Return to KLUBNL.PL main page

rsgb_lf_group
[Top] [All Lists]

R: LF: Re: I: Fw: For today the FETs survived...

To: <[email protected]>
Subject: R: LF: Re: I: Fw: For today the FETs survived...
From: "[email protected]" <[email protected]>
Date: Mon, 29 May 2017 19:48:20 +0200 (CEST)
Cc: <[email protected]>
Reply-to: [email protected]
Sender: [email protected]
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  the PA can give higher satisfaction :-)

Hopefully the FETs will survive and this time I'm ready to burnout the 
antenna hi

Will keep you both updated, thank you once more Andy

73 Marco, IK1HSS
----Messaggio originale----
Da: [email protected]
Data: 28-mag-2017 21.18
A: "[email protected]"<[email protected]>, 
<[email protected]>
Cc: <[email protected]>
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 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, [email protected] <[email protected]> 
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: "[email protected]" [email protected]
> Date: Sun, 28 May 2017 17:01:33 +0200
> To: [email protected]
> 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: "[email protected]" <[email protected]>
> To: <[email protected]>
> 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.
>
>
>
>




<Prev in Thread] Current Thread [Next in Thread>