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authorpriyanka2015-06-24 15:03:17 +0530
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+//Example 9.15.refer fig.9.59.
+clc
+beta=100
+VCC=10
+R2=4.7*10^3
+R1=27*10^3
+RE=680
+RC=3.3*10^3
+RS=600
+RL=15*10^3
+disp("Referring to fig.9.59(a),")
+format(5)
+VB=(10*4.7*10^3)/((27*10^3)+(4.7*10^3))
+disp(VB," VB(V) = (R2 / R1+R2)*VCC =") // answer in textbook is wrong
+VE=1.39-0.7
+disp(VE," VE(V) = 1.39 - 0.7 =")
+format(4)
+IE=VE/RE
+x1=IE*10^3
+disp(x1," IE(mA) = VE / RE =")
+re=0.026/IE
+disp(re," re(ohm) = 0.026/IE =")
+x=beta*(re+RE)
+format(5)
+Zi=(R1*R2*x)/((R2*x)+(R1*x)+(R1+R2)) // answer in textbook is wrong
+x2=Zi*10^-3
+disp(x2," Zi(k-ohm) = R1 || R2 || beta*(re+RE) =")
+format(4)
+y=(-RC/(RE+re))*(Zi/(RS+Zi))*(RL/(RC+RL))
+disp(y,"The overall voltage gain is VL/VS = (-RC/RE+re)*(Zi/RS+Zi)*(RL/RC+RL) =")
+disp("Referring to fig.9.59(b),")
+format(5)
+u=beta*re
+Zi=(R1*R2*u)/((R2*u)+(R1*u)+(R1*R2))
+x3=Zi*10^-3
+disp(x3," Zi(k-ohm) = R1 || R2 || betare =")
+z=(-RC/re)*(Zi/(RS+Zi))*(RL/(RC+RL)) // answer in textbook is wrong
+disp(z," VL/VS = (-RC/re)*(Zi/RS+Zi)*(RL/RC+RL) =") \ No newline at end of file