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

---
 1691/CH2/EX2.18/exmp2_18.sce | 28 ++++++++++++++++++++++++++++
 1 file changed, 28 insertions(+)
 create mode 100755 1691/CH2/EX2.18/exmp2_18.sce

(limited to '1691/CH2/EX2.18')

diff --git a/1691/CH2/EX2.18/exmp2_18.sce b/1691/CH2/EX2.18/exmp2_18.sce
new file mode 100755
index 000000000..1b0d57d82
--- /dev/null
+++ b/1691/CH2/EX2.18/exmp2_18.sce
@@ -0,0 +1,28 @@
+//Example 2.18
+clc
+disp("Refering to equation(1) of section 4.5.3, the input impedance is given by,")
+disp("R''_i = R1 || R2 || h_ie")
+disp("Now  R1 = 25 k-ohm,  R2 = 47 k-ohm,  and  h_ie = 2 k-ohm")
+format(7)
+ri=(25*47*2)/((47*2)+(25*2)+(25*47))  // in k-ohm
+disp(ri,"Therefore,  R''_i(in k-ohm) =")
+disp("  K = R_C / R")
+disp("Now  R_C = 10 k-ohm        ...given")
+disp("Now  f = 1 / 2*pi*R*C*sqrt(6+4K)")
+disp("Therefore,  R*sqrt(6+4K) = 31830.989")
+disp("Now  K = R_C / R = 10*10^3 / R")
+disp("Therefore,  R*sqrt(6+(40*10*10^3/R)) = 31830.989")
+disp("Therefore,  R^2*(6+(40*10*10^3/R)) = (31830.989)^2")
+R=poly(0,'R')
+p1=6*R^2+(40*10^3)*R-(31830.989)^2
+t1=roots(p1)
+ans1=t1(1)
+format(6)
+disp((-ans1)*10^-3,"Therefore,  R(in k-ohm)=            Neglecting negative value")
+k=10/16.74
+format(7)
+disp(k,"Therefore,  K = R_C / R =")
+disp("Therefore,  h_fe >= 4K + 23 + 29/K")
+hfe=(4*0.5973)+23+(29/0.5973)
+format(6)
+disp(hfe,"  h_fe >=")
-- 
cgit