Understanding the Basics of VNA Measurements with an SMA Connector

[Thomas Rigby]

TL;DR Summary

Conceptual question for intuitive understanding

I have a VNA with an SMA connector for measuring reflection. If I connect only the center conductor of the SMA to a piece of metal, I get a spectrum. My question is, what should I be doing with the outer connector?

I am trying to learn EE by reading and doing. I don't have a lot of intuition yet, and I am asking this question in the context of very basic physics in terms of currents, voltages, conductors, etc. I don't know how to "google" this question, so I am asking humans to help me understand.

 

[Baluncore]

The VNA is designed to analyse a network at the other end of a coaxial cable.
You have no cable, and no matching network for the wire, which is like a whip antenna.

The outer part of the SMA connector is supposed to be ground. The analysis signal is transmitted and received on the centre electrode, referenced to the outer electrode.

 

[Thomas Rigby]

Thank you, I believe I understand. Since the outer connector is just a reference point, I don't need to connect it to anything, unless I have some reason for preferring a different reference.

 

[sophiecentaur]

Thank you, I believe I understand. Since the outer connector is just a reference point, I don't need to connect it to anything, unless I have some reason for preferring a different reference.

This is too simple, I'm afraid. If you consider the current flowing 'out' of the centre pin of the connector, equal and opposite current will be flowing 'in' to the screen / outer conductor. Its path can be back down the outer of the coax, out along an Earth plane (if that's the way it's mounted) or a path through anything else the connector is mounted to.

The DC 'reference point' you refer to is not very relevant, imo but the mean potential will depend on the ratio of the impedances of the two legs of the circuit (wire and undefined base connection).

We have slid seamlessly into how antennae are fed and the impedances they present to the feeder. How far didi you want to go?

 

[Thomas Rigby]

I connect this to my NanoVNA. The reason I sent this image is to show you that there is no connection between the inner conductor and the outer conductor. My mental picture is this: a signal is transmitted out along the inner conductor, hits whatever is at the other end (in this image, nothing) and the NanoVNA looks at the reflection. No different than jiggling the end of a string, and looking at what comes back from whatever the other end of the string is attached to (possibly attached to nothing).

If my mental picture is an oversimplification, then I need to understand what I am missing.

 

[sophiecentaur]

My mental picture is this: a signal is transmitted out along the inner conductor,

A popular misconception. Inside a coax cable, there is equal and opposite current on the inner and back down the (inside of the) outer. At the open end, that outer current has to go somewhere. That may go down on the outside of the screen or anywhere else it may be connected to. Think of the Kirchhoff 1 Law. The sum of the currents is zero. So you can't ignore the coax screen - you have to define how it's connected before you can predict what's happening. The signal 'on the wire' is affected by the waves elsewhere.

 

[Thomas Rigby]

I don't attach a coax cable to the little wire. I attach a solid 14 AWG wire to it. There is no co-axial wire.

 

[berkeman]

I connect this to my NanoVNA. The reason I sent this image is to show you that there is no connection between the inner conductor and the outer conductor. My mental picture is this: a signal is transmitted out along the inner conductor, hits whatever is at the other end (in this image, nothing) and the NanoVNA looks at the reflection. No different than jiggling the end of a string, and looking at what comes back from whatever the other end of the string is attached to (possibly attached to nothing).

If my mental picture is an oversimplification, then I need to understand what I am missing.

I don't attach a coax cable to the little wire. I attach a solid 14 AWG wire to it. There is no co-axial wire.

Sorry to be blunt, but why are you experimenting with a Vector Network Analyzer if you don't understand the basics of transmission lines? IMO, you are wasting your time right now.

Instead, please learn the basics of transmission lines and then about the transfer functions of 2-port networks. Then learn the basics of S-parameter measurements, and then finally learn more (now with an adequate technical background) about how to use a VNA...

https://en.wikipedia.org/wiki/Transmission_line

https://en.wikipedia.org/wiki/Two-port_network

https://www.rfpage.com/s-parameters-in-rf-testing-simple-explanation/

https://en.wikipedia.org/wiki/Network_analyzer_(electrical)#VNA

 

[DaveE]

OK, key point here in all electronics, which you seem to have missed. Voltages are ALWAYS measured between two points. Currents always have a return path. Granted for things like radio antennas it seems like it's just a wire, but, trust me, the guy that designed that knew and cared a lot about the ground side of that circuit. Maybe the earth, maybe the chassis of a car. There is ALWAYS something we can call ground.

The people that designed your VNA decided that it would be a lot easier for you to understand the instrument and to set up a good test if they gave you that ground in a convenient, standard, predictable way. That is the outer conductor of your connector, in this case. You'll want to learn about this and use it to create a test set up that you understand. Because the ground or return path IS THERE whether you know it or not. It is essential to what you are actually measuring. It's easy to collect garbage and think it's data. I've done a lot of that. Good test setups either require that you use a standard configuration that others have verified or that you put nearly as much work (often more) into verifying your test so that you're sure the data you get represents what you intended.

 

[DaveE]

OK, sorry in advance. I know this is harsh but...

Elon Musk could buy a da Vinci surgical robot, but he couldn't do anything useful with it without a lot of studying. The robot isn't enough, you also want a surgeon.

While they are cheap on eBay now, VNAs are intended to address some pretty complex EE issues. They are really easy to use incorrectly. So, it can be a great toy, and I would encourage you to play with it. However, if you really want to understand it you'll need to understand it. As @berkeman said, time to study up a bit. This isn't an entry level instrument.

 

[Thomas Rigby]

Sorry to be blunt, but why are you experimenting with a Vector Network Analyzer if you don't understand the basics of transmission lines? IMO, you are wasting your time right now.

Instead, please learn the basics of transmission lines and then about the transfer functions of 2-port networks. Then learn the basics of S-parameter measurements, and then finally learn more (now with an adequate technical background) about how to use a VNA...

https://en.wikipedia.org/wiki/Transmission_line

https://en.wikipedia.org/wiki/Two-port_network

https://www.rfpage.com/s-parameters-in-rf-testing-simple-explanation/

https://en.wikipedia.org/wiki/Network_analyzer_(electrical)#VNA

Transmission lines require two conductors. I am just using one conductor. But thanks for participating.

 

[Thomas Rigby]

OK, key point here in all electronics, which you seem to have missed. Voltages are ALWAYS measured between two points. Currents always have a return path. Granted for things like radio antennas it seems like it's just a wire, but, trust me, the guy that designed that knew and cared a lot about the ground side of that circuit. Maybe the earth, maybe the chassis of a car. There is ALWAYS something we can call ground.

The people that designed your VNA decided that it would be a lot easier for you to understand the instrument and to set up a good test if they gave you that ground in a convenient, standard, predictable way. That is the outer conductor of your connector, in this case. You'll want to learn about this and use it to create a test set up that you understand. Because the ground or return path IS THERE whether you know it or not. It is essential to what you are actually measuring. It's easy to collect garbage and think it's data. I've done a lot of that. Good test setups either require that you use a standard configuration that others have verified or that you put nearly as much work (often more) into verifying your test so that you're sure the data you get represents what you intended.

I don't understand what you mean when you say "currents always have a return path". Suppose I have a two conducting spheres; one charged and one uncharged. If I then connect a single wire between them, a current will flow until the equilibrium is reached. There doesn't appear to be any "return path" here.

 

[DaveE]

I don't understand what you mean when you say "currents always have a return path". Suppose I have a two conducting spheres; one charged and one uncharged. If I then connect a single wire between them, a current will flow until the equilibrium is reached. There doesn't appear to be any "return path" here.

That return path would be through the capacitor created when you put the two spheres in place. You could search for "Displacement Current" for the advanced level description.

 

[berkeman]

Transmission lines require two conductors. I am just using one conductor. But thanks for participating.

That's nonsense. Last chance -- please read the reference links that I listed before posting further in this thread. You are wasting your time and ours. PM me when you are ready for this thread to be reopened. (click on my avatar and Start a Conversation to send me a PM)