RF attenuator formula. I am confused ...

[JacquesBBB]

Dear All,

I do not understand the RF attenuator formula that I find in many places  on the Web.
As for example in http://www.qsl.net/4f5aww/module5p.htm

If I compute the resulting voltage divider  ratio, I find

R1/(R1+R2) = 2k/(k^2+1) ~ 2/k

and not 1/k  as one would expect for a k factor attenuation.

I am sure I am missing something, but what ?

Thanks

[w2aew]

What you're missing is the fact that both ends of this attenuator are terminated in 50 ohms when used in RF applications.

[JacquesBBB]

Thanks, w2aew, for the comment.  This was missing in the web sites I looked to.
It is also missing on your video, which is nevertheless very interesting.

So I have added a Z  impedance in parallel to the R1 resistors.
Now the  ratio of the voltages is  (using the formula given above)

Vo/Vi = R1*Z/(R1*R2 +(R1+R2)*Z) = 1/k

So exactly what is needed.

This means that the impedance of  the attenuator needs to be adjusted to the impedance  of the
source and of the receiver.

Thanks again.

[w2aew]

Thanks, w2aew, for the comment.  This was missing in the web sites I looked to.
It is also missing on your video, which is nevertheless very interesting.

So I have added a Z  impedance in parallel to the R1 resistors.
Now the  ratio of the voltages is  (using the formula given above)

Vo/Vi = R1*Z/(R1*R2 +(R1+R2)*Z) = 1/k

So exactly what is needed.

This means that the impedance of  the attenuator needs to be adjusted to the impedance  of the
source and of the receiver.

Thanks again.

Yes, this is one of those *given* or *assumed* things in RF applications.

[LukeW]

Yep, it is assumed that the attenuator you're designing is going into a system with some impedance given, say 50 ohms. And if you build that attenuator and then move it to a 75 ohm system it will not work as intended.