.2 kohm ",R6);R6=8.2;
+Ic2=1000*Ve2/R6;
+Ic1=1;//mA
+VR3=Vcc-Vc1;
+R3=VR3/Ic1;
+printf("\nR3=%d kohm,use standard value 5.6 kohm ",R3);R3=5.6;
+Ic1=VR3/R3;
+R4=Ve1/Ic1;
+printf("\nR4 is approximately %.1f kohm which is standard value",R4);
+Vb1=Ve1+0.7;
+I2=round(1000*Ic1/10);
+R2=1000*Vb1/I2;
+printf("\nR2=%.1f kohm,use standard value 47 kohm ",R2);R2=47;
+I21=1000*Vb1/R2;
+R1=1000*(Vcc-Vb1)/I21;
+printf("\nR1=%d kohm which is standard value ",R1);
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