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author | priyanka | 2015-06-24 15:03:17 +0530 |
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committer | priyanka | 2015-06-24 15:03:17 +0530 |
commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
tree | ab291cffc65280e58ac82470ba63fbcca7805165 /181/CH6/EX6.3 | |
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initial commit / add all books
Diffstat (limited to '181/CH6/EX6.3')
-rwxr-xr-x | 181/CH6/EX6.3/example6_3.sce | 26 | ||||
-rwxr-xr-x | 181/CH6/EX6.3/example6_3.txt | 2 |
2 files changed, 28 insertions, 0 deletions
diff --git a/181/CH6/EX6.3/example6_3.sce b/181/CH6/EX6.3/example6_3.sce new file mode 100755 index 000000000..b426dbc8d --- /dev/null +++ b/181/CH6/EX6.3/example6_3.sce @@ -0,0 +1,26 @@ +// Find the value of Rd
+// Basic Electronics
+// By Debashis De
+// First Edition, 2010
+// Dorling Kindersley Pvt. Ltd. India
+// Example 6-3 in page 275
+
+clear; clc; close;
+
+// Given data
+Ids=12*10^-3; // Drain current in mA
+Vp=-4; // Peak voltage in V
+Rs=0; // Source resistance in ohms
+Vds=0.1; // Drain-source voltage in V
+Vgg=0; // Gate voltage in V
+
+// Calculation
+id=Ids*(50*10^-3-625*10^-6);
+Rd=(15-Vds)/id;
+
+printf("(a)i_d = %0.3e A\n",id);
+printf("(b)Rd = %0.3e ohm",Rd);
+
+// Result
+// (a) i_d = 592.6 mu-A
+// (b) Rd = 25.15 k-ohm
\ No newline at end of file diff --git a/181/CH6/EX6.3/example6_3.txt b/181/CH6/EX6.3/example6_3.txt new file mode 100755 index 000000000..4a1e5ac27 --- /dev/null +++ b/181/CH6/EX6.3/example6_3.txt @@ -0,0 +1,2 @@ + (a)i_d = 5.925e-004 A
+(b)Rd = 2.515e+004 ohm
\ No newline at end of file |