[Lowfer] Epson Question

[Stewart Nelson]

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:ve2iq@magma.ca] 
Sent: Sunday, April 18, 2004 2:37 PM
To: lowfer@mailman.qth.net
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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