Return-Path: <owner-rsgb_lf_group@blacksheep.org> Received: from mtain-mg10.r1000.mx.aol.com (mtain-mg10.r1000.mx.aol.com [172.29.96.210]) by air-de03.mail.aol.com (v126.13) with ESMTP id MAILINDE034-5ead4b71c2c3257; Tue, 09 Feb 2010 15:17:07 -0500 Received: from post.thorcom.com (post.thorcom.com [193.82.116.20]) by mtain-mg10.r1000.mx.aol.com (Internet Inbound) with ESMTP id CED06380002D4; Tue, 9 Feb 2010 15:17:04 -0500 (EST) Received: from majordom by post.thorcom.com with local (Exim 4.14) id 1NewUc-0004w2-Cm for rs_out_1@blacksheep.org; Tue, 09 Feb 2010 20:15:30 +0000 Received: from [193.82.116.32] (helo=relay1.thorcom.net) by post.thorcom.com with esmtp (Exim 4.14) id 1NewUb-0004vt-Jw for rsgb_lf_group@blacksheep.org; Tue, 09 Feb 2010 20:15:29 +0000 Received: from mail.gmx.net ([213.165.64.20]) by relay1.thorcom.net with smtp (Exim 4.63) (envelope-from <cpaul@gmx.net>) id 1NewUW-0004Id-WA for rsgb_lf_group@blacksheep.org; Tue, 09 Feb 2010 20:15:29 +0000 Received: (qmail invoked by alias); 09 Feb 2010 20:15:05 -0000 Received: from p5DD353DF.dip.t-dialin.net (EHLO Clemens04) [93.211.83.223] by mail.gmx.net (mp058) with SMTP; 09 Feb 2010 21:15:05 +0100 X-Authenticated: #17214767 X-Provags-ID: V01U2FsdGVkX1/FWoJhfCHF5julsdfSi/yUbUWDy2A7gRDNsoQwem 0WQ1xNUPrz3qR3 Message-ID: <008a01caa9c4$92d24b60$0201a8c0@Clemens04> From: "Clemens Clemens" <cpaul@gmx.net> To: <rsgb_lf_group@blacksheep.org> References: <fc7eccec1002090910s5fc7d42aybfce15ed093b032a@mail.gmail.com> Date: Tue, 9 Feb 2010 21:14:37 +0100 MIME-Version: 1.0 X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 6.00.2900.3598 X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.3350 X-Y-GMX-Trusted: 0 X-FuHaFi: 0.5,0.5 X-Spam-Score: 0.0 (/) X-Spam-Report: autolearn=disabled,HTML_MESSAGE=0.001 Subject: LF: Re: PA matching oddity Content-Type: multipart/alternative; boundary="----=_NextPart_000_0084_01CAA9CC.E2D01270" 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 x-aol-global-disposition: G x-aol-sid: 3039ac1d60d24b71c2c02400 X-AOL-IP: 193.82.116.20 ------=_NextPart_000_0084_01CAA9CC.E2D01270 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable I've no practical experience with Mosfet PAs. Nonetheless according to your diagram this is definitely a 1:3 transfo= rmer. Ther's no way how an autotransformer mode could come into play somehow= . >The secondary winding is made of insulated coax, two turns are full= screened as they pass=20 >through the cores / >tubing, but each turn has the braid cut at the= hot end and *joined to the=20 >two ends of the secondary*, You mean...joined to the two ends of the *primary* ? It seems that those connections serve to increase the effective conduc= tor cross section of the primary to carry the heavy currents involved. The braids of the coax ar eused as parallel conductors of the brass tu= bing =3D primary winding.. 73 Clemens DL4RAJ ----- Original Message -----=20 From: Andy Talbot To: rsgb_lf_group@blacksheep.org ; ukmicrowaves@yahoogroups.com Sent: Tuesday, February 09, 2010 6:10 PM Subject: LF: PA matching oddity Has anyone got practical experience of the output matching transform= ers used on MOSFET PAs -=20 I've got a confusing one here? I recently acquired some big HF PA modules, each rated at over 1kW= out, and made up from 8=20 MOSFETS, RFPP53 types, roughly equivalent to MRF140. It runs from wh= at is more than likely a=20 50V rail. The modules were part of an industrial RF heater running at= 13.56MHz, but the design=20 is wideband(ish) with the normal ferrite matching transformers at inpu= t and output. Which is=20 where I may be missing something - they may not be quite so normal... The output transformer has a slightly different topology to designs= seen before - such as=20 those given in the Motorola handbook. The secondary winding is made= of insulated coax, two=20 turns are full screened as they pass through the cores / tubing, but= each turn has the braid cut=20 at the hot end and joined to the two ends of the secondary, with the= third turn consisting just=20 the inner conductor with no braid over it. All three turns (two of= coax plus the single core)=20 sit inside the usual single turn primary made up from brass tube, surr= ounded by a pair of=20 ferrite cores with a connection at the far end. A diagram can be se= en at=20 http://www.g4jnt.com/pamatch.gif Now, the bit that doesn't seem right... the impedances don't work out properly... Assuming it is designed to run into 50 ohms, a 1:3 transformer will= present a load of 5.56=20 ohms to the push pull devices. From a 50V rail this should result= in a maximum power output=20 of 2*(50^2)/5.56 =3D 900 Watts. (Sanity check, a single ended desi= gn at half the Rload =3D=20 (50^2)/ 2 / 2.78 =3D 450 Watts each- normal push pull PA calculation).= Which is not 1kW and is=20 only an absolute theoretical maximum, anyway. BUT, if the transformer were 1:4 instead, , Rload would be 3.125= ohms, Pout max would be=20 1600 watts which is exactly the sort of value I'd expect to see on a= real 1kW rated PA module. Has anyone met that winding configuration before? Is it really th= e 1:3 turns ratio it=20 intuitively looks like, or is there some way the windings could have= have become an=20 auto-transformer and be giving 1:4 turns ratio ? If it really is 1:3 will have to assume the voltage supply may be hi= gher. But for a 10 year=20 old design, sounds very unlikely. A few pictures of the O/P transformer: http://www.g4jnt.com/PAtfmr= 1.jpg=20 http://www.g4jnt.com/PAtfmr2.jpg http://www.g4jnt.com/PAtfmr3.jpg On a quick test on the module today, running from a 10A supply, it= delivered nearly 150 Watts=20 with the PSU current limiting and dragging the supply volts down to 17= V. Now, plugging these=20 values into the matching equation 2 * (17^2) / 5.556 =3D 100 Watts max= possible, - but I was=20 seeing more power. ........ Extra support to the possibility of it being 1:4 - BUT HOW= ? Andy www.g4jnt.com ----------------------------------------------------------------------= -------- Eingehende eMail ist virenfrei. Von AVG =FCberpr=FCft - www.avg.de Version: 9.0.733 / Virendatenbank: 271.1.1/2677 - Ausgabedatum: 02/0= 9/10 08:35:00 ------=_NextPart_000_0084_01CAA9CC.E2D01270 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <HTML><HEAD> <META http-equiv=3DContent-Type content=3D"text/html; charset=3Diso-88= 59-1"> <META content=3D"MSHTML 6.00.2900.3660" name=3DGENERATOR> <STYLE></STYLE> </HEAD> <BODY bgColor=3D#ffffff> <DIV><FONT face=3DArial size=3D2>I've no practical experience with Mos= fet=20 PAs.</FONT></DIV> <DIV><FONT face=3DArial size=3D2>Nonetheless according to your diagram= this is=20 definitely a 1:3 transformer.</FONT></DIV> <DIV><FONT face=3DArial size=3D2>Ther's no way how an autotransformer= mode could=20 come into play somehow.</FONT></DIV> <DIV><FONT face=3DArial size=3D2></FONT> </DIV> <DIV>>The secondary winding is made of insulated coax, two turns ar= e full=20 screened as they pass through the cores / >tubing, but each turn&nb= sp;has the=20 braid cut at the hot end and *joined to the two ends of the secondary*= ,</DIV> <DIV> </DIV> <DIV><FONT face=3DArial size=3D2>You mean...joined to the two ends of= the *primary*=20 ?</FONT></DIV> <DIV><FONT face=3DArial size=3D2></FONT> </DIV> <DIV><FONT face=3DArial size=3D2>It seems that those connections serve= to increase=20 the effective conductor cross section</FONT></DIV> <DIV><FONT face=3DArial size=3D2>of the primary to carry the= heavy=20 currents involved.</FONT></DIV> <DIV><FONT face=3DArial size=3D2>The braids of the coax ar eused as pa= rallel=20 conductors of the brass tubing =3D primary winding..</FONT></DIV> <DIV><FONT face=3DArial size=3D2></FONT> </DIV> <DIV><FONT face=3DArial size=3D2>73<BR>Clemens<BR>DL4RAJ</FONT></DIV> <DIV><FONT face=3DArial size=3D2></FONT> </DIV> <BLOCKQUOTE style=3D"PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-L= EFT: 5px; BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px"> <DIV style=3D"FONT: 10pt arial">----- Original Message ----- </DIV> <DIV style=3D"BACKGROUND: #e4e4e4; FONT: 10pt arial; font-color:= black"><B>From:</B>=20 <A title=3Dandy.g4jnt@googlemail.com href=3D"mailto:andy.g4jnt@g= ooglemail.com">Andy Talbot</A> </DIV> <DIV style=3D"FONT: 10pt arial"><B>To:</B> <A title=3Drsgb_lf_group@= blacksheep.org href=3D"mailto:rsgb_lf_group@blacksheep.org">rsgb_l= f_group@blacksheep.org</A> ;=20 <A title=3Dukmicrowaves@yahoogroups.com href=3D"mailto:ukmicrowa= ves@yahoogroups.com">ukmicrowaves@yahoogroups.com</A>=20 </DIV> <DIV style=3D"FONT: 10pt arial"><B>Sent:</B> Tuesday, February 09,= 2010 6:10=20 PM</DIV> <DIV style=3D"FONT: 10pt arial"><B>Subject:</B> LF: PA matching oddi= ty</DIV> <DIV><BR></DIV> <DIV>Has anyone got practical experience of the output matching tran= sformers=20 used on MOSFET PAs - I've got a confusing one here?</DIV> <DIV> </DIV> <DIV>I recently acquired some big HF PA modules, each rated at= over 1kW=20 out, and made up from 8 MOSFETS, RFPP53 types, roughly equival= ent to=20 MRF140. It runs from what is more than likely a 50V rail. = ; The=20 modules were part of an industrial RF heater running at 13.56MHz, bu= t the=20 design is wideband(ish) with the normal ferrite matching= =20 transformers at input and output. Which is where I may be miss= ing=20 something - they may not be quite so normal...</DIV> <DIV> </DIV> <DIV>The output transformer has a slightly different topology to des= igns seen=20 before - such as those given in the Motorola handbook. Th= e=20 secondary winding is made of insulated coax, two turns are full scre= ened as=20 they pass through the cores / tubing, but each turn has the bra= id cut at=20 the hot end and joined to the two ends of the secondary, with= the third=20 turn consisting just the inner conductor with no braid over it.= =20 All three turns (two of coax plus the single core) sit inside= the usual=20 single turn primary made up from brass tube, surrounded by a pair of= ferrite=20 cores with a connection at the far end. A diagram= can be=20 seen at <A href=3D"http://www.g4jnt.com/pamatch.gif">http://www.= g4jnt.com/pamatch.gif</A> =20 </DIV> <DIV> </DIV> <DIV>Now, the bit that doesn't seem right...</DIV> <DIV>the impedances don't work out properly...</DIV> <DIV> </DIV> <DIV>Assuming it is designed to run into 50 ohms, a 1:3 transformer= will=20 present a load of 5.56 ohms to the push pull devices. &nb= sp; From a=20 50V rail this should result in a maximum power output of 2*(50^2)/5.= 56 =3D 900=20 Watts. (Sanity check, a single ended design= at half=20 the Rload =3D (50^2)/ 2 / 2.78 =3D 450 Watts each- norma= l push pull PA=20 calculation). Which is not 1kW and is only an absolute= theoretical=20 maximum, anyway.</DIV> <DIV> </DIV> <DIV>BUT, if the transformer were 1:4 instead, , Rload= would be=20 3.125 ohms, Pout max would be 1600 watts which is exactly the sort= of value=20 I'd expect to see on a real 1kW rated PA module.</DIV> <DIV> </DIV> <DIV>Has anyone met that winding configuration before? &n= bsp; Is it=20 really the 1:3 turns ratio it intuitively looks like, or is there so= me way the=20 windings could have have become an auto-transformer and be givi= ng=20 1:4 turns ratio ?</DIV> <DIV>If it really is 1:3 will have to assume the voltage supply may= be=20 higher. But for a 10 year old design, sounds very unlikely.&nb= sp;</DIV> <DIV> </DIV> <DIV>A few pictures of the O/P transformer: <A href= =3D"http://www.g4jnt.com/PAtfmr1.jpg">http://www.g4jnt.com/PAtfmr1.jpg= </A> =20 <A href=3D"http://www.g4jnt.com/PAtfmr2.jpg">http://www.g4jnt.co= m/PAtfmr2.jpg</A> =20 <A href=3D"http://www.g4jnt.com/PAtfmr3.jpg">http://www.g4jnt.co= m/PAtfmr3.jpg</A></DIV> <DIV> </DIV> <DIV>On a quick test on the module today, running from a = 10A=20 supply, it delivered nearly 150 Watts with the PSU current limiting= and=20 dragging the supply volts down to 17V. Now, plugging these val= ues into=20 the matching equation 2 * (17^2) / 5.556 =3D 100 Watts max possible,= -=20 but I was seeing more power. </DIV> <DIV> ........ Extra support to the possibility of it being 1:4= - BUT HOW=20 ?</DIV> <DIV><BR clear=3Dall>Andy<BR><A href=3D"http://www.g4jnt.com">ww= w.g4jnt.com</A><BR></DIV> <P> <HR> <P></P><BR>Eingehende eMail ist virenfrei.<BR>Von AVG =FCberpr=FCft= - www.avg.de=20 <BR>Version: 9.0.733 / Virendatenbank: 271.1.1/2677 - Ausgabedatum:= 02/09/10=20 08:35:00 <BR></BLOCKQUOTE></BODY></HTML> ------=_NextPart_000_0084_01CAA9CC.E2D01270--