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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/CH7/EX7.26/example7_26.sce | |
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| -rwxr-xr-x | 181/CH7/EX7.26/example7_26.sce | 27 |
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diff --git a/181/CH7/EX7.26/example7_26.sce b/181/CH7/EX7.26/example7_26.sce new file mode 100755 index 000000000..c10795e84 --- /dev/null +++ b/181/CH7/EX7.26/example7_26.sce @@ -0,0 +1,27 @@ +// Calculate quiescent values of Id,Vds,Vgs
+// Basic Electronics
+// By Debashis De
+// First Edition, 2010
+// Dorling Kindersley Pvt. Ltd. India
+// Example 7-26 in page 330
+
+clear; clc; close;
+
+// Given data
+Ids=20*10^-3; // Drain-souce current in mA
+Vp=-4; // Pinch off voltage in volts
+
+// Calculation
+printf("We get the equation:\n0.3125*Id^2-6*Id+20=0\n");
+Id1=(6+sqrt(6^2-4*0.3125*20))/(2*0.3125);
+Id2=(6-sqrt(6^2-4*0.3125*20))/(2*0.3125);
+printf("Id = %0.1f mA and %0.1f mA\n",Id1,Id2);
+printf("We consider only %0.1f mA\n",Id2);
+Vgs=-Id2*0.5;
+Vds=30-(Id2*(5+0.5));
+printf("Vgs = %0.2f V\n(c)Vds = %0.2f V",Vgs,Vds);
+
+// Result
+// Id = 4.3 mA
+// Vgs = -2.15 V
+// Vds = 6.35 V
\ No newline at end of file |