[R-390] Ballastubes (was inrush current limiters)

[Heinz und Hannelore Breuer]

Hi,

could somebody please give me the mathematical expression for a half
sinewave. I don't get it that the RMS value of a half sinewave should be
0.5 of the peak value. 
As I understand it we use a diode to cut off one halfwave (i.e. the
negative). So all we have is a positive halfwave in the first half
period and nothing in the second half period. At 60 Hz that is a
positive halfwave in the first 8.33 ms and nothing at all in the second
8.33 ms. To get a RMS value of 0.5 the waveform must be a squarewave and
not a sinus. Take a piece of paper and draw it up.
What I am missing here?

Merry Christmas
Heinz DH2FA, KM5VT



Drew Papanek wrote:
> 
> On Mon, 23 Dec 2002 21:58:34 -0600, Cecil Acuff wrote:
> 
> <snip>
> 
> >An AC waveform is a Peak to Peak waveform.  Moving equal amounts above and
> >below Zero.  (in this case)
> >
> >You arrived at peak value by half wave rectifying the Peak to Peak sine
> >wave.
> 
> Half wave will posess the same (neglecting diode drop) peak value as the
> symmetrical sinewave whence it came, but a half sinewave is not the same
> waveform as a full sinewave.
> 
> >You now multiply that value by .707 to get the RMS value.  To go from
> >RMS back to Peak you multiply by 1.414 and then double that to get Peak to
> >Peak values.
> 
> This applies to a full sine wave, but not a half sine wave.  Yes, the RMS
> value of a sine wave is .707 times its peak value.  However, the RMS value
> of a half sinewave is .5 times its peak value.
> 
> Drew
> 
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