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 v4VIhQkf005638 for ; Wed, 31 May 2017 20:43:27 +0200 Received: from majordom by post.thorcom.com with local (Exim 4.14) id 1dG8Wn-0002tg-9q for rs_out_1@blacksheep.org; Wed, 31 May 2017 19:39:29 +0100 Received: from [195.171.43.32] (helo=relay1.thorcom.net) by post.thorcom.com with esmtp (Exim 4.14) id 1dG8Wm-0002tX-G5 for rsgb_lf_group@blacksheep.org; Wed, 31 May 2017 19:39:28 +0100 Received: from mail-lf0-x232.google.com ([2a00:1450:4010:c07::232]) by relay1.thorcom.net with esmtps (TLSv1.2:ECDHE-RSA-AES256-GCM-SHA384:256) (Exim 4.89) (envelope-from ) id 1dG8Wi-0002Fu-WA for rsgb_lf_group@blacksheep.org; Wed, 31 May 2017 19:39:27 +0100 Received: by mail-lf0-x232.google.com with SMTP id c184so9612131lfe.2 for ; Wed, 31 May 2017 11:39:24 -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=VfTglZTL7fdbUSI2+bQv6W6M65VZ+/dyurn8GHcCSVI=; b=hut2/AC7vKV0I+m9v6Z/yup5r0D2MgY/UqAo2D3FQSJsaWrw2UDzwddC7x2hPfALGP HcShSBra9DLyZav0+CKMJK+D057FPDG/YVfIitxgCZOZ16fhjiwqYMpkY0XTCrxKzGLE d803KNCXl9GaBlZlTGFqvdHV/HASau2VaqHBQZUDy3UE78tcmAAskZ2nmE7tcTZsS12Q QjGSjGaDeWIRV7IbQkYKFL47VCZkhikGfrujqHSPVjKWWu4B/tHRMINb8RnclK4LqTrJ y3WglY3841/ac4kT1x4jWhKDmqdmJIvELXAiiBVLEAZPNfi4DSXk8IJ4HHZuEAcZDubm AF4A== 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=VfTglZTL7fdbUSI2+bQv6W6M65VZ+/dyurn8GHcCSVI=; b=icP6Q39S4hbTq/HeJRI2RzB8PBtUK6WwM/TJYsNq9I+4r/UKk/wJuipZfTa3fLB+Fp x1L7zXivOLUIsDjs8p//LeVUaPTn/JZc0P7AUCE4USTq3hMd9N7Cj9hrH8Q4Y1cnBiIR yvnmqbx325ZdiIWFKBANxiQXD64B8s3CDLRzYvMKgCMnpI07l/kzS1/p3NxgIeyEC5rB IODUqAK5iPL1h7kekrh2Mky/SCCbt0VqQihCxNuaU96952Cj5/CNmGxWpSyYBtlDlXHU 4wodKJdHi1D5eTp2OJrHhcfPF/NpzWpZOMhAPiEamIm5Qe6Ik2kug7Fv7Ty1rIVev0hb C38g== X-Gm-Message-State: AODbwcCQr0OMPo0K0FC4z6L7xIb5DUOUu/eZ1TfM6FfroeF0IDqoamVv 45J7BZ95am5wSpo2abBEo4DwPwqDIA== X-Received: by 10.46.69.212 with SMTP id s203mr8245849lja.8.1496255963574; Wed, 31 May 2017 11:39:23 -0700 (PDT) MIME-Version: 1.0 Received: by 10.25.156.13 with HTTP; Wed, 31 May 2017 11:39:22 -0700 (PDT) In-Reply-To: <15c5fa191e2.marcocadeddu@tin.it> References: <15c5fa191e2.marcocadeddu@tin.it> From: Andy Talbot Date: Wed, 31 May 2017 19:39:22 +0100 Message-ID: To: rsgb_lf_group@blacksheep.org Cc: dead.fets@gmail.com X-Scan-Signature: 599cc1adb8e3cb65882b46c3fd2fd2b6 Subject: Re: Re: LF: Re: I: Fw: For today the FETs survived... Content-Type: multipart/alternative; boundary="001a114b159233bbc30550d63d69" 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: 11842 --001a114b159233bbc30550d63d69 Content-Type: text/plain; charset="UTF-8" 7:19 turns (assuming 50R output) means you have a load resistance of 6.8 ohms which for 180V DC (81 V RSM fund sine) is nearly 1kW I don't think you really mean to go that extreme do you? 13 ohms is more realistic. As for the tank resonance changing as power increases, that is very wrong. I wonder if the transformer is saturating. Not sure of your core Ae, but lets assume 200mm square, a core of about 16mm diameter. V = 4.44.F.N.A.B Plugging in 137kHz 7 turns, 200 mm^2 and a Bmax of 0.1 that suggests 85V RMS. Which is exactlyly what you have. I suggest more primary turns . Before a transformer ratio of 1:2 was suggested, for Rload = 13 ohms Is the guard circuit in place ? Don't forget, it has to be customised to you exact currents and coil Q. Get teh PA operating to its proper settings foirst - that you can do at low voltage power, it scales perfectly. Only when it it working properly can you add and set up the guard circuit. When I did teh 700W PA, I had a complete workign (albeit unreliable) unit before even thinking of teh guard circuitry. Andy On 31 May 2017 at 18:50, marcocadeddu@tin.it wrote: > Hi Andy... me again... > > I was so curious to see what could happen thatI had a very quick > dinner and connected all, but... > > now the output xfmr has 7T/19T here my > readings/calculations: > (see attached picture) > again the power increase from 10 to 30Vcc then from 30 to 50Vcc after > an initial burst it start to fall down.. > I checked also the resonance of the LC: till 30Vcc is tuned on 137 kHz > with a 3dB bandwidth of 20 kHz, when I move to 40 and 50Vcc the > "maximum" output shifts to 165 kHz... > > mumble mumble > > I tempted to have roasted FETs for dessert and see what happens at > 180V! > > Marco, IK1HSS > > > ----Messaggio originale---- > Da: andy.g4jnt@gmail.com > Data: 30-mag-2017 23.50 > A: > Cc: > Ogg: Re: LF: Re: I: Fw: For today the FETs survived... > > 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. > >> > > >> > > >> > > >> > > >> > >> > >> > >> > >> > > > > > --001a114b159233bbc30550d63d69 Content-Type: text/html; charset="UTF-8" Content-Transfer-Encoding: quoted-printable
7:19 turns (assuming 50R output) =C2=A0means you have a lo= ad resistance of 6.8 ohms which for 180V DC (81 V RSM fund sine) is nearly = 1kW =C2=A0I don't think you really mean to go that extreme do you? =C2= =A0 =C2=A013 ohms is more realistic.

As for the tank res= onance changing as power increases, that is very wrong.=C2=A0 I wonder if t= he transformer is saturating. =C2=A0 Not sure of your core Ae, but lets ass= ume 200mm square, a core of about 16mm diameter.

V= =3D 4.44.F.N.A.B =C2=A0 =C2=A0Plugging in 137kHz 7 turns, 200 mm^2 and a B= max of 0.1 that suggests 85V RMS.
Which is exactlyly what you hav= e.=C2=A0 I suggest more primary turns . =C2=A0 Before a transformer ratio o= f 1:2 was suggested, for Rload =3D 13 ohms

Is the = guard circuit in place ? =C2=A0 Don't forget, it has to be customised t= o you exact currents and coil Q.=C2=A0 Get teh PA operating to its proper s= ettings foirst - that you can do at low voltage power, it scales perfectly.= =C2=A0 Only when it it working properly can you add and set up the guard c= ircuit.

When I did teh 700W PA, I had a complete w= orkign (albeit unreliable) unit before even thinking of teh guard circuitry= .

Andy





On 31 May 2017 at 18:50, = marcocadeddu@tin.it <marcocadeddu@tin.it> wrote:
Hi Andy... me again...

I was so curious to see what could happen thatI had a very quick
dinner and connected all, but...

now the output xfmr has 7T/19T here my
readings/calculations:
(see attached picture)
again the power increase from 10 to 30Vcc then from 30 to 50Vcc after
an initial burst it start to fall down..
I checked also the resonance of the LC: till 30Vcc is tuned on 137 kHz
with a 3dB bandwidth of 20 kHz, when I move to 40 and 50Vcc the
"maximum" output shifts to 165 kHz...

mumble mumble

I tempted to have roasted FETs for dessert and see what happens at
180V!

Marco, IK1HSS


----Messaggio originale----
Da: andy.g4jnt@gmail.com
Data: 30-mag-2017 23.50
A: <rsgb_lf_group@blacks= heep.org>
Cc: <dead.fets@gmail.com><= br> Ogg: Re: LF: Re: I: Fw: For today the FETs survived...

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 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=C2=A0 G4JNT

On 29 May 2017 at 19:07, Andy Talbot <andy.g4jnt@gmail.com> wrote:

> Yes.
> As you'll see in my original write up, I originally forgot that th= e
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.=C2=A0 The RMS of th= e
> resulting sine=C2=A0 is SQRT(2) less than this giving a Peak square to= RMS-
sine
> ratio of=C2=A0 0.9..=C2=A0 =C2=A0If you specifye peak-peak of the squa= re wave, a
further
> factor of 2 applies, leading to the 0.45 ratio described before.
>
> Incidentally, this same ratio appears in that equation for=C2=A0 flux = in a
> magnetic code,=C2=A0 =C2=A0V =3D 4.44.F.N.A.B
> The magic number 4.44=C2=A0 is actually SQRT(2) * pi=C2=A0 =C2=A0 =C2= =A0and comes about
from
> the same sort of sine to square transform.
>
> Andy
>
> On 29 May 2017 at 18:48, marcoc= adeddu@tin.it <marcocadeddu@t= in.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 bur= nout
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.i= t"<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 seconda= ry of
12
>> 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<= br> >>
>> 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 yo= u 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 RM= S
(fundamental)
>> 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 watt= s=C2=A0 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=C2=A0 G4JNT
>>
>>
>>
>> On 28 May 2017 at 19:34, ma= rcocadeddu@tin.it <marcocaded= du@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 an= d
>> > suggestion are welcome!
>> >
>> > 73, Marco IK1HSS
>> >
>> >
>> > -----Original message-----
>> >
>> > From: "marcocaded= du@tin.it" marcocadeddu@tin= .it
>> > Date: Sun, 28 May 2017 17:01:33 +0200
>> > To: rsgb_lf_g= roup@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 bridg= e 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 believ= ed to
be
>> > clean.
>> >
>> >
>> >
>> >
>> >
>> >
>> >
>> > ---------- Forwarded message ----------
>> > From: "marcocaded= du@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 bridg= e 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 believ= ed to
be
>> > clean.
>> >
>> >
>> >
>> >
>>
>>
>>
>>
>>
>



--001a114b159233bbc30550d63d69--