Simple 50R I/O RF attenuator (1-30MHz)

[neazoi]

Hi I would like to build a simple RF attenuator for 1-30MHz that will be able to maintain 50R in it's input and output ports.
I know I can use a stepped attenuator, but I want to use a simple potentiometer for the purpose.

I am thinking of having a 50R 10db 50R fixed attenuator pad and then a simple potentiometer as a potential divider. Then another 10db 50R fixed attenuator connected to the potentiometer wiper and to the output port.
There will be impedance mismatch at the points where the potentiometer will be connected to the two attenuator pads, but should I care, since this is a variable attenuator after all?

I do not care about the 20db fixed loss in this attenuator, what I care is the I/O ports of the attenuator to be 50R when I set the potentiometer wiper at all settings.

Can it be done that way?

 

[FvM]

I believe it's not possible to keep the impedance matching with a single variable resistor or potentiometer.

 

[volker@muehlhaus]

I agree with FvM. For these small attenuation values, there is a lot of variation in shunt resistance required, to keep the 50 Ohm input/output impedance.

If you look at voltage variable attenuators, they adjust both series and shunt path resistance.
https://www.minicircuits.com/pdfs/ZX73-2500.pdf
https://www.electronics-notes.com/a...de-variable-voltage-controlled-attenuator.php

 

[BigBoss]

This company manufactures some interesting components which you may be interested in..
http://en.yantel-corp.com/pro_show_id_19.html

 

[vfone]

Your idea it will work somehow, and the explanation is this:
A 10dB attenuator will have the input return loss (S11) better than -20dB whatever the output terminated impedance is (anywhere between infinite ohms and zero ohms).

For example, a single series variable resistor placed between two 10dB attenuators (50Ω in/out) will give you the overall attenuation as in the list below (first number is the value of the series resistor and second is the overall attenuation in dB):
- 200Ω > 29.5dB
- 300Ω > 32dB
- 400Ω > 34dB
- 500Ω > 35.5dB
- 600Ω > 36.9dB
- 700Ω > 38dB
- 800Ω > 39dB
- 900Ω > 40dB
- 1000Ω > 40.8dB

So, for 800Ω series resistance variation you get 11.3dB attenuation variation, while keeping the input/output return loss of the group (S11 and S22) better than -20dB, which is a reasonable value.

 

[neazoi]

I agree with FvM. For these small attenuation values, there is a lot of variation in shunt resistance required, to keep the 50 Ohm input/output impedance.[/url]

Hm... that makes me think. How about this one?
It uses a stereo potentiometer of 50R. The tracks are part of the input and output Pi attenuators and their resistance is constant.
The input/output fixed pads will be happy as far as concern the impedances, it is just that less signal is "coupled" from one pad to the other by the pots.
In the extreme case where the wiper is on top, it will be like cascading two 50R pads (node impedance 25R). In the extreme case where the wiper is at the ground end, it will be like two disconnected pads.

Will that work?

- - - Updated - - -

Your idea it will work somehow

Will the circuit on post #6 work better?

- - - Updated - - -

With my previous circuit (right side image) I worry a bit about the series resistance that couples the two fixed pads. This resistance is formed by the top track side of the first potentiometer to it's wiper, which is in series with the resistance from the wiper of the second potentiometer to the top of it's track. This resistance will not be constant through out the different potentiometer settings.

So I have revised the circuit like this (left side image). The two tracks of the stereo potentiometer are connected in different polarity (noted by the letters a and b). So as the wiper is moved, the series resistance is increased in one of the potentiometers and decreased by the same amount in the other. So the series resistance stays always at 50R, constant as the wiper is moved!
Apart from the extreme cases (wiper to GND or wiper to top) it will effectively be like a third fixed pad between the two outside pads. However, because of the potential divided created by the first track, the coupled signal to the right pad will be less. hence attenuation without impedance changing.

How do you judge my attenuator?

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I think both circuits the left and the right will work ok. the left will form a T pad at the middle point, whereas the right will form a Pi pad.

It is important to know your thoughts on these circuits, as this is something I have not seen in the literature.

 

[FvM]

You should be able to calculate the impedance mismatch involved with the suggested circuits. You didn't yet give a specification of acceptable mismatch of your attenuator design.

Another point to consider. Depending on the design, a 50 ohm potentiometer may expose considerable series inductance and not work as resistive attenuator, even over the small 30 MHz range.

Stepwise variable attenuator with GHz bandwidth are available on the market, unfortunately expensive.

E.g. https://www.apitech.com/globalasset...attenuators/continuously-variable/wmod940.pdf

 

[volker@muehlhaus]

Your idea it will work somehow, and the explanation is this:
A 10dB attenuator will have the input return loss (S11) better than -20dB whatever the output terminated impedance is (anywhere between infinite ohms and zero ohms).

Yes, sorry, I misunderstood the initial question and didn't realize that it is a series resistor in between two fixed 10dB attenuators. I understood it's about implementation of those 10dB attentuators. Sorry for the confusion!

 

[neazoi]

You should be able to calculate the impedance mismatch involved with the suggested circuits. You didn't yet give a specification of acceptable mismatch of your attenuator design.

Another point to consider. Depending on the design, a 50 ohm potentiometer may expose considerable series inductance and not work as resistive attenuator, even over the small 30 MHz range.[/url]

I am not much into calculations Is there a tool I can use?

I do not know what will the acceptable impedance mismatch be. This attenuator will be connected after a DDS with 50R shunt resistor at it's output (that of the dds). The other end of the attenuator will be connected to a spectrum analyzer which has an internal 50R shunt to ground. So what will the acceptable impedance mismatch be?

I've been warned about the inductive component of the potentiometers before. I mostly use miniature types like these plastic tiny Panasonic ones and never experienced weird behaviour. I have used them in all kinds of RF experiments on up to 30MHz.

In my attenuator I am also not sure if I must include additional 50R shunt resistors parallel to the potentiometer tracks (i.e to separate the tracks completely from the fixed pads and present a 25R node impedance). Any ideas?

 

[vfone]

volker, no problem. I was sure that was a misunderstanding.

neazoi, your circuit in post #6 will not do the job, because the overall input and output return loss of the group is not very good.
You have to start from using two 10dB (or higher) pad attenuators (50 ohms input/output) and place between them whatever resistive network you want, but don't modify any of the resistor values part of the pad attenuator structures.

 

[neazoi]

volker, no problem. I was sure that was a misunderstanding.

neazoi, your circuit in post #6 will not do the job, because the overall input and output return loss of the group is not very good.
You have to start from using two 10dB (or higher) pad attenuators (50 ohms input/output) and place between them whatever resistive network you want, but don't modify any of the resistor values part of the pad attenuator structures.

You refer to the right hand side circuit right?

I am not sure if I must include additional 50R shunt resistors parallel to the potentiometer tracks (i.e to separate the tracks completely from the fixed pads and present a 25R node impedance between them, like they where three separate pads cascaded). Or if it is ok as it is, with the track of the potentiometers shared between the fixed pads and the central variable pad?

 

[vfone]

I didn't say that you have to include 50 ohms shunt resistors.
The attenuators should have 50 ohms input/output impedance (Zo).
Here you can find few options how to design these attenuators, and after that use any resistive network you want between them.
https://www.microwaves101.com/calculators/858-attenuator-calculator

 

[neazoi]

and after that use any resistive network you want between them.

Even a single potential divider?
Based on your replies, I think if the stereo potentiometer approach is of any better compared to the single potential divider on post #1?

 

[betwixt]

Just a thought - it is possible to use pre-set potentiometers with plastic center 'screws', either a slot or hex hole? You won't get perfect matching but you might get something close enough if you link three together on a single shaft and wire them so the outer resistances drop as the inner one increases. Sort of like a pi network with all three legs adjustable together. The middle pot can be a different value to the outer ones but all should be linear tracks.

Brian.

 

[neazoi]

Just a thought - it is possible to use pre-set potentiometers with plastic center 'screws', either a slot or hex hole? You won't get perfect matching but you might get something close enough if you link three together on a single shaft and wire them so the outer resistances drop as the inner one increases. Sort of like a pi network with all three legs adjustable together. The middle pot can be a different value to the outer ones but all should be linear tracks.

Brian.

I think that will complicate things. i would use a switched stepped attenuator better then.
I wanted a quick way of varying a marker on a spectrum analyzer screen without (attenuation of a DDS generator) so I can measure the level of the marker later on.
The measurement equipment has a 50R shunt on it's input, like the analyzer does, so the marker level won't change on the measurement equipment, compared to what is displayed on the analyzer screen.
Attenuation steps will make the min step at 1db, which is not good and complex.