This is a simple question from microelectronics by sedra smith 5e page
Question: Refer to the figure below:
Let Vi have a peak value of 10V and r=1k Ohms. Find the peak value if Id (current through diode) and the dc component of Vo.
Answer: 10mA; 3.18V
They are assuming a perfect diode otherwise I (peak) is approximately 9.3 ma not 10 ma. With a perfect diode I (peak) = V(in) / R à 10V/1K = 10 ma.
Dc component of Vout is simply Vavg /2 = (10 * 0.63)/2. We're dividing by 2 because we're passing only one alternation (positive) per cycle (half-wave rectification).
Hope this helps
Dorian
Peak current:
Ipeak = Vpeak/R = 10V/1K = 10mA
DC component of Vo is the average, so averaging the voltage over a complete cycle (2pi radians) and noting that due to the action of the diode the output voltage will be zero when the angle is pi < q < 2pi we have:
Vdc = (1/2pi)*INT[0-->pi,Vi*sin(q)dq] = 3.183V
Assuming an ideal diode (no voltage drop in the forward direction) the peak resistor current is the peak signal voltage divided by the resistance, or 10 / 1000 = 0.010 ==> 10 mA
The DC component of Vo is the average value of Vo. To find the average value of a waveform, divide "the area between the curve and the horizontal axis" by the period of the waveform. The area of one alternation of a sine wave is 2 times the peak amplitude. The distance along one period of any sine wave is 2 times pi. So the average value is (2 * 10) / (2 * pi) = 20 / 6.28 = 3.18.
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