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

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 181/CH4/EX4.7/example4_7.sce | 40 ++++++++++++++++++++++++++++++++++++++++
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 create mode 100755 181/CH4/EX4.7/example4_7.sce

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+// Calculate Labeled Currents and Voltages
+// Basic Electronics
+// By Debashis De
+// First Edition, 2010
+// Dorling Kindersley Pvt. Ltd. India
+// Example 4-7 in page 210
+
+clear; clc; close;
+
+// Given Data
+beta_bjt=100; // beta gain of BJT
+Vbe=0.7; // Base-Emitter voltage of BJT in V
+
+//Calculation
+Vcc1=10; 
+Vee1=-10; 
+Ve1=-0.7; 
+R1=10*10^3; 
+Ie1=(Vcc1-Vbe)/R1; 
+Ib1=Ie1/(beta_bjt+1);
+Vc1=Vcc1-R1*(Ie1-Ib1); 
+Vcc2=10; 
+Vee2=-15; 
+Ve2=-0.7; 
+R2=5*10^3; 
+Ie2=(Vcc2-Vbe)/R2; 
+Ic2=(beta_bjt/(beta_bjt+1))*Ie2; 
+Vc2=Vee2+R2*(Ie2); 
+printf("Circuit 1:\n(a)Emitter Current=%0.2e A\n(b)Base Current=%0.2e A\n(c)Collector Voltage=%0.3f V\n\n",Ie1,Ib1,Vc1);
+printf("Circuit 2:\n(a)Emitter Current=%0.2e A\n(b)Collector Current=%0.3e A\n(c)Collector Voltage=%0.3f V\n",Ie2,Ic2,Vc2);
+
+// Results
+// (a) Circuit 1 : Emitter Current = 0.93 mA
+// (b) Circuit 1 : Base Current = 9.2 mu-A
+//(c) Circuit 1 : Collector Voltage = 0.792 V
+
+//(a) Circuit 2 : Emitter Current = 1.86 mA
+//(b) Circuit 2 : Collector Current = 1.842 mA
+//(c) Circuit 2 : Collector Voltage : -5.7 V
+ 
\ No newline at end of file
-- 
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