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From: N1BUG <paul@n1bug.com>
Message-ID: <bd8dadd0-0309-9798-27ea-8c8e0bb9d3e5@n1bug.com>
Date: Sun, 24 Dec 2017 18:14:31 -0500
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Subject: Re: LF: TXing 2200m WSPR
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The plot thickens!

The square wave isn't exactly symmetrical but by that calculation it 
makes no more than 8 watts into the dummy load without the filter! 
It's likely a bit more because it isn't symmetrical, but nowhere 
near the 25 watts I get with the filter. Hmm...

Paul N1BUG


On 12/24/2017 05:42 PM, Andy Talbot wrote:
> Well ...    Using exactly the values in the filter circuit diagram, 
> 50R transforms through the filter to 48.5 - j2.86 (Ret Loss = 30dB,  
> VSWR = 1.07)   [Using GM3SEK's original Netcalc prog.]
> 
> So that's pretty conclusive the ideal filter values will not be 
> upsetting things at the fundamental frequency.
> According to Google, the  T106-2 has a stated Al value of  13.5nH 
> /turn^2  so 72 turns does indeed give 70uH.   So IF your core is 
> correct, the filter should be OK.
> 
> It's a bit difficult from now on, at a distance, to try to work out 
> what is happening.
> Anyone else, any suggestions ?
> 
> BTW ...
> 
> Peak to peak of a { symetrical }square wave needs to be multiplied 
> by 4/pi to get the peak-to-peak of the fundamental component.   So 
> the amploitude you see will be lower by about 1.3 times for teh same 
> fundamental power component.
> 
> Andy  G4JNT
> 
> 
> 
> On 24 December 2017 at 21:22, N1BUG <paul@n1bug.com 
> <mailto:paul@n1bug.com>> wrote:
> 
>     Hi Andy,
> 
>     Thank you very much for that. I am learning and knowledge
>     sometimes comes together from bits and pieces in discussions
>     like this.
> 
>     Anything is possible with this amp or the filter.
> 
>     I tried without the filter but as it is now a somewhat spiky
>     square wave I have no idea how it compares power-wise. It did
>     appear to be lower. Except for a short spike at the leading edge
>     of the positive pulses the pk-pk amplitude was much less than
>     with the filer.
> 
>     This is the filer, for what it's worth:
> 
>     http://n1bug.com/n1debug/ASB_LF_LPF-20171224.jpg
>     <http://n1bug.com/n1debug/ASB_LF_LPF-20171224.jpg>
> 
>     You may have helped me understand something I saw the other day.
>     One of the .01 capacitors on the output end of the filter became
>     disconnected and I saw more power out of the amp. Perhaps that
>     was due to an impedance change.
> 
>     Paul N1BUG
> 
> 
> 
>     On 12/24/2017 03:44 PM, Andy Talbot wrote:
> 
>         The scopematch circuit looks reasonable.   Arithmetic is correct
>         (as an aside, I always group constants together for
>         calibrations like that.    Measuring peak to peak across a
>         50R load, power then becomes  Vpk-pk ^ 2 / 400
>         When my 30dB power attenuator is in circuit this becomes P =
>         2.5 Vp-p ^ 2    You could do the same grouping and
>         simplifying for your scopematch dividers and Vp-p readings)
> 
>         Back to your amp ...
> 
>         You get 25W into 50R going via the low pass filter
>         You don't show your low pass filter circuit.
>         It is possible that has the wrong values for 50R and is
>         providing an impedance transformation from your 50R true
>         load to present the amplifier with a lower RL, thus allowing
>         more output
> 
>         Terminate the amplifier directly with  the 50 ohs load and
>         see what you get - so eliminating the filter
> 
>         As Sherlock Holmes said  "Once you have eliminated the
>         impossible, whatever remains, however improbable, must be so"
> 
>         Andy  G4JNT
> 
> 
>         On 24 December 2017 at 20:12, N1BUG <paul@n1bug.com
>         <mailto:paul@n1bug.com> <mailto:paul@n1bug.com
>         <mailto:paul@n1bug.com>>> wrote:
> 
>              Hi Andy,
> 
>              This is very interesting, because 12 watts is what I
>         think I
>              have consistently been able to generate into the
>         antenna before
>              problems creep in.
> 
>              I can easily accept that this amplifier may only be
>         capable of
>              12 or 13 watts.
> 
>              What I don't understand is why I think I am seeing a
>         clean 25
>              watts into a pure resistive 50 ohm load. Can someone please
>              check my math?
> 
>              I'm running the output of the amplifier through a low pass
>              filter, then a scopematch, then into a high quality 50
>         ohm load.
> 
>              My scopematch circuit is here:
> 
>         http://n1bug.com/n1debug/LF_ScopeMatch-20171224.jpg
>         <http://n1bug.com/n1debug/LF_ScopeMatch-20171224.jpg>
>              <http://n1bug.com/n1debug/LF_ScopeMatch-20171224.jpg
>         <http://n1bug.com/n1debug/LF_ScopeMatch-20171224.jpg>>
> 
>              Note that it is configured such that on the current sense
>              output, 1V=1A and for voltage, 1V=50V.
> 
>              Running into a pure 50 ohm resistive load I am seeing
>         exactly 4
>              divisions peak to peak on the scope (2 divisions above
>         center, 2
>              below).
> 
>              500 mV/div * 4 divs = 2.0V peak-peak or 0.707V RMS.
> 
>              0.707 * 50 (1V=50V on the scopematch) = 35.4V RMS.
> 
>              35.4^2 / 50 ohms = 25 watts.
> 
>              Where am I going wrong?
> 
>              As a check on scopematch calibration I set my HP 3325B
>         to 10V
>              peak-peak, ran it through the scopematch into the same
>         50 ohm
>              dummy load at 137.5 kHz. I set the scope to 50 mV/div
>         and it
>              read exactly 4 divisions peak-peak = .2V * 50 = 10V
>         which seems
>              to verify. Admittedly this verification is at a much
>         lower power
>              level.
> 
>              73,
>              Paul N1BUG