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SWR meter how its works?

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asrock70

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I found a SWR meter scheme and I do not understand it.

T1 is current transformer primar is coaxial with measured SWR, ses is 7 an 38 turns
C8/R5 is voltage divider connected over C9 to center of T1.

I do not understand how it affects T1 output and why windings are in proportion 1:5.5?

Thanks you

SWR.png
 

Thanks for information.
I understand it correctly
C8/R5 is voltage divider make for example prom 20V RMS on line 8.9V (for 15MHz) no C9
over D1 go on R2 negative voltage as forward wave size
over D2 go on R2 positive voltage as reflected wave size
Because sec of T1 have ratio 1:5.5 Voltage on R2 will be
negative if forward wave is 5.5 bigger than reflected wave
0V if forward wave = 5.5 x reflected wave
it is true?

If yes my idea is redesign this for AD8302
it means delete everything at the secondary and connect AD8302
AD8302 have max input -60 - 0dBm (1mW)
On measures line is for example 0-100W
my question is
what is the T1 suppression (CT) and how it counts?
 

Look on tjis **broken link removed**
respectively on **broken link removed**
 
Using AD8302, you don't need a bridge circuit, just connect voltage and current channel to the detector.
 
Thanks all for help ,
I have already understood the issue of SWR
I have a question about the problem of this problem, but for the sake of clarity I will create a new thread
 

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