Return-Path: X-Spam-Checker-Version: SpamAssassin 3.4.0 (2014-02-07) on lipkowski.org X-Spam-Level: X-Spam-Status: No, score=-2.3 required=5.0 tests=FREEMAIL_FORGED_FROMDOMAIN, FREEMAIL_FROM,HEADER_FROM_DIFFERENT_DOMAINS,HTML_MESSAGE,RCVD_IN_DNSWL_MED, SPF_PASS,T_DKIM_INVALID autolearn=ham autolearn_force=no version=3.4.0 X-Spam-DCC: : mailn 1480; Body=2 Fuz1=2 Fuz2=2 Received: from post.thorcom.com (post.thorcom.com [195.171.43.25]) by lipkowski.org (8.14.4/8.14.4/Debian-8+deb8u1) with ESMTP id v4ULwRKC001429 for ; Tue, 30 May 2017 23:58:28 +0200 Received: from majordom by post.thorcom.com with local (Exim 4.14) id 1dFp3t-0005RY-GN for rs_out_1@blacksheep.org; Tue, 30 May 2017 22:52:21 +0100 Received: from [195.171.43.32] (helo=relay1.thorcom.net) by post.thorcom.com with esmtp (Exim 4.14) id 1dFp1x-0005PS-Fv for rsgb_lf_group@blacksheep.org; Tue, 30 May 2017 22:50:21 +0100 Received: from mail-lf0-x231.google.com ([2a00:1450:4010:c07::231]) by relay1.thorcom.net with esmtps (TLSv1.2:ECDHE-RSA-AES256-GCM-SHA384:256) (Exim 4.89) (envelope-from ) id 1dFp1s-0007X9-L1 for rsgb_lf_group@blacksheep.org; Tue, 30 May 2017 22:50:18 +0100 Received: by mail-lf0-x231.google.com with SMTP id h4so56192249lfj.3 for ; Tue, 30 May 2017 14:50:16 -0700 (PDT) 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 :cc; bh=vjRMLdc+3MpVCkUh+fkA3szeFk5rzqM9ROniaNr12uA=; b=YiUskCP3cKHTdSNY4xxtz6TWWenmsBgHpplALPmRyLMAt29f0JfeXmVGCHIR/P6mRX 7EgiTmvoKTT0JN/Gni9fiGq8d9kON77peVANSP1+Do0379v6CAn8Nua9/m8Mz0MakZGZ o+8P6AoBldhVB4jL2Nut2j+jpvRhp+5tjB/7WxjZjuF5hHBwO32W7aso2rhvN524LNMk 3X1Cdf6s5CbFRzaIYHsj+zTz4wDjeNELjD4SXl8lvouBc8qipQVXjYzG9WeowafeBq9u wkIIIKNR9jc3IWivHlgXZj9zKO4VXAMQLj6I1Ze55gRgwOrsW7lDGFBoDTiWv6FvSDwC x4Aw== 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:cc; bh=vjRMLdc+3MpVCkUh+fkA3szeFk5rzqM9ROniaNr12uA=; b=s7a0RNpc353nuzRdjpRjzUHUyNxFDT3bNfTywM0PBeURTifMbkrrrww2F3tt8RH3Qf QoHAtmDTurNgs33kK3njleHNbH2L4Tof996K2j4nzC7F+GZZM+MZ2XThR8SA/tkVojo6 np8OddS65VRIPDck86wJx/qcmITVAHO4VSvV/uucfPlM212FxXrZ8YD4hM9WNpvTB3Za jMleVKXnJEezLkMXFBjh0ShWJkPt79rPynF3jHgdH+WQFWN0PAJXXp1oxG+HPn/Hsw/j hBHbEMY3o3JfzM9OFyKJTzrF/GP+Ra4cbG/sSIJCgyl8DTyPHmQob/bP6Q279apfmiBM qeUg== X-Gm-Message-State: AODbwcBHTLO8cc+PSp8oTQb/5Ca/pQPV97p7Okcp+b08g19azFJzDHLA qupYC6P1mb6IMlSXVGUMIdN3vw+3xA== X-Received: by 10.25.155.7 with SMTP id d7mr6628765lfe.74.1496181014507; Tue, 30 May 2017 14:50:14 -0700 (PDT) MIME-Version: 1.0 Received: by 10.25.156.13 with HTTP; Tue, 30 May 2017 14:50:13 -0700 (PDT) In-Reply-To: References: <15c5552ac1b.marcocadeddu@tin.it> From: Andy Talbot Date: Tue, 30 May 2017 22:50:13 +0100 Message-ID: To: rsgb_lf_group@blacksheep.org Cc: dead.fets@gmail.com X-Scan-Signature: 852e9a9ee930056348c0102e6df31911 Subject: Re: LF: Re: I: Fw: For today the FETs survived... Content-Type: multipart/alternative; boundary="001a114020b2e3b92f0550c4c988" 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 X-Scanned-By: MIMEDefang 2.75 Status: RO X-Status: X-Keywords: X-UID: 11838 --001a114020b2e3b92f0550c4c988 Content-Type: text/plain; charset="UTF-8" I've just looked again at the circuit diagram you sent - on there the values are different from your statement in the email. It shows primary 5 turns, secondary 12 turns so a load resistance in the order of 9 ohms which is rather low if you are intending a Vdd of 180V - but closer to the ideal Rl The tank components have a reactance of 130 ohms which is too high a Q is used with that 9 ohms Rload, You should be aiming for a Q in the region of 6. Even with the optimum load R of 13 ohms described last time for 500 Watts from 180V rail the resulting Q of 10 is a bit too high - you will end up with high voltage and critical tuning Andy G4JNT On 29 May 2017 at 19:07, Andy Talbot wrote: > Yes. > As you'll see in my original write up, I originally forgot that the peak > of the fundamental sine component of a square wave is GREATER than the peak > by a factor of 4 / pi and initially my PA delivered a lot more power (1.6 > times) than it was supposed to. > > So if the square wave has a peak value of 1, its fundamental sine > component has a peak value of 4/pi or around 1.27. The RMS of the > resulting sine is SQRT(2) less than this giving a Peak square to RMS-sine > ratio of 0.9.. If you specifye peak-peak of the square wave, a further > factor of 2 applies, leading to the 0.45 ratio described before. > > Incidentally, this same ratio appears in that equation for flux in a > magnetic code, V = 4.44.F.N.A.B > The magic number 4.44 is actually SQRT(2) * pi and comes about from > the same sort of sine to square transform. > > Andy > > On 29 May 2017 at 18:48, marcocadeddu@tin.it wrote: > >> >> uhuh... a slightly silly misleading assumption... Vdc are the same of >> Vrms before FETs make their work! >> >> Thank you Andy for pointing out it!! >> With this approach calculation changes a bit and probably with the >> right Xfmr the PA can give higher satisfaction :-) >> >> Hopefully the FETs will survive and this time I'm ready to burnout the >> antenna hi >> >> Will keep you both updated, thank you once more Andy >> >> 73 Marco, IK1HSS >> ----Messaggio originale---- >> Da: andy.g4jnt@gmail.com >> Data: 28-mag-2017 21.18 >> A: "marcocadeddu@tin.it", >> >> Cc: >> Ogg: LF: Re: I: Fw: For today the FETs survived... >> >> First thing I noticed is that your turns ratio on the output >> transformer >> doesn't look right. >> You quote "* ... with primary winding of 15 turns and secondary of 12 >> turns...*" >> >> 180V DC in a half bridge is 180V peak-peak square wave. >> The fundamental sine part of that is 4/pi * 180 = 229V pk-pk >> so is 229V /[2.SQRT(2)] = 81V RMS >> >> To a good approximation RMS(fund) from a half bridge is Vrms(fund) = >> 0.45VDC >> >> For 500 Watts out, Rload = 81 ^ 2 / 500 = 13 ohms >> >> So to match to 50 ohms you need a turns ratio of SQRT(50/13) = 1.9: >> 1 so >> call it 2:1 Keeping 12 turns on the secondary means you need 6 turns >> on >> the primary >> >> When operating at reduced voltage, the power out will vary exactly as >> the >> square of the voltage. >> Recalculating from first principles for a 12V supply: >> >> 12V DC = 12V pk-pk = 12 / [2.SQRT(2)] * 4/pi = 5.4V RMS (fundamental) >> in 13 ohms should give 5.4^2/13 = 2.2 Watts >> >> check using ratio of voltages, squared : >> >> (12V/180V) ^ 2 * 500W = 2.2 Watts which is the same as above. >> QED >> >> Your 15:12 ratio result sin a load impedance of (15/12)^2 * 50 = 78 >> ohms >> >> At 40V DC == 18V RMS(fund) that will give 18^2/78 = 4.1 watts which is >> actually LESS that you are seeing - the 2* discrepancy is odd, but the >> low >> power is in the area of what you measured.. >> >> Andy G4JNT >> >> >> >> On 28 May 2017 at 19:34, marcocadeddu@tin.it >> wrote: >> >> > Hi Chris, >> > >> > I tried to post this message on the reflector but apparently I had no >> > success.. >> > As promised I keep you updated but as you can read in the >> > attachment the first trials were not enocouraging... >> > Andy, may I ask you to read my report? your interpretation and >> > suggestion are welcome! >> > >> > 73, Marco IK1HSS >> > >> > >> > -----Original message----- >> > >> > From: "marcocadeddu@tin.it" marcocadeddu@tin.it >> > Date: Sun, 28 May 2017 17:01:33 +0200 >> > To: rsgb_lf_group@blacksheep.org >> > Subject: For today the FETs survived... >> > >> > Hi LF, >> > >> > hope that also the toroids of Chris survived! >> > My FETs survived, but they are not working as expected :-( >> > Attached the report on my attempt to duplicate the half bridge of >> > Andy.. >> > Has anyone suggestions before I try to cook all connecting to the >> > 180Vdc supply? >> > >> > Thank you >> > 73 Marco IK1HSS >> > >> > >> > -- >> > This message has been scanned by E.F.A. Project and is believed to be >> > clean. >> > >> > >> > >> > >> > >> > >> > >> > ---------- Forwarded message ---------- >> > From: "marcocadeddu@tin.it" >> > To: >> > Cc: >> > Bcc: >> > Date: Sun, 28 May 2017 17:01:33 +0200 (CEST) >> > Subject: For today the FETs survived... >> > Hi LF, >> > >> > hope that also the toroids of Chris survived! >> > My FETs survived, but they are not working as expected :-( >> > Attached the report on my attempt to duplicate the half bridge of >> > Andy.. >> > Has anyone suggestions before I try to cook all connecting to the >> > 180Vdc supply? >> > >> > Thank you >> > 73 Marco IK1HSS >> > >> > >> > -- >> > This message has been scanned by E.F.A. Project and is believed to be >> > clean. >> > >> > >> > >> > >> >> >> >> >> > --001a114020b2e3b92f0550c4c988 Content-Type: text/html; charset="UTF-8" Content-Transfer-Encoding: quoted-printable
I've just looked again at the circuit diagram you sent= - on there the values are different from your statement in the email.=C2= =A0 It shows primary 5 turns, secondary 12 turns so a load resistance in th= e order of 9 ohms which is rather low if you are intending a Vdd of 180V - = but closer to the ideal Rl

The tank components have a re= actance of 130 ohms which is too high a Q is used with that 9 ohms Rload, Y= ou should be aiming for a Q in the region of 6.

Ev= en with the optimum load R of 13 ohms described last time for 500 Watts fro= m 180V rail the resulting Q of 10 is a bit too high - you will end up with = high voltage and critical tuning

Andy =C2=A0G4JNT<= /div>

On 29 M= ay 2017 at 19:07, Andy Talbot <andy.g4jnt@gmail.com> wrot= e:
Yes.
As you'l= l see in my original write up, I originally forgot that the peak of the fun= damental sine component of a square wave is GREATER than the peak by a fact= or of 4 / pi and initially my PA delivered a lot more power (1.6 times) tha= n it was supposed to.

So if the square wave has a = peak value of 1, its fundamental sine component has a peak value of 4/pi or= around 1.27.=C2=A0 The RMS of the resulting sine =C2=A0is SQRT(2) less tha= n this giving a Peak square to RMS-sine ratio of =C2=A00.9.. =C2=A0 If you = specifye peak-peak of the square wave, a further factor of 2 applies, leadi= ng to the 0.45 ratio described before.

Incidentall= y, this same ratio appears in that equation for =C2=A0flux in a magnetic co= de, =C2=A0 V =3D 4.44.F.N.A.B
The magic number 4.44 =C2=A0is actu= ally SQRT(2) * pi =C2=A0 =C2=A0 and comes about from the same sort of sine = to square transform.

Andy

On 29 May 2017 at 18:48, marcocadeddu@tin.it= <marcocadeddu@tin.it> wrote:

uhuh... a slightly silly misleading assumption... Vdc are the same of
Vrms before FETs make their work!

Thank you Andy for pointing out it!!
With this approach calculation changes a bit and probably with the
right Xfmr=C2=A0 the PA can give higher satisfaction :-)

Hopefully the FETs will survive and this time I'm ready to burnout the<= br> antenna hi

Will keep you both updated, thank you once more Andy

73 Marco, IK1HSS
----Messaggio originale----
Da: andy.g4jnt@gm= ail.com
Data: 28-mag-2017 21.18
A: "marcocade= ddu@tin.it"<marcocadeddu@tin.it>,
<rsgb_= lf_group@blacksheep.org>
Cc: <dead.fets@= gmail.com>
Ogg: LF: Re: I: Fw: For today the FETs survived...

First thing I noticed is that your turns ratio on the output
transformer
doesn't look right.
You quote "* ... with primary winding of 15 turns and secondary of 12<= br> turns...*"

180V DC in a half bridge is 180V peak-peak square wave.
The fundamental sine part of that is=C2=A0 4/pi * 180 =3D 229V pk-pk
so is 229V /[2.SQRT(2)] =3D 81V RMS

To a good approximation RMS(fund) from a half bridge is Vrms(fund) =3D
0.45VDC

For 500 Watts out, Rload =3D=C2=A0 81 ^ 2 / 500 =3D=C2=A0 13 ohms

So to match to 50 ohms you need a turns ratio of SQRT(50/13) =3D 1.9:
1=C2=A0 =C2=A0 =C2=A0so
call it 2:1=C2=A0 Keeping 12 turns on the=C2=A0 secondary means you need 6 = turns
on
the primary

When operating at reduced voltage, the power out will vary exactly as
the
square of the voltage.
Recalculating from first principles for a 12V supply:

12V=C2=A0 DC =3D 12V pk-pk =3D 12 / [2.SQRT(2)] * 4/pi =3D 5.4V RMS (fundam= ental)
in 13 ohms should give 5.4^2/13 =3D 2.2 Watts

check using ratio of voltages, squared :

(12V/180V) ^ 2 * 500W =3D 2.2 Watts which is the same as above.
QED

Your 15:12 ratio result sin a load impedance of (15/12)^2 * 50 =3D 78
ohms

At 40V DC =3D=3D 18V RMS(fund) that will give 18^2/78 =3D 4.1 watts=C2=A0 w= hich is
actually LESS that you are seeing - the 2* discrepancy is odd, but the
low
power is in the area of what you measured..

Andy=C2=A0 G4JNT



On 28 May 2017 at 19:34, marcocadeddu@tin.it <marcocadeddu@tin.it>
wrote:

> Hi Chris,
>
> I tried to post this message on the reflector but apparently I had no<= br> > success..
> As promised I keep you updated but as you can read in the
> attachment the first trials were not enocouraging...
> Andy, may I ask you to read my report? your interpretation and
> suggestion are welcome!
>
> 73, Marco IK1HSS
>
>
> -----Original message-----
>
> From: "m= arcocadeddu@tin.it" marcocadeddu@tin.it
> Date: Sun, 28 May 2017 17:01:33 +0200
> To: = rsgb_lf_group@blacksheep.org
> Subject: For today the FETs survived...
>
> Hi LF,
>
> hope that also the toroids of Chris survived!
> My FETs survived, but they are not working as expected :-(
> Attached the report on my attempt to duplicate the half bridge of
> Andy..
> Has anyone suggestions before I try to cook all connecting to the
> 180Vdc supply?
>
> Thank you
> 73 Marco IK1HSS
>
>
> --
> This message has been scanned by E.F.A. Project and is believed to be<= br> > clean.
>
>
>
>
>
>
>
> ---------- Forwarded message ----------
> From: "m= arcocadeddu@tin.it" <marcocadeddu@tin.it>
> To: <rsgb_lf_group@blacksheep.org>
> Cc:
> Bcc:
> Date: Sun, 28 May 2017 17:01:33 +0200 (CEST)
> Subject: For today the FETs survived...
> Hi LF,
>
> hope that also the toroids of Chris survived!
> My FETs survived, but they are not working as expected :-(
> Attached the report on my attempt to duplicate the half bridge of
> Andy..
> Has anyone suggestions before I try to cook all connecting to the
> 180Vdc supply?
>
> Thank you
> 73 Marco IK1HSS
>
> > --
> This message has been scanned by E.F.A. Project and is believed to be<= br> > clean.
>
>
>
>






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