[R-390] RMS power and voltage (was BallasTubes...)

[Drew Papanek]

Hello Heinz,

RMS POWER in a square wave is directly proportional to duty cycle and 
proportional to the square of peak voltage or current.

RMS VOLTAGE (or current ) in a zero-referenced square wave of peak voltage 
Vp (or Ip as applicable) and duty cycle d is given by:  VRMS=((Vp^2)*d)^.5  
(math formulas can be cumbersome in ASCII).  Thus, a 50% duty cycle square 
wave would have an RMS voltage of .707 times Vp and an RMS current of .707 
time Ip..

A half wave rectified sine wave has an RMS voltage value of half its peak 
voltage.

A half wave rectified sine wave has an RMS POWER value of .5 times RMS power 
of the whole sine wave.

Resistance of 2 seriesed 6BA6 heaters is 12.6V/300mA or 42 ohms

With a whole sine wave of 12.6 VRMS applied to a 42-ohm load, RMS 
power=(12.6^2) /42 or 3.78 watts.
With a whole sine wave of 25.2 VRMS applied to a 42-ohm load, RMS 
power=(25.2^2) /42 or 15.12 watts.
Half wave rectify the 25.2 VRMS, apply it to a 42 ohm load, RMS 
power=.5*15.12 or 7.56 watts.

Double the voltage, power goes up 4 times.  Half wave rectify, the power 
goes down to half.  4 times one half equals twice the original power.

Those 6BA6 heaters glow brighter for a reason!  In actuality, as they get 
brighter, their resistance increases so the power increase is less than 2 
times, but power is still higher than when they were powered by 12.6 VRMS.

Yes, your R-390A will still hear faint flea flatulence from Fiji, but life 
of those 2 6BA6's will be reduced.

In past postings I should have been more specific that the RMS values to 
which I was referring were for voltage or current, as opposed to power.  
Sorry for any confusion I caused.

Have a happy new year,

Drew

Heinz wrote:

>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


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