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+//Ex8_4
+clc
+RL=5*10^3
+Rs=1.2*10^3
+hre=2.5*10^-4
+hie=1.1*10^3
+hfe=100
+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
+//calculation for current gain:
+Ai=-hfe/(1+(hoe*RL))
+disp("Ai = -hfe/(1+(hoe*RL)) = "+string(abs(Ai)))
+//calculation for input resistance:
+Ri = hie+(hre*Ai*RL)
+disp("Ri = hie+(hre*Ai*RL) = "+string(Ri)+"ohm")
+//calculation for voltage gain:
+Av = Ai*RL/Ri
+disp("Av = Ai*RL/Ri = "+string(Av))
+//calculation for output resistance:
+Go=hoe-((hre*hfe)/(hie+Rs))
+Ro = 1/Go
+disp("Ro = 1/Go")
+disp("Go = hoe-((hre*hfe)/(hie+Rs)) = "+string(Go)+"mho")
+disp("Ro = "+string(Ro)+"ohm")
+
+//note : in the textbook, above problem has given two values for "hfe" and no value for "hre"... 
+//        thus assuming value for "hre = 2.5*10^-4" as taken in previous example 8_2
+//        and "hfe=100" 
+
+//note : in text LOAD RESISTANCE is noted as Rc in question, but RL in solution.
+//       I have work with Load Resistance with notification RL.