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Date: Tue, 9 Feb 2010 17:10:05 +0000
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From: Andy Talbot <andy.g4jnt@googlemail.com>
To: rsgb_lf_group@blacksheep.org, ukmicrowaves@yahoogroups.com
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Subject: LF: PA matching oddity
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Has anyone got practical experience of the output matching transformers used
on MOSFET PAs - 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 MOSFETS,  RFPP53 types, roughly equivalent to MRF140.  It
runs from what is more than likely a 50V rail.  The modules were part of an
industrial RF heater running at 13.56MHz, but the design is wideband(ish)
with the normal ferrite matching transformers at input and output.  Which is
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 those given in the Motorola handbook.  The secondary
winding is made of insulated coax, two turns are full screened as they pass
through the cores / tubing, but each turn has the braid cut at the hot end
and joined to the two ends of the secondary, with the third turn consisting
just the inner conductor with no braid over it.    All three turns (two of
coax plus the single core) sit inside the usual single turn primary made up
from brass tube, surrounded by a pair of ferrite cores with a connection at
the far end.    A diagram can be seen at 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 ohms to the push pull devices.    From a 50V rail this should
result in a maximum power output of 2*(50^2)/5.56 = 900 Watts.    (Sanity
check,  a single ended design at half the Rload =   (50^2)/ 2 / 2.78 = 450
Watts each- normal push pull PA calculation).   Which is not 1kW and is only
an absolute theoretical maximum, anyway.

BUT,   if the transformer were 1:4 instead, , Rload would be 3.125 ohms,
Pout max would be 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 the 1:3
turns ratio it intuitively looks like, or is there some way the windings
could have have become an auto-transformer and be giving  1:4 turns ratio ?
If it really is 1:3 will have to assume the voltage supply may be higher.
But for a 10 year old design, sounds very unlikely.

A few pictures of the O/P transformer:   http://www.g4jnt.com/PAtfmr1.jpg
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 with the PSU current limiting and dragging the
supply volts down to 17V.  Now, plugging these values into the matching
equation 2 * (17^2) / 5.556 = 100 Watts max possible,  - but I was seeing
more power.
 ........ Extra support to the possibility of it being 1:4 - BUT HOW ?

Andy
www.g4jnt.com

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<div>Has anyone got practical experience of the output matching transforme=
rs used on MOSFET PAs - I&#39;ve got a confusing one here?</div>
<div>=A0</div>
<div>I recently acquired some=A0big HF PA modules, each rated at over 1kW=
 out, and made up from 8 MOSFETS,=A0 RFPP53 types, roughly equivalent to=
 MRF140.=A0 It runs from what is more than likely a 50V rail.=A0 The modul=
es were part of an industrial RF heater running at 13.56MHz, but=A0the des=
ign=A0is wideband(ish) with the=A0normal ferrite matching transformers at=
 input and output.=A0 Which is where I may be missing something - they may=
 not be quite so normal...</div>

<div>=A0</div>
<div>The output transformer has a slightly different topology to designs=
 seen before - such as those given in the Motorola handbook.=A0=A0The seco=
ndary winding is made of insulated coax, two turns are full screened as th=
ey pass through the cores / tubing, but each turn=A0has the braid cut at=
 the hot end and joined to the two ends of the secondary,=A0with the third=
 turn consisting just the inner conductor=A0with no braid over it.=A0 =A0=
 All three turns (two of coax plus the single core) sit inside the usual=
 single turn primary made up from brass tube, surrounded by a pair of ferr=
ite cores with a connection at the far end.=A0=A0=A0 A diagram can be seen=
 at <a href=3D"http://www.g4jnt.com/pamatch.gif">http://www.g4jnt.com/pama=
tch.gif</a>=A0=A0 </div>

<div>=A0</div>
<div>Now, the bit that doesn&#39;t seem right...</div>
<div>the impedances don&#39;t work out properly...</div>
<div>=A0</div>
<div>Assuming it is designed to run into 50 ohms, a 1:3 transformer will=
 present a load of 5.56 ohms to the push pull devices.=A0=A0=A0 From a 50V=
 rail this should result in a maximum power output of 2*(50^2)/5.56 =3D 90=
0 Watts.=A0=A0=A0 (Sanity check, =A0a single ended design at half the Rloa=
d=A0=3D =A0 (50^2)/ 2 / 2.78 =3D 450 Watts each- normal push pull PA calcu=
lation).=A0=A0 Which is not 1kW and is only an absolute theoretical maximu=
m, anyway.</div>

<div>=A0</div>
<div>BUT,=A0=A0 if the transformer were 1:4 instead, , Rload would be 3.12=
5 ohms, Pout max would be 1600 watts which is exactly the sort of value I&=
#39;d expect to see on a real 1kW rated PA module.</div>
<div>=A0</div>
<div>Has anyone met that winding configuration=A0 before?=A0=A0 Is it real=
ly the 1:3 turns ratio it intuitively looks like, or is there some way the=
 windings could have have become an auto-transformer and=A0be giving =A01:=
4 turns ratio=A0?</div>

<div>If it really is 1:3 will have to assume the voltage supply may be hig=
her.=A0 But for a 10 year old design, sounds very unlikely.=A0</div>
<div>=A0</div>
<div>A few pictures of the O/P transformer:=A0=A0 <a href=3D"http://www.g4=
jnt.com/PAtfmr1.jpg">http://www.g4jnt.com/PAtfmr1.jpg</a>=A0=A0=A0 <a href=
=3D"http://www.g4jnt.com/PAtfmr2.jpg">http://www.g4jnt.com/PAtfmr2.jpg</a>=
=A0=A0 <a href=3D"http://www.g4jnt.com/PAtfmr3.jpg">http://www.g4jnt.com/P=
Atfmr3.jpg</a></div>

<div>=A0</div>
<div>On=A0a quick test on the module today, running from a=A0 10A supply,=
 it delivered nearly 150 Watts with the PSU current limiting and dragging=
 the supply volts down to 17V.=A0 Now, plugging these values into the matc=
hing equation 2 * (17^2) / 5.556 =3D 100 Watts max possible, =A0- but=A0I=
 was seeing more power.=A0=A0=A0</div>

<div>=A0........ Extra support to the possibility of it being 1:4 - BUT HO=
W ?</div>
<div><br clear=3D"all">Andy<br><a href=3D"http://www.g4jnt.com">www.g4jnt.=
com</a><br></div>

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