Return-Path: Received: from post.thorcom.com (post.thorcom.com [195.171.43.25]) by klubnl.pl (8.14.4/8.14.4/Debian-8+deb8u2) with ESMTP id vBOMlSd2020729 for ; Sun, 24 Dec 2017 23:47:31 +0100 Received: from majordom by post.thorcom.com with local (Exim 4.14) id 1eTEyN-00052F-DR for rs_out_1@blacksheep.org; Sun, 24 Dec 2017 22:42:23 +0000 Received: from [195.171.43.32] (helo=relay1.thorcom.net) by post.thorcom.com with esmtp (Exim 4.14) id 1eTEyM-000526-G4 for rsgb_lf_group@blacksheep.org; Sun, 24 Dec 2017 22:42:22 +0000 Received: from mail-wm0-x232.google.com ([2a00:1450:400c:c09::232]) by relay1.thorcom.net with esmtps (TLSv1.2:ECDHE-RSA-AES256-GCM-SHA384:256) (Exim 4.89) (envelope-from ) id 1eTEyI-0006as-Ui for rsgb_lf_group@blacksheep.org; Sun, 24 Dec 2017 22:42:21 +0000 Received: by mail-wm0-x232.google.com with SMTP id f140so29995757wmd.2 for ; Sun, 24 Dec 2017 14:42:18 -0800 (PST) X-DKIM-Result: Domain=gmail.com Result=Good and Known Domain DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=gmail.com; s=20161025; h=mime-version:in-reply-to:references:from:date:message-id:subject:to; bh=ruyJNPSlKMsBy05ARRJ1s6WrjMf7zuGRBcFUBU+MeFk=; b=P0H56d88HZQGp5sKAZAHvAJD4Xo0R313HeQYaqAV6UzV38F1AlFn6q7B+RqHj6nQ5I wOBtFSY1HpQ47bnk/O0uf+P8DqRTiotrYJBnFtFgz08BzBJhRLoioXQ3LeOxPKaQbTdC /SSkHwn4hNLAw9X401Whc1mzLs3u1XzUxYkLFTJgmo1NjguNP+OfGgIppvfcWHrby5yu qlWipujg0bQ/KG/NCSyypHmZJio8SvSEHDsX416WtZjXaUvsRQXxuwM3W6KC8vAiFGw7 B6uT32ACR1EjwE3qiY6Eeo7u3SrrSAF9TC8AsFcVHyBPxwGqGo/goeUlR1Swb+fjU0jG nnzg== X-Google-DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=1e100.net; s=20161025; h=x-gm-message-state:mime-version:in-reply-to:references:from:date :message-id:subject:to; bh=ruyJNPSlKMsBy05ARRJ1s6WrjMf7zuGRBcFUBU+MeFk=; b=rVc9wvkz6sB8u/72rJcLlkidS7HiF5dF01TiPVH3U3FdN/GhUWBTVgtHDcgFw2utnK uwmNxC19tPwPZbvcR3j+RYlH+8h7NR2JjxLoWWnJhYEpn4H478wrEMZo6xzVRzH5MZMW KejFvMaa771C3BCN4x/5E/+MwTDmpzSyEvROOkwulvqCBneM0rUELNWwp7wNAMtN4IUk HliWMGZlnkb2wwChsIlitx/DZem7NP0sC2DwUW1fuh/UWoy7ExoWScWI/BOtSFOUbyY0 GjJUz2Z8d+beGdaj4+BaqdOh62klHLKZJYHolTB/0ULcRmGrd0Dox+DfbLdRcb2C/JmM YbLw== X-Gm-Message-State: AKGB3mJslxGy0+dkyc97IubmXhmbq2YlymIFWzLkkIvb5UqJVaSoK2gT qcUxACIWj7A2KRk+VS4Gog5SjS4F6eRXR9Ab+8Q= X-Google-Smtp-Source: ACJfBouhnb3eSedC2mKR0fPYOMYqLIXQ5OgpnSyCKB5rdmHf1xiOGq7Q1JOIULYvMti2YJXAyF6IMhkhoJ9RdVrBKm4= X-Received: by 10.80.170.87 with SMTP id p23mr25426096edc.289.1514155337327; Sun, 24 Dec 2017 14:42:17 -0800 (PST) MIME-Version: 1.0 Received: by 10.80.177.54 with HTTP; Sun, 24 Dec 2017 14:42:16 -0800 (PST) In-Reply-To: References: <4ffc7958-810a-e4d2-7745-258a58338934@n1bug.com> <5A3AE81B.9010205@posteo.de> <5A3B8FB1.1020403@posteo.de> <7ee833a8-04b4-4712-d927-7eccf37eec14@n1bug.com> <5A3D9BC1.1040609@posteo.de> <6fa587ef-d5b0-6a67-b604-ba94144444f7@n1bug.com> <5A3E77DD.4090908@posteo.de> <103A2C60B87B4930A99ED33442E477AF@StevePC> <792271d9-34bb-3fb1-140c-c322015e2e9b@n1bug.com> <5A3F9D04.90800@posteo.de> <6be50d62-c33c-a781-1138-268803e0b6f1@n1bug.com> From: Andy Talbot Date: Sun, 24 Dec 2017 22:42:16 +0000 Message-ID: To: rsgb_lf_group@blacksheep.org X-Spam-Score: 0.0 (/) X-Spam-Report: Spam detection software, running on the system "relay1.thorcom.net", has NOT identified this incoming email as spam. The original message has been attached to this so you can view it or label similar future email. If you have any questions, see the administrator of that system for details. Content preview: 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. [...] Content analysis details: (0.0 points, 5.0 required) pts rule name description ---- ---------------------- -------------------------------------------------- 0.0 FREEMAIL_FROM Sender email is commonly abused enduser mail provider (andy.g4jnt[at]gmail.com) 0.0 HTML_MESSAGE BODY: HTML included in message 0.0 T_DKIM_INVALID DKIM-Signature header exists but is not valid X-Scan-Signature: 3ecbf20a510e9846bb1bdaf7e0334c65 Subject: Re: LF: TXing 2200m WSPR Content-Type: multipart/alternative; boundary="94eb2c0df40e04278205611dc3fc" X-Spam-Checker-Version: SpamAssassin 2.63 (2004-01-11) on post.thorcom.com X-Spam-Level: X-Spam-Status: No, hits=0.0 required=5.0 tests=HTML_MESSAGE autolearn=no version=2.63 X-SA-Exim-Scanned: Yes Sender: owner-rsgb_lf_group@blacksheep.org Precedence: bulk Reply-To: rsgb_lf_group@blacksheep.org X-Listname: rsgb_lf_group X-SA-Exim-Rcpt-To: rs_out_1@blacksheep.org X-SA-Exim-Scanned: No; SAEximRunCond expanded to false --94eb2c0df40e04278205611dc3fc Content-Type: text/plain; charset="UTF-8" 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 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 > > 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>> 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 >> > > --94eb2c0df40e04278205611dc3fc Content-Type: text/html; charset="UTF-8" Content-Transfer-Encoding: quoted-printable
Well ...=C2=A0 =C2=A0 Using exactly the values in the filt= er circuit diagram, 50R transforms through the filter to 48.5 - j2.86 (Ret = Loss =3D 30dB,=C2=A0 VSWR =3D 1.07)=C2=A0 =C2=A0[Using GM3SEK's origina= l Netcalc prog.]

So that's pretty conclusive the ide= al filter values will not be upsetting things at the fundamental frequency.= =C2=A0 =C2=A0=C2=A0
According to Google, the=C2=A0 T106-2 has a s= tated Al value of=C2=A0 13.5nH /turn^2=C2=A0 so 72 turns does indeed give 7= 0uH.=C2=A0 =C2=A0So 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.=C2=A0 =C2=A0So the amploitude you se= e will be lower by about 1.3 times for teh same fundamental power component= .=C2=A0

Andy=C2=A0 G4JNT

=

On 24 D= ecember 2017 at 21:22, N1BUG <paul@n1bug.com> wrote:
Hi Andy,

Thank you very much for that. I am learning and knowledge sometimes comes t= ogether 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 ampl= itude was much less than with the filer.

This is the filer, for what it's worth:

http://n1bug.com/n1debug/ASB_LF_LPF-20171224.jp= g

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 s= aw more power out of the amp. Perhaps that was due to an impedance change.<= br>
Paul N1BUG



On 12/24/2017 03:44 PM, Andy Talbot wrote:
The scopematch circuit looks reasonable.=C2=A0 =C2=A0Arithmetic is correct<= br> (as an aside, I always group constants together 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
When my 30dB 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 scope= match 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 impeda= nce transformation from your 50R true load to present the amplifier with a = lower RL, thus allowing more output

Terminate the amplifier directly with=C2=A0 the 50 ohs load and see what yo= u get - so eliminating the filter

As Sherlock Holmes said=C2=A0 "Once you have eliminated the impossible= , whatever remains, however improbable, must be so"

Andy=C2=A0 G4JNT


On 24 December 2017 at 20:12, N1BUG <paul@n1bug.com <mailto:paul@n1bug.com>> wrote:

=C2=A0 =C2=A0 Hi Andy,

=C2=A0 =C2=A0 This is very interesting, because 12 watts is what I think I<= br> =C2=A0 =C2=A0 have consistently been able to generate into the antenna befo= re
=C2=A0 =C2=A0 problems creep in.

=C2=A0 =C2=A0 I can easily accept that this amplifier may only be capable o= f
=C2=A0 =C2=A0 12 or 13 watts.

=C2=A0 =C2=A0 What I don't understand is why I think I am seeing a clea= n 25
=C2=A0 =C2=A0 watts into a pure resistive 50 ohm load. Can someone please =C2=A0 =C2=A0 check my math?

=C2=A0 =C2=A0 I'm running the output of the amplifier through a low pas= s
=C2=A0 =C2=A0 filter, then a scopematch, then into a high quality 50 ohm lo= ad.

=C2=A0 =C2=A0 My scopematch circuit is here:

=C2=A0 =C2=A0 http://n1bug.com/n1debug/LF_Sc= opeMatch-20171224.jpg
=C2=A0 =C2=A0 <http://n1bug.com/n1debug/LF_ScopeMatch-20171224.jpg>

=C2=A0 =C2=A0 Note that it is configured such that on the current sense
=C2=A0 =C2=A0 output, 1V=3D1A and for voltage, 1V=3D50V.

=C2=A0 =C2=A0 Running into a pure 50 ohm resistive load I am seeing exactly= 4
=C2=A0 =C2=A0 divisions peak to peak on the scope (2 divisions above center= , 2
=C2=A0 =C2=A0 below).

=C2=A0 =C2=A0 500 mV/div * 4 divs =3D 2.0V peak-peak or 0.707V RMS.

=C2=A0 =C2=A0 0.707 * 50 (1V=3D50V on the scopematch) =3D 35.4V RMS.

=C2=A0 =C2=A0 35.4^2 / 50 ohms =3D 25 watts.

=C2=A0 =C2=A0 Where am I going wrong?

=C2=A0 =C2=A0 As a check on scopematch calibration I set my HP 3325B to 10V=
=C2=A0 =C2=A0 peak-peak, ran it through the scopematch into the same 50 ohm=
=C2=A0 =C2=A0 dummy load at 137.5 kHz. I set the scope to 50 mV/div and it<= br> =C2=A0 =C2=A0 read exactly 4 divisions peak-peak =3D .2V * 50 =3D 10V which= seems
=C2=A0 =C2=A0 to verify. Admittedly this verification is at a much lower po= wer
=C2=A0 =C2=A0 level.

=C2=A0 =C2=A0 73,
=C2=A0 =C2=A0 Paul N1BUG


--94eb2c0df40e04278205611dc3fc--