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From: Andy Talbot <andy.g4jnt@gmail.com>
Date: Sun, 24 Dec 2017 20:44:29 +0000
Message-ID: <CAA8k23TnyE0NUGA9XQsmDhOxn81dqNTC3pQh2H7VfEoQHH=mXw@mail.gmail.com>
To: rsgb_lf_group@blacksheep.org
Subject: Re: LF: TXing 2200m WSPR
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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> 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
>
> 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
>
>>
>>

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<div dir=3D"ltr"><div>The scopematch circuit looks reasonable.=C2=A0 =C2=A0=
Arithmetic is correct</div><div>(as an aside, I always group constants toge=
ther for calibrations like that.=C2=A0 =C2=A0 Measuring peak to peak across=
 a 50R load, power then becomes=C2=A0 Vpk-pk ^ 2 / 400</div><div>When my 30=
dB power attenuator is in circuit this becomes P =3D 2.5 Vp-p ^ 2=C2=A0 =C2=
=A0 You could do the same grouping and simplifying for your scopematch divi=
ders and Vp-p readings)</div><div><br></div><div>Back to your amp ...</div>=
<div><br></div><div>You get 25W into 50R going via the low pass filter</div=
>You don&#39;t show your low pass filter circuit.=C2=A0 =C2=A0=C2=A0<div>It=
 is possible that has the wrong values for 50R and is providing an impedanc=
e transformation from your 50R true load to present the amplifier with a lo=
wer RL, thus allowing more output</div><div><br></div><div>Terminate the am=
plifier directly with=C2=A0 the 50 ohs load and see what you get - so elimi=
nating the filter</div><div><br></div><div>As Sherlock Holmes said=C2=A0 &q=
uot;Once you have eliminated the impossible, whatever remains, however impr=
obable, must be so&quot;</div><div><br></div><div>Andy=C2=A0 G4JNT</div><di=
v><br></div><div class=3D"gmail_extra"><br><div class=3D"gmail_quote">On 24=
 December 2017 at 20:12, N1BUG <span dir=3D"ltr">&lt;<a href=3D"mailto:paul=
@n1bug.com" target=3D"_blank">paul@n1bug.com</a>&gt;</span> wrote:<br><bloc=
kquote class=3D"gmail_quote" style=3D"margin:0 0 0 .8ex;border-left:1px #cc=
c solid;padding-left:1ex">Hi Andy,<br>
<br>
This is very interesting, because 12 watts is what I think I have consisten=
tly been able to generate into the antenna before problems creep in.<br>
<br>
I can easily accept that this amplifier may only be capable of 12 or 13 wat=
ts.<br>
<br>
What I don&#39;t understand is why I think I am seeing a clean 25 watts int=
o a pure resistive 50 ohm load. Can someone please check my math?<br>
<br>
I&#39;m running the output of the amplifier through a low pass filter, then=
 a scopematch, then into a high quality 50 ohm load.<br>
<br>
My scopematch circuit is here:<br>
<br>
<a href=3D"http://n1bug.com/n1debug/LF_ScopeMatch-20171224.jpg" rel=3D"nore=
ferrer" target=3D"_blank">http://n1bug.com/n1debug/LF_Sc<wbr>opeMatch-20171=
224.jpg</a><br>
<br>
Note that it is configured such that on the current sense output, 1V=3D1A a=
nd for voltage, 1V=3D50V.<br>
<br>
Running into a pure 50 ohm resistive load I am seeing exactly 4 divisions p=
eak to peak on the scope (2 divisions above center, 2 below).<br>
<br>
500 mV/div * 4 divs =3D 2.0V peak-peak or 0.707V RMS.<br>
<br>
0.707 * 50 (1V=3D50V on the scopematch) =3D 35.4V RMS.<br>
<br>
35.4^2 / 50 ohms =3D 25 watts.<br>
<br>
Where am I going wrong?<br>
<br>
As a check on scopematch calibration I set my HP 3325B to 10V peak-peak, ra=
n 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 =3D .=
2V * 50 =3D 10V which seems to verify. Admittedly this verification is at a=
 much lower power level.<br>
<br>
73,<br>
Paul N1BUG<span class=3D""><br></span><blockquote class=3D"gmail_quote" sty=
le=3D"margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><span =
class=3D""><br></span></blockquote></blockquote></div></div></div>

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