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 <1325698854.20171224230351@gmail.com>
From: Andy Talbot <andy.g4jnt@gmail.com>
Date: Sun, 24 Dec 2017 23:16:40 +0000
Message-ID: <CAA8k23RVOd0+c2eppnFfEeNOzKahBgBPmcydXuDOuR8SPZJ-+Q@mail.gmail.com>
To: rsgb_lf_group@blacksheep.org
Cc: dead.fets@gmail.com
Subject: Re: LF: TXing 2200m WSPR
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Yes

Try anything and try to work out what is happening.  But first of all, he
needs to know what power being deliverfed into a 50 ohmsload is seemingly
higher than it 'can' be.    The Vdd^2/2/Rl cannot be violated.
If it appears to be so, then Rl or Vdd or the measurement  has to be wrong.

I once had to go though a similar
try-everything-and-believe-nothing-until-tested process to work out why my
700Watt 137kHz Tx appeared to be more than 100% efficient!
It turned out my Bird 30dB 1kW attenuator only had 29.4dB attenuation at
DC.  And for a 1GHz jobbie, 137kHz is DC.

On 24 December 2017 at 23:03, Chris Wilson <dead.fets@gmail.com> wrote:

>
>
> Hello Andy,
>
> Is  there  any point in Paul putting his dummy load on the end of his
> 75  odd meters of feeder co-ax outside? And then on the secondary of
> his  matching  transformer when tapped one to one? In case the co-ax is
> iffy or the impedance matching transfomer's doing something odd?
>
>
> Sunday, December 24, 2017, 10:42:16 PM, you 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
>
>
>
>
>
> --
> Best regards,
>  Chris                            mailto:dead.fets@gmail.com
>
>
>

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<div dir=3D"ltr">Yes<div><br></div><div>Try anything and try to work out wh=
at is happening.=C2=A0 But first of all, he needs to know what power being =
deliverfed into a 50 ohmsload is seemingly higher than it &#39;can&#39; be.=
=C2=A0 =C2=A0 The Vdd^2/2/Rl cannot be violated.</div><div>If it appears to=
 be so, then Rl or Vdd or the measurement=C2=A0 has to be wrong.</div><div>=
<br></div><div>I once had to go though a similar try-everything-and-believe=
-nothing-until-tested process to work out why my 700Watt 137kHz Tx appeared=
 to be more than 100% efficient!</div><div>It turned out my Bird 30dB 1kW a=
ttenuator only had 29.4dB attenuation at DC.=C2=A0 And for a 1GHz jobbie, 1=
37kHz is DC.</div></div><div class=3D"gmail_extra"><br><div class=3D"gmail_=
quote">On 24 December 2017 at 23:03, Chris Wilson <span dir=3D"ltr">&lt;<a =
href=3D"mailto:dead.fets@gmail.com" target=3D"_blank">dead.fets@gmail.com</=
a>&gt;</span> wrote:<br><blockquote class=3D"gmail_quote" style=3D"margin:0=
 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><br>
<br>
Hello Andy,<br>
<br>
Is=C2=A0 there=C2=A0 any point in Paul putting his dummy load on the end of=
 his<br>
75=C2=A0 odd meters of feeder co-ax outside? And then on the secondary of<b=
r>
his=C2=A0 matching=C2=A0 transformer when tapped one to one? In case the co=
-ax is<br>
iffy or the impedance matching transfomer&#39;s doing something odd?<br>
<div class=3D"HOEnZb"><div class=3D"h5"><br>
<br>
Sunday, December 24, 2017, 10:42:16 PM, you wrote:<br>
<br>
&gt; Well ...=C2=A0 =C2=A0 Using exactly the values in the filter circuit d=
iagram,<br>
&gt; 50R transforms through the filter to 48.5 - j2.86 (Ret Loss =3D 30dB,=
=C2=A0<br>
&gt; VSWR =3D 1.07)=C2=A0 =C2=A0[Using GM3SEK&#39;s original Netcalc prog.]=
<br>
<br>
<br>
&gt; So that&#39;s pretty conclusive the ideal filter values will not be<br=
>
&gt; upsetting things at the fundamental frequency.=C2=A0 =C2=A0=C2=A0<br>
&gt; According to Google, the=C2=A0 T106-2 has a stated Al value of=C2=A0 1=
3.5nH<br>
&gt; /turn^2=C2=A0 so 72 turns does indeed give 70uH.=C2=A0 =C2=A0So IF you=
r core is correct, the filter should be OK.<br>
<br>
<br>
&gt; It&#39;s a bit difficult from now on, at a distance, to try to work ou=
t what is happening.<br>
&gt; Anyone else, any suggestions ?<br>
<br>
<br>
&gt; BTW ...<br>
<br>
<br>
&gt; Peak to peak of a { symetrical }square wave needs to be multiplied<br>
&gt; by 4/pi to get the peak-to-peak of the fundamental component.=C2=A0 =
=C2=A0So<br>
&gt; the amploitude you see will be lower by about 1.3 times for teh same f=
undamental power component.=C2=A0<br>
<br>
<br>
&gt; Andy=C2=A0 G4JNT<br>
<br>
<br>
<br>
<br>
<br>
</div></div><span class=3D"HOEnZb"><font color=3D"#888888">--<br>
Best regards,<br>
=C2=A0Chris=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 mailto:<a href=3D"mailto:dead.fets@gmail=
.com">dead.fets@gmail.com</a><br>
<br>
<br>
</font></span></blockquote></div><br></div>

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