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authorpriyanka2015-06-24 15:03:17 +0530
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+//Ex8_3
+clc
+RL = 8*10^3
+Rs= 500
+hie=1.0*10^3
+hre=2.5*10^-4
+hfe=50
+hoe=25*10^-6
+disp("RL = "+string(RL)+"ohm")//load resistance
+disp("Rs = "+string(Rs)+"ohm")//source resistance
+//h-parameters for CE transistor amplifier are as follows:
+disp("hie = "+string(hie)+"ohm")//input resistance of CE transistor
+disp("hre = "+string(hre))//voltage gain of CE transistor
+disp("hfe = "+string(hfe))//current gain of CE transistor
+disp("hoe = "+string(hoe)+"mho")//output conductance of CE transistor
+
+Ai=-hfe/(1+(hoe*RL))
+disp("Ai = -hfe/(1+(hoe*RL)) = "+string(Ai))//calculation for current gain
+
+Ri = hie+(hre*Ai*RL)
+disp("Ri = hie+(hre*Ai*RL) = "+string(Ri)+"ohm")//calculation for input resistance
+
+Ais=(Ai*Rs)/(Ri+Rs)
+disp("Ais = (Ai*Rs)/(Ri+Rs)= "+string(Ais))//current gain with source resistance
+
+Avs = Ai*RL/Ri
+disp("Avs = Ai*RL/Ri = "+string(Avs))//voltage gain with source resistance
+
+//note : in the textbook above problem has given two values for hie BUT no value for hfe ...
+// thus assuming hie=50 as hfe =50, as given in the previous example 8_2
+
+//note : answer in the textbook is not accuratly calculated.