Numerical Question

[Irfan]

This is a simple question from microelectronics by sedra smith 5e page
143. But i am not clear how they calculate it. kindly explain it to me
i.e the method n the formulas used.

Question:
Refer to the figure below:
http://i5.tinypic.com/153bk1k.jpg

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

 

[Dorian McIntire]

This is a simple question from microelectronics by sedra smith 5e page
143. But i am not clear how they calculate it. kindly explain it to me
i.e the method n the formulas used.

Question:
Refer to the figure below:
http://i5.tinypic.com/153bk1k.jpg

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

 

[Greg Neill]

This is a simple question from microelectronics by sedra smith 5e page
143. But i am not clear how they calculate it. kindly explain it to me
i.e the method n the formulas used.

Question:
Refer to the figure below:
http://i5.tinypic.com/153bk1k.jpg

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

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

 

[Bob Penoyer]

This is a simple question from microelectronics by sedra smith 5e page
143. But i am not clear how they calculate it. kindly explain it to me
i.e the method n the formulas used.

Question:
Refer to the figure below:
http://i5.tinypic.com/153bk1k.jpg

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

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.