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 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
(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
>
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: "
>> <rsgb_lf_group@= blacksheep.org>
>> 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
>>
>>
>>
>> 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...
>> > suggestion are welcome!
>> >
>> > 73, Marco IK1HSS
>> >
>> >
>> > -----Original message-----
>> >
>> > 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 ----------
>> > 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--