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Diffstat (limited to '2459/CH9/EX9.2')
| -rw-r--r-- | 2459/CH9/EX9.2/Ex9_2.PNG | bin | 0 -> 7416 bytes | |||
| -rw-r--r-- | 2459/CH9/EX9.2/Ex9_2.sce | 23 | ||||
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3 files changed, 23 insertions, 0 deletions
diff --git a/2459/CH9/EX9.2/Ex9_2.PNG b/2459/CH9/EX9.2/Ex9_2.PNG Binary files differnew file mode 100644 index 000000000..831404ea0 --- /dev/null +++ b/2459/CH9/EX9.2/Ex9_2.PNG diff --git a/2459/CH9/EX9.2/Ex9_2.sce b/2459/CH9/EX9.2/Ex9_2.sce new file mode 100644 index 000000000..99a44add5 --- /dev/null +++ b/2459/CH9/EX9.2/Ex9_2.sce @@ -0,0 +1,23 @@ +//chapter9
+//example9.2
+//page145
+
+Vi_p=20 // V
+rf=10 // ohm
+Rl=500 // ohm
+Vo=0.7 // V
+Vin=20 // V
+
+// peak current through diode will occur when Vin=Vf so
+Vf=Vin
+// since Vf=Vo+If_peak(rf+Rl) making If_peak as subject we get
+If_peak1=(Vf-Vo)/(rf+Rl) // in ampere
+Vout_peak1=If_peak1*Rl
+
+// for ideal diode, Vo=0 and rf=0 so
+// Vf=If_peak*Rl so we get
+If_peak2=Vf/Rl // in ampere
+Vout_peak2=If_peak2*Rl
+
+printf("peak current through given diode = %.3f mA and peak output voltage = %.3f V \n",If_peak1*1000,Vout_peak1)
+printf("peak current through ideal diode = %.3f mA and peak output voltage = %.3f V \n",If_peak2*1000,Vout_peak2)
diff --git a/2459/CH9/EX9.2/Figure9_2.JPG b/2459/CH9/EX9.2/Figure9_2.JPG Binary files differnew file mode 100644 index 000000000..967505ea7 --- /dev/null +++ b/2459/CH9/EX9.2/Figure9_2.JPG |
