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| author | priyanka | 2015-06-24 15:03:17 +0530 |
|---|---|---|
| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /181/CH6/EX6.11/example6_11.sce | |
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| -rwxr-xr-x | 181/CH6/EX6.11/example6_11.sce | 22 |
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diff --git a/181/CH6/EX6.11/example6_11.sce b/181/CH6/EX6.11/example6_11.sce new file mode 100755 index 000000000..dbc2f577b --- /dev/null +++ b/181/CH6/EX6.11/example6_11.sce @@ -0,0 +1,22 @@ +// Determine drain-source resistance
+// Basic Electronics
+// By Debashis De
+// First Edition, 2010
+// Dorling Kindersley Pvt. Ltd. India
+// Example 6-11 in page 287
+
+clear; clc; close;
+
+// Given data
+Ids=10*10^-3; // Drain current in mA
+Vp=-2; // Peak voltage in V
+Vgs=[1 2]; // Gate-source voltage in V
+
+// Calculation
+for i=1:2
+ rds=Vp^2/(2*Ids*(Vgs(i)-Vp));
+ printf("Rds = %0.1f ohm\n",rds);
+end
+
+// Result
+// Rds = 66.7 ohm, 50 ohm
\ No newline at end of file |