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In-Reply-To: <15c505653f7.marcocadeddu@tin.it>
References: <15c505653f7.marcocadeddu@tin.it>
From: Andy Talbot <andy.g4jnt@gmail.com>
Date: Sun, 28 May 2017 20:18:12 +0100
Message-ID: <CAA8k23SYBUuh0grja+f8nqDcCLFb+HOzoRu16T0tWGBdCp__Jw@mail.gmail.com>
To: "marcocadeddu@tin.it" <marcocadeddu@tin.it>, rsgb_lf_group@blacksheep.org
Cc: dead.fets@gmail.com
Subject: LF: Re: I: Fw: For today the FETs survived...
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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 <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" <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
> clean.
>
>
>
>

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<div dir=3D"ltr">First thing I noticed is that your turns ratio on the outp=
ut transformer doesn&#39;t look right. =C2=A0=C2=A0<div>You quote &quot;<i>=
<b> ... with primary winding of 15 turns and secondary of 12 turns...</b></=
i>&quot;<br></div><div><div><br></div><div>180V DC in a half bridge is 180V=
 peak-peak square wave. =C2=A0=C2=A0</div><div>The fundamental sine part of=
 that is =C2=A04/pi * 180 =3D 229V pk-pk =C2=A0=C2=A0</div><div>so is 229V =
/[2.SQRT(2)] =3D 81V RMS</div><div><br></div><div>To a good approximation R=
MS(fund) from a half bridge is Vrms(fund) =3D 0.45VDC</div><div><br></div><=
div>For 500 Watts out, Rload =3D =C2=A081 ^ 2 / 500 =3D =C2=A013 ohms</div>=
<div><br></div><div>So to match to 50 ohms you need a turns ratio of SQRT(5=
0/13) =3D 1.9:1 =C2=A0 =C2=A0 so call it 2:1 =C2=A0Keeping 12 turns on the =
=C2=A0secondary means you need 6 turns on the primary</div><div><br></div><=
div>When operating at reduced voltage, the power out will vary exactly as t=
he square of the voltage. =C2=A0=C2=A0</div><div>Recalculating from first p=
rinciples for a 12V supply:</div><div><br></div><div>12V =C2=A0DC =3D 12V p=
k-pk =3D 12 / [2.SQRT(2)] * 4/pi =3D 5.4V RMS (fundamental)</div><div>in 13=
 ohms should give 5.4^2/13 =3D 2.2 Watts</div><div><br></div><div>check usi=
ng ratio of voltages, squared :</div><div><br></div><div>(12V/180V) ^ 2 * 5=
00W =3D 2.2 Watts which is the same as above.</div><div>QED</div><div><br><=
/div><div>Your 15:12 ratio result sin a load impedance of (15/12)^2 * 50 =
=3D 78 ohms</div><div><br></div><div>At 40V DC =3D=3D 18V RMS(fund) that wi=
ll give 18^2/78 =3D 4.1 watts =C2=A0which is actually LESS that you are see=
ing - the 2* discrepancy is odd, but the low power is in the area of what y=
ou measured..</div><div><br></div><div>Andy =C2=A0G4JNT<br></div><div><br><=
/div><div>=C2=A0</div></div></div><div class=3D"gmail_extra"><br><div class=
=3D"gmail_quote">On 28 May 2017 at 19:34, <a href=3D"mailto:marcocadeddu@ti=
n.it">marcocadeddu@tin.it</a> <span dir=3D"ltr">&lt;<a href=3D"mailto:marco=
cadeddu@tin.it" target=3D"_blank">marcocadeddu@tin.it</a>&gt;</span> wrote:=
<br><blockquote class=3D"gmail_quote" style=3D"margin:0 0 0 .8ex;border-lef=
t:1px #ccc solid;padding-left:1ex">Hi Chris,<br>
<br>
I tried to post this message on the reflector but apparently I had no<br>
success..<br>
As promised I keep you updated but as you can read in the<br>
attachment the first trials were not enocouraging...<br>
Andy, may I ask you to read my report? your interpretation and<br>
suggestion are welcome!<br>
<br>
73, Marco IK1HSS<br>
<br>
<br>
-----Original message-----<br>
<br>
From: &quot;<a href=3D"mailto:marcocadeddu@tin.it">marcocadeddu@tin.it</a>&=
quot; <a href=3D"mailto:marcocadeddu@tin.it">marcocadeddu@tin.it</a><br>
Date: Sun, 28 May 2017 17:01:33 +0200<br>
To: <a href=3D"mailto:rsgb_lf_group@blacksheep.org">rsgb_lf_group@blackshee=
p.org</a><br>
Subject: For today the FETs survived...<br>
<br>
Hi LF,<br>
<br>
hope that also the toroids of Chris survived!<br>
My FETs survived, but they are not working as expected :-(<br>
Attached the report on my attempt to duplicate the half bridge of<br>
Andy..<br>
Has anyone suggestions before I try to cook all connecting to the<br>
180Vdc supply?<br>
<br>
Thank you<br>
73 Marco IK1HSS<br>
<br>
<br>
--<br>
This message has been scanned by E.F.A. Project and is believed to be<br>
clean.<br>
<br>
<br>
<br>
<br>
<br>
<br><br>---------- Forwarded message ----------<br>From:=C2=A0&quot;<a href=
=3D"mailto:marcocadeddu@tin.it">marcocadeddu@tin.it</a>&quot; &lt;<a href=
=3D"mailto:marcocadeddu@tin.it">marcocadeddu@tin.it</a>&gt;<br>To:=C2=A0&lt=
;<a href=3D"mailto:rsgb_lf_group@blacksheep.org">rsgb_lf_group@blacksheep.o=
rg</a>&gt;<br>Cc:=C2=A0<br>Bcc:=C2=A0<br>Date:=C2=A0Sun, 28 May 2017 17:01:=
33 +0200 (CEST)<br>Subject:=C2=A0For today the FETs survived...<br>Hi LF,<b=
r>
<br>
hope that also the toroids of Chris survived!<br>
My FETs survived, but they are not working as expected :-(<br>
Attached the report on my attempt to duplicate the half bridge of<br>
Andy..<br>
Has anyone suggestions before I try to cook all connecting to the<br>
180Vdc supply?<br>
<br>
Thank you<br>
73 Marco IK1HSS<br>
<br>
<br>
--<br>
This message has been scanned by E.F.A. Project and is believed to be clean=
.<br>
<br>
<br>
<br></blockquote></div><br></div>

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