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From: N1BUG <paul@n1bug.com>
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Date: Sun, 24 Dec 2017 15:12:06 -0500
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Subject: Re: LF: TXing 2200m WSPR
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Hi Andy,

This is very interesting, because 12 watts is what I think I have 
consistently been able to generate into the antenna before problems 
creep in.

I can easily accept that this amplifier may only be capable of 12 or 
13 watts.

What I don't understand is why I think I am seeing a clean 25 watts 
into a pure resistive 50 ohm load. Can someone please check my math?

I'm running the output of the amplifier through a low pass filter, 
then a scopematch, then into a high quality 50 ohm load.

My scopematch circuit is here:

http://n1bug.com/n1debug/LF_ScopeMatch-20171224.jpg

Note that it is configured such that on the current sense output, 
1V=1A and for voltage, 1V=50V.

Running into a pure 50 ohm resistive load I am seeing exactly 4 
divisions peak to peak on the scope (2 divisions above center, 2 below).

500 mV/div * 4 divs = 2.0V peak-peak or 0.707V RMS.

0.707 * 50 (1V=50V on the scopematch) = 35.4V RMS.

35.4^2 / 50 ohms = 25 watts.

Where am I going wrong?

As a check on scopematch calibration I set my HP 3325B to 10V 
peak-peak, ran it through the scopematch into the same 50 ohm dummy 
load at 137.5 kHz. I set the scope to 50 mV/div and it read exactly 
4 divisions peak-peak = .2V * 50 = 10V which seems to verify. 
Admittedly this verification is at a much lower power level.

73,
Paul N1BUG


On 12/24/2017 01:06 PM, Andy Talbot wrote:
> The other thing about that circuit that is wrong is the power out 
> specification for the turns ratio / loading resistance.
> 
> For a single ended PA, Pout max = Vdd ^2 / 2 / RL.   With a 9:1 (3:1 
> turns) transformer RL = 50 / 9 = 5.6 ohms
> So for a 13.8V suply (ignoring Vsaturation) that means  an absolute 
> maximum power output possible of 13.8^  / 2 / 5.6 = 17 Watts into a 
> true 50 ohms load on the transformer output.
> 
> In practice the device will saturate with a volt or so across it, 
> lowering the voltage swing and keeping power out to a lower value.  
>   If the peak swing is 12V, Pmax now drops to 12^2/2/5.6 = 13 watts
> 
>   Riccardo, I assume you mean 1:4 turns ratio,  or 1:16 impedance 
> transformation.   That Means RL = 50/16 = 3.1 ohms and at 12V swing 
> gives 23 watts out as you state
> 
> I made a push-pull PA using a similar  type of FET, but used a 28V 
> supply and 1+1:2 transformer.  It gives 40 - 50 Watts over the 
> frequency range  100kHz to 2MHz
> 
> http://www.g4jnt.com/Linear_LF_PA3.pdf
> 
> Andy  G4JNT
> 
> On 24 December 2017 at 17:33, Riccardo Zoli 
> <riccardozoli80@gmail.com <mailto:riccardozoli80@gmail.com>> wrote:
> 
> 
>     Hi Paul, Stefan
> 
>     my first PA for MF and LF, based on a IRF520, was similar.
>     However, I used a most appropriate 1:4 output transformer
>     instead a 1:9. With 13.8V, a 50mA bias current is more than
>     adequate for AB class. You could obtain about 20-22W max on output.
>     Obiviously, more supply voltage it means more power output.
> 
>     Merry XMAS & HNY!
> 
>     All the best
> 
> 
>     73 de Riccardo IW4DXW
> 
> 
>     Il 24/Dic/2017 13:30, "DK7FC" <selberdenken@posteo.de
>     <mailto:selberdenken@posteo.de>> ha scritto:
> 
>         Hi Paul,
> 
>         Am 23.12.2017 23 <tel:23.12.2017%2023>:36, schrieb N1BUG:
> 
>             For anyone trying to follow along, this is my current
>             version:
> 
>             http://n1bug.com/ASB_LF_Amp-20171223.jpg
>             <http://n1bug.com/ASB_LF_Amp-20171223.jpg>
> 
> 
>         The circuit looks good to me now. Maybe it helps to use a
>         low pass filter in T configuration instead of a PI config.
>         The T config starts with a series L. The PI config starts
>         with a parallel C. That could make a significant difference
>         for the harmonics.
> 
>         73 and Merry christmas, Stefan
>