From b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b Mon Sep 17 00:00:00 2001
From: priyanka
Date: Wed, 24 Jun 2015 15:03:17 +0530
Subject: initial commit / add all books

---
 1583/CH5/EX5.3/HFAAGC_Ex_5_3.sce             |  18 ++++++++++++++++++
 1583/CH5/EX5.3/Result_of_Chapter_5_Ex5_3.jpg | Bin 0 -> 171467 bytes
 2 files changed, 18 insertions(+)
 create mode 100755 1583/CH5/EX5.3/HFAAGC_Ex_5_3.sce
 create mode 100755 1583/CH5/EX5.3/Result_of_Chapter_5_Ex5_3.jpg

(limited to '1583/CH5/EX5.3')

diff --git a/1583/CH5/EX5.3/HFAAGC_Ex_5_3.sce b/1583/CH5/EX5.3/HFAAGC_Ex_5_3.sce
new file mode 100755
index 000000000..56bc553e4
--- /dev/null
+++ b/1583/CH5/EX5.3/HFAAGC_Ex_5_3.sce
@@ -0,0 +1,18 @@
+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
new file mode 100755
index 000000000..18899eed5
Binary files /dev/null and b/1583/CH5/EX5.3/Result_of_Chapter_5_Ex5_3.jpg differ
-- 
cgit