[Lowfer] Epson Question

Mon, 19 Apr 2004 17:24:35 +1200

I think it all comes right if the DC term IS included under the square root
sign when figuring the RMS  value-which of course it should be.
73
 Dave, ZL3FJ

----- Original Message ----- 
From: "Stewart Nelson" <[email protected]>
To: <[email protected]>
Sent: Monday, April 19, 2004 11:41 AM
Subject: RE: [Lowfer] Epson Question

> Hi all,
>
> There is no need to argue -- both Bills are right, but are describing
> different quantities.  Bill de Carle's formula of V^2/2*R gives the total
> power delivered to a load, when the signal switches between V and ground,
> with a 50% duty cycle.  However, half of that power is DC, and would not
> contribute to radiation, drive power for an amp, etc.  Bill Ashlock's
> formula of V^2/4*R gives the total AC power in the square wave.  However,
> that also includes the harmonics, which are not useful HiFER output.
IIRC,
> the p-p amplitude of the fundamental is ~1.275 times the p-p of a square
> wave.  Then, applying V^2/8*R, where R is the effective load resistance at
> the fundamental frequency, should give a reasonably accurate result.
>
> 73,
>
> Stewart KK7KA
>
>
> -----Original Message-----
> From: Bill de Carle [mailto:[email protected]]
> Sent: Sunday, April 18, 2004 2:37 PM
> To: [email protected]
> Subject: RE: [Lowfer] Epson Question
>
> At 03:49 PM 4/18/2004 -0400, Bill Ashlock wrote:
> >>What is the power output of an the Epson oscillators we use for HiFER in
> mW
> >>or dB?  I don't see specs on this in any of my catalogs.
> >
> >Kurt the output power depends on the voltage out and the load
> resistance.The
> >basic formula for power in a load of course is P= V^2/R  where V = RMS
> >voltage applied to a resistive load R. The tricky part is converting the
> >waveform to RMS.  For a sine wave the above formula becomes P = V^2/8*R
> >where V is measured peak to peak. For a square wave it's P= V^2/4*R where
V
>
> >is also in peak to peak volts.
>
> Something doesn't look right, Bill.
>
> The formula for sinewave makes sense, but the one for a squarewave
doesn't,
> at least not to me.  Let's take an example:  say our squarewave is
switching
> between ground (0 volts) and +5V and that it's a perfect squarewave with
> 50 percent duty cycle.  Clearly, when the input voltage is 5V, the power
> delivered to a 1-ohm load would be 5*5/1 = 25 watts.  Now, the peak to
peak
> voltage in this case is 5 volts, right?  And that power is only delivered
> 50 percent of the time.  So the formula ought to be V^2/2*R, not V^2/4*R.
>
> I suppose you could argue you were talking about a squarewave that
> alternates
> polarity - i.e. a balanced line squarewave, in which case "peak-to-peak"
> doesn't
> apply because the absolute value of the voltage across the resistor is
> constant, V.
> Power delivered is then V^2/R, assuming the switching time for polarity
> reversal
> is negligible.
>
> This has the feel of a big argument coming up :-)
>
> Bill VE2IQ
>
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