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.

--001a114b159233bbc30550d63d69--

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 <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

readings/calculations:

(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

>

> On 29 May 2017 at 18:48, marcoc= adeddu@tin.it <marcocadeddu@t= in.it>

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: "marcocadeddu@tin.i= t"<marcocadeddu@tin=
.it >,

>> <rsgb_lf_group@= blacksheep.org>

>> Cc: <dead.fets@gmail.com= >

>> 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

>>

>>

>>

>> On 28 May 2017 at 19:34, ma= rcocadeddu@tin.it <marcocaded= du@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 an= d

>> > suggestion are welcome!

>> >

>> > 73, Marco IK1HSS

>> >

>> >

>> > -----Original message-----

>> >

>> > From: "marcocaded= du@tin.it" marcocadeddu@tin= .it

>> > 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 ----------

>> > From: "marcocaded= du@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 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.

>> >

>> >

>> >

>> >

>>

>>

>>

>>

>>

>

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

readings/calculations:

(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

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

> 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

>

> On 29 May 2017 at 18:48, marcoc= adeddu@tin.it <marcocadeddu@t= in.it>

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: "marcocadeddu@tin.i= t"<marcocad

>> <rsgb_lf_group@= blacksheep.org>

>> Cc: <dead.fets@gmail.com= >

>> 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

>>

>>

>>

>> On 28 May 2017 at 19:34, ma= rcocadeddu@tin.it <marcocaded= du@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 an= d

>> > suggestion are welcome!

>> >

>> > 73, Marco IK1HSS

>> >

>> >

>> > -----Original message-----

>> >

>> > From: "marcocaded= du@tin.it" marcocadeddu@tin= .it

>> > 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 ----------

>> > From: "marcocaded= du@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 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.

>> >

>> >

>> >

>> >

>>

>>

>>

>>

>>

>