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authorpriyanka2015-06-24 15:03:17 +0530
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diff --git a/1583/CH5/EX5.3/HFAAGC_Ex_5_3.sce b/1583/CH5/EX5.3/HFAAGC_Ex_5_3.sce
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+clc
+//Chapter 5:High Frequency Amplifiers and Automatic Gain Control
+//example 5.3 page no 153
+//given
+gm=2*10^-3//transconductance
+Cgs=5*10^-12//equivalent Miller's input capacitance
+Cgd=1*10^-12//equivalent Miller's output capacitance
+Cds=1*10^-12
+rd=13*10^3
+R=5*10^3//source resistance
+RL=(6*10^3*13*10^3)/(6*10^3+13*10^3)//total load resistance
+Av=-gm*RL//voltage gain
+R_L=RL*rd/(RL+rd)
+CT=Cgs+Cgd*(1+gm*R_L)//total capacitance
+Co=Cds+(Cgd*(1+gm*R_L)/(gm*R_L))//output capacitance
+w1=(R*CT)^-1//pole due to input circuit
+w2=(RL*Co)^-1//pole due to output circuit
+mprintf('the voltage gain is %f \n the total capacitance is %3.2e pF \n the output capacitance is %3.2e pF \n the pole due to input circuit is %3.2e rad/s \n the pole due to output circuit is %3.2e rad/s ',Av,CT,Co,w1,w2)
diff --git a/1583/CH5/EX5.3/Result_of_Chapter_5_Ex5_3.jpg b/1583/CH5/EX5.3/Result_of_Chapter_5_Ex5_3.jpg
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