Re: LF: TXing 2200m WSPR

[Andy Talbot <andy.g4jnt@gmail.com>]

Yes

Try anything and try to work out what is happening.  But first of all, he needs to know what power being deliverfed into a 50 ohmsload is seemingly higher than it 'can' be.    The Vdd^2/2/Rl cannot be violated.

If it appears to be so, then Rl or Vdd or the measurement  has to be wrong.

I once had to go though a similar try-everything-and-believe-nothing-until-tested process to work out why my 700Watt 137kHz Tx appeared to be more than 100% efficient!

It turned out my Bird 30dB 1kW attenuator only had 29.4dB attenuation at DC.  And for a 1GHz jobbie, 137kHz is DC.

On 24 December 2017 at 23:03, Chris Wilson <dead.fets@gmail.com> wrote:

Hello Andy,

Is  there  any point in Paul putting his dummy load on the end of his
75  odd meters of feeder co-ax outside? And then on the secondary of
his  matching  transformer when tapped one to one? In case the co-ax is
iffy or the impedance matching transfomer's doing something odd?

Sunday, December 24, 2017, 10:42:16 PM, you wrote:

> Well ...    Using exactly the values in the filter circuit diagram,
> 50R transforms through the filter to 48.5 - j2.86 (Ret Loss = 30dB, 
> VSWR = 1.07)   [Using GM3SEK's original Netcalc prog.]


> So that's pretty conclusive the ideal filter values will not be
> upsetting things at the fundamental frequency.    
> According to Google, the  T106-2 has a stated Al value of  13.5nH
> /turn^2  so 72 turns does indeed give 70uH.   So IF your core is correct, the filter should be OK.


> It's a bit difficult from now on, at a distance, to try to work out what is happening.
> Anyone else, any suggestions ?


> BTW ...


> Peak to peak of a { symetrical }square wave needs to be multiplied
> by 4/pi to get the peak-to-peak of the fundamental component.   So
> the amploitude you see will be lower by about 1.3 times for teh same fundamental power component. 


> Andy  G4JNT

--
Best regards,
 Chris                            mailto:dead.fets@gmail.com