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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+//chapter 18
+//example 18.9
+//page 782
+printf("\n")
+printf("given")
+I2=1*10^-3;Vr2=7.15;Vref=Vr2;Vo=10;Pdmax=1000*10^-3;
+R2=Vref/I2
+R2=6.8*10^3;//use standard value and recalculate the I2
+I2=Vref/R2
+R1=(Vo-Vref)/I2
+Vs=Vo+5//for satisfactory operation of series pass transistor
+Iint=25*10^-3;//internal circuit current
+Pi=Vs*Iint
+disp("maximum power dissipated in series pass transistor")
+Pd=Pdmax-Pi
+disp("maximum load current is ")
+Il=Pd/(Vs-Vo)