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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/CH3/EX3.15/example3_15.sce | |
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initial commit / add all books
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| -rwxr-xr-x | 181/CH3/EX3.15/example3_15.sce | 28 |
1 files changed, 28 insertions, 0 deletions
diff --git a/181/CH3/EX3.15/example3_15.sce b/181/CH3/EX3.15/example3_15.sce new file mode 100755 index 000000000..44644e7db --- /dev/null +++ b/181/CH3/EX3.15/example3_15.sce @@ -0,0 +1,28 @@ +// Find current,dc voltage,voltage across load
+// Basic Electronics
+// By Debashis De
+// First Edition, 2010
+// Dorling Kindersley Pvt. Ltd. India
+// Example 3-15 in page 150
+
+clear; clc; close;
+
+// Given data
+Vm=50; // Maximum voltage in V
+f=50; // Frequency in Hz
+Rf=20; // Internal resistance in ohms
+Rl=5000; // Load resistance in ohms
+
+// Calculation
+Im=Vm/(Rl+Rf);
+printf("Since diode conducts only during pisitive half of the input,Im = %0.0e A\n",Im);
+printf("(a)Hence i = 10*sin100*pi*t\n");
+Vdc=(Im/%pi)*Rl;
+printf("(b)V_dc = %0.1f V\n",Vdc);
+printf("Hence V_0v=15.9sin100*pi*t\n");
+printf("(c)When diode is reverse biased,voltage across diode = %0.1f*sin100*pi*t for 0<100*pi*t<pi and 0 for pi,100*pi*t<2*pi",Vdc);
+
+// Result
+// (a) Current in the circuit = 10sin100*pi*t
+// (b) DC output voltage across load = 15.9sin100*pi*t
+// (c) Voltage across diode = 15.9sin100*pi*t for 0<100*pi*t and 0 for pi<100*pi*t<2*pi
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