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 w43GHeex008098 for ; Thu, 3 May 2018 18:17:41 +0200 Received: from majordom by post.thorcom.com with local (Exim 4.14) id 1fEGlO-0004qj-1P for rs_out_1@blacksheep.org; Thu, 03 May 2018 17:07:22 +0100 Received: from [195.171.43.32] (helo=relay1.thorcom.net) by post.thorcom.com with esmtp (Exim 4.14) id 1fEGjw-0004qY-5V for rsgb_lf_group@blacksheep.org; Thu, 03 May 2018 17:05:52 +0100 Received: from mail-wm0-x235.google.com ([2a00:1450:400c:c09::235]) by relay1.thorcom.net with esmtps (TLSv1.2:ECDHE-RSA-AES256-GCM-SHA384:256) (Exim 4.89) (envelope-from ) id 1fEGjq-0001v5-3d for rsgb_lf_group@blacksheep.org; Thu, 03 May 2018 17:05:49 +0100 Received: by mail-wm0-x235.google.com with SMTP id j4so29307110wme.1 for ; Thu, 03 May 2018 09:05:46 -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; bh=7HM5W+WFR8mDXwfWgyhcIWd720g5I9iwnSsoX2xqj9Y=; b=alhwamzn86vFs/YAH5Ue3oJeqzOqfC9MU1QWiHLLl1zv+spBYCtQ83TTR284OTo38E G7EBunXdHl166ph6PiZd21i/Byq/L2F8y97x5VQvgzUTTKrR62uBdlAvLJ+pIVBNc3ue PLpdPJrrbIXmy5eNtybITa9KKAaUV8NzNIdiZdpEXKQhha3yF5wqKZHHmRZmimFfkb2J B0PyEHQ7ru+LT5EoS3PmGanpQc02K/OjoNSb2Q2MS/RZgN2XOrisClMk1LSHEgDHJxX7 K6FzjbuKNwRtT4mozFgRuLka9Y1CSJF0SFTku/pm1dxFRcGr3fRcTOrN3y4j21cg6Ki6 iirQ== 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=7HM5W+WFR8mDXwfWgyhcIWd720g5I9iwnSsoX2xqj9Y=; b=Yj8iE7E1ivrCM0xR+A8thptB1G9B9FdnKyrw2Sd34e4M9Pgq1qFcAtfM/pZl5ESIga Oozb25/2PZfBFAUXPz2v32un0fgLImMq79OXNqxGQyU0xFPzwc+71cajCz/bzNqftPm1 uNibdotUuMEdeVaPvG2CGDmosOm4QbPjc9qL95WUiw28SEy5v5iUOkEFknqylw5PipDx Rp6uVkHvY5mQzH7vEBAUzWM96MAqG2ciaqldvHhV8qW5P3E4kIUkpnpu9Ee2DMZPjLa0 z7x1pgcyEDclsDzodtod1cANeEata3xoOadD9EgWE4L8mkCc1sMyiH4eBWkCEGVS5Z1d adKw== X-Gm-Message-State: ALQs6tCSomNY60xTE2BUwvkQaCO6dbAfr3FogbiEvu3/rYscSIas07hz Ac2hq0ugc8lRJ9/muiAiBCbO/kJgjmjaNiMyK4k= X-Google-Smtp-Source: AB8JxZq6Qx6Qnoz52iu9jC2NXKkHv03d3WzOKQZIy2rD3IsQoL4V+Sacxk5115IDODzhC9so9xMKkhafZLC+63Lr4ms= X-Received: by 2002:a50:ab82:: with SMTP id u2-v6mr31205059edc.163.1525363543570; Thu, 03 May 2018 09:05:43 -0700 (PDT) MIME-Version: 1.0 Received: by 10.80.159.130 with HTTP; Thu, 3 May 2018 09:05:43 -0700 (PDT) In-Reply-To: References: <376794419.20180503144640@gmail.com> From: Andy Talbot Date: Thu, 3 May 2018 17:05:43 +0100 Message-ID: To: LineOne X-Spam-Score: 1.8 (+) 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: The peak of the fundamental sine component in a square wave is 4/pi times the square wave amplitude (yes, it is bigger). The RMS of the sine is Peak / SQRT(2) . Take the two together and the RMS of your sine is therefore 0.9 * Peak of the square wave [...] Content analysis details: (1.8 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) 0.1 URIBL_SBL_A Contains URL's A record listed in the Spamhaus SBL blocklist [URIs: n1bug.com] 0.6 URIBL_SBL Contains an URL's NS IP listed in the Spamhaus SBL blocklist [URIs: n1bug.com] 0.0 HTML_MESSAGE BODY: HTML included in message 0.0 T_DKIM_INVALID DKIM-Signature header exists but is not valid 1.0 FREEMAIL_REPLY From and body contain different freemails X-Scan-Signature: ecf102e6d15c1e466f357d5e6d956ac3 Subject: Re: LF: How can you tell if a Class D amp output transformer is starting to saturate? Content-Type: multipart/alternative; boundary="0000000000002afc4d056b4f6027" X-Spam-Checker-Version: SpamAssassin 2.63 (2004-01-11) on post.thorcom.com X-Spam-Level: X-Spam-Status: No, hits=0.5 required=5.0 tests=HTML_20_30,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 --0000000000002afc4d056b4f6027 Content-Type: text/plain; charset="UTF-8" The peak of the fundamental sine component in a square wave is 4/pi times the square wave amplitude (yes, it is bigger). The RMS of the sine is Peak / SQRT(2) . Take the two together and the RMS of your sine is therefore 0.9 * Peak of the square wave So if you had a Vdd of 50V you would have 45V rms across each half of the winding, or 90V rms across the two halves. Which answers your other question Actually, thinking about what I've just said, you are putting the square wave though the transformer, not a sine. So in that case you have a bit more leeway in that it can go to the same peak value as the sine would have given you. Unless I'm building SMPSUs, I don't put square waves though transformers in transmitters - teh RF is always filtered beforehand. There is an equivalent equation for square waves and used in SMPSU design, it is V.t = N.A.B Now V is input voltage, t is the on-time , and N.A.B as before And yes, your area is correct. 1.58cm^2 = 0.000158m^2 Andy www.g4jnt.com On 3 May 2018 at 16:36, N1BUG wrote: > I am still not sure whether I am using that equation correctly. > > Chris, if you will forgive me for tagging along on your post, I'm > trying to get this equation sorted in my head. Put no faith in > anything I say with regard to this! ;-) > > Andy, could you please check me out on this? > > Let's use the case Chris asks about as an example. > > Suppose I want to calculate the maximum RMS voltage to stay out of > saturation. > > The manufacturer of the core states its Ae (equivalent cross > sectional area) is 1.58 cm^2. I believe that works out to 0.000158 m^2. > > F = 137000 > > I do not know if I should be using 4 turns or 8 turns for the > primary. It's 4+4. I will assume 8. > > Vrms = 4.44 * 137000 * 8 * 0.000158 * 0.1 > > Vrms = ~77V > > Does that look right? > > As for practical application I have no idea what this means. The > equation uses Vrms but we're talking about something closer to a > square wave than sine wave. But then it's not really a square wave > either... > > Paul > > > > On 05/03/2018 10:00 AM, Andy Talbot wrote: > > By measuring the voltage across the winding using a scope, then > > seeing what Bmax is using teh equation > > > > Vrms = 4.44.F.N.A.B > > > > F Hz N turns A core cross-sectional area in metres^2 B tesla > > > > If B works out higher than about 0.1 for a ferrite core, then you > > are approaching saturation > > > > Andy > > www.g4jnt.com > > > > > > On 3 May 2018 at 14:46, Chris Wilson > > wrote: > > > > > > > > Hello LF'ers > > > > In say a W1VD power amp, or in general, how can you test if the > > output transformer is > > starting to saturate please? Thanks. > > > > http://www.w1vd.com/137-500-KWTX.html > > > > > > > > > > -- > > Best regards, > > Chris mailto:dead.fets@gmail.com > > > > --0000000000002afc4d056b4f6027 Content-Type: text/html; charset="UTF-8" Content-Transfer-Encoding: quoted-printable
The peak of the fundamental sine component in a square wav= e is 4/pi times the square wave amplitude (yes, it is bigger).=C2=A0 =C2=A0= The RMS of the sine is Peak / SQRT(2) .=C2=A0 Take the two together and the= RMS of your sine is therefore 0.9 * Peak of the square wave

<= /div>
So if you had a Vdd of 50V you would have 45V rms across each hal= f of the winding, or 90V rms across the two halves.=C2=A0 Which answers you= r other question

Actually, thinking about what I&#= 39;ve just said, you are putting the square wave though the transformer, no= t a=C2=A0 sine.=C2=A0 So in that case you have a bit more leeway in that it= can go to the same peak value as the sine would have given you.
=
Unless I'm building SMPSUs, I don't put square waves= though transformers in transmitters - teh RF is always filtered beforehand= .=C2=A0 =C2=A0There is an equivalent equation for square waves and used in = SMPSU design,=C2=A0 it is=C2=A0 =C2=A0V.t =3D N.A.B=C2=A0 =C2=A0 Now V is i= nput voltage,=C2=A0 t is the on-time , and N.A.B as before

And yes, your area is correct.=C2=A0 1.58cm^2 =3D 0.000158m^2


On 3 May 2018 at 16:36, N1BUG <paul@n1bug.com<= /a>> wrote:
I am still not sure= whether I am using that equation correctly.

Chris, if you will forgive me for tagging along on your post, I'm
trying to get this equation sorted in my head. Put no faith in
anything I say with regard to this! ;-)

Andy, could you please check me out on this?

Let's use the case Chris asks about as an example.

Suppose I want to calculate the maximum RMS voltage to stay out of
saturation.

The manufacturer of the core states its Ae (equivalent cross
sectional area) is 1.58 cm^2. I believe that works out to 0.000158 m^2.

F =3D 137000

I do not know if I should be using 4 turns or 8 turns for the
primary. It's 4+4. I will assume 8.

Vrms =3D 4.44 * 137000 * 8 * 0.000158 * 0.1

Vrms =3D ~77V

Does that look right?

As for practical application I have no idea what this means. The
equation uses Vrms but we're talking about something closer to a
square wave than sine wave. But then it's not really a square wave
either...

Paul



On 05/03/2018 10:00 AM, Andy Talbot wrote:
> By measuring the voltage across the winding using a scope, then
> seeing what Bmax is using teh equation
>
> Vrms =3D 4.44.F.N.A.B
>
> F Hz=C2=A0 =C2=A0 =C2=A0N turns=C2=A0 =C2=A0A core cross-sectional are= a in metres^2=C2=A0 =C2=A0 B tesla
>
> If B works out higher than about 0.1 for a ferrite core, then you
> are approaching saturation
>
> Andy
> www.g4jnt.com <http://www.g4jnt.com>
>
>
> On 3 May 2018 at 14:46, Chris Wilson <dead.fets@gmail.com
> <mailto:dead.fets@gmail.com>> wrote:
>
>
>
>=C2=A0 =C2=A0 =C2=A0Hello LF'ers
>
>=C2=A0 =C2=A0 =C2=A0In=C2=A0 say a W1VD power amp, or in general, how c= an you test if the
>=C2=A0 =C2=A0 =C2=A0output transformer is
>=C2=A0 =C2=A0 =C2=A0starting to saturate please? Thanks.
>
>=C2=A0 =C2=A0 =C2=A0http://www.w1vd.com/137-500-KWTX.h= tml
>=C2=A0 =C2=A0 =C2=A0<http://www.w1vd.com/137-500-KW= TX.html>
>
>
>
>=C2=A0 =C2=A0 =C2=A0--
>=C2=A0 =C2=A0 =C2=A0Best regards,
>=C2=A0 =C2=A0 =C2=A0=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 mailto:dead.fets@gmail.com
>=C2=A0 =C2=A0 =C2=A0<mailto:dead.fets@gmail.com>


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