i need the derivations that how these formulae comes
Vavg=Vpeak/pi for half wave rectifier
Vavg=2Vpeak/pi for full wave
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A:
Full wave is double of half wave because both 'bumps' are used. That is easy.
Now derive the halfwave:
V(t) = Vpeak*sin(t);
2 PI is a full circle and we only use half of that : 0->PI
Vavg = Vpeak*integral(sin(t),0 -> PI)/2*PI
indefinite integral of sin(x) = -cos(x)
= Vpeak*(-cos(PI) + cos(0))/2*PI
= Vpeak*(-(-1) + 1) / 2*PI
= Vpeak ( 1 + 1) / 2*PI
= Vpeak / PI
that's it.
Peter Tillemans
2010-08-18 16:46:55
oh thanks a lot. just 1 thing is not clear as my physics is not so good. how we know that we should multiply Vpeak to sin(t) to get the voltage???
Sweety Khan
2010-08-18 17:17:47
Because that is the definition of Vpeak : it is the amplitude of the sine wave. A sine wave goes from 0 to 1 to -1 to 0 again. So the peak voltage occurs at 1 and -1 points. To scale to the real voltage you multiply by Vpeak. (for more details see http://en.wikipedia.org/wiki/Alternating_current)
Peter Tillemans
2010-08-18 17:26:58
@Sweety Khan A rectifier takes as input a wave and outputs a DC voltage.
James
2010-08-18 17:27:29
thank u soo much Peter and James for helping me instead of abstaining my question like others . thanx a lot. my best wishes r for u
Sweety Khan
2010-08-18 17:40:18