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 --- 181/CH7/EX7.35/example7_35.sce | 34 ++++++++++++++++++++++++++++++++++ 181/CH7/EX7.35/example7_35.txt | 5 +++++ 2 files changed, 39 insertions(+) create mode 100755 181/CH7/EX7.35/example7_35.sce create mode 100755 181/CH7/EX7.35/example7_35.txt (limited to '181/CH7/EX7.35') diff --git a/181/CH7/EX7.35/example7_35.sce b/181/CH7/EX7.35/example7_35.sce new file mode 100755 index 000000000..d8c01fc68 --- /dev/null +++ b/181/CH7/EX7.35/example7_35.sce @@ -0,0 +1,34 @@ +// Find values of R2,Vdd,Vds +// Basic Electronics +// By Debashis De +// First Edition, 2010 +// Dorling Kindersley Pvt. Ltd. India +// Example 7-35 in page 338 + +clear; clc; close; + +// Given data +Vp=-5; // Pinch off voltage in V +Idss=12*10^-3; // Drain-source current in mA +Vdd=18; // Drain voltage in V +Rs=2*10^3; // Source resistance in K-ohms +Rd=2*10^3; // Drain resistance in K-ohms +R2=90*10^3; // Original value of R2 in K-ohms + +// Calculation +Vgs1=(-5.3+sqrt(5.3^2-(4*0.48*10.35)))/(2*0.48); +Vgs2=(-5.3-sqrt(5.3^2-(4*0.48*10.35)))/(2*0.48); +printf("Vgs = %0.2f V or %0.2f V\nTherefore Vgs = -2.53 V\n",Vgs1,Vgs2); +Id=(3.306-Vgs2)/2; +Vds=18-(Id*Rd)-(Id*Rs); +r2=(13.47*400)/4.53; +vdd=((16-2.53)*(400+90))/90; +vds=vdd-16-16; +printf("(a)The new value of R2 is %0.1f K-ohm\n",r2); +printf("(b)The new value of Vdd = %0.2f V\n",vdd); +printf("(c)The new value of Vds = %0.2f V",vds); + +// Result +// (a) R2 = 1189.4 K-ohm +// (b) Vdd = 73.34 V +// (c) Vds = 41.34 V \ No newline at end of file diff --git a/181/CH7/EX7.35/example7_35.txt b/181/CH7/EX7.35/example7_35.txt new file mode 100755 index 000000000..b54b22778 --- /dev/null +++ b/181/CH7/EX7.35/example7_35.txt @@ -0,0 +1,5 @@ + Vgs = -2.53 V or -8.51 V +Therefore Vgs = -2.53 V +(a)The new value of R2 is 1189.4 K-ohm +(b)The new value of Vdd = 73.34 V +(c)The new value of Vds = 41.34 V \ No newline at end of file -- cgit