Can a microwave VSWR meter be used to measure Power?

[wirefree]

G'day.

Recently we conducted an experiment in the microwave engineering lab where we were tasked to measure power in, what's called, a Directional Coupler. Here you'll find much on it.

My concern with the setup is that we took our measurements on a VSWR meter by noting off the db scale. So, for example, we match-terminated the coupled and receiver port, balanced the bench to get VSWR equal to one, which is zero db on the db scale, and then used that to obtain power on port one:

0 = -20 log P₁ (mW)

As another instance, we found the VSWR for port two by terminating all other ports, then found power on port two as follows:

if the db scale read VSWR as 4.6, then power in mW on port two:
4.6 = -20 log P₂ (mW)
=> P₂ = .588 mWI am perplexed how this could be.
In need of your guidance,

with warm regards,
wirefree

 

[tech99]

Actually I am confused about your port numbering - can you clarify exactly please.
You seem to be measuring reflected power using a VSWR reading. To do this, you must first convert VSWR to Reflection Coefficient. Ref Coeff = (V-1)/(V+1). Then if you take 20 log Ref Coeff that will give you the Reflection loss in dB, called the Return Loss. If you then know the sending power then you can find the reflected power in mW or in dBm.

 

[wirefree]

Much relieved by your response.
I study in a small university in the Himalayas, and although I didn't expect Ivy League education, I suspected something was off.

Actually I am confused about your port numbering - can you clarify exactly please.

My regrets - I assumed conventional use of numbering for ports of a Directional Coupler:

Your suggestion of the recommended procedure of proceeding with Reflection Coefficient is very helpful. I couldn't reconcile how our laboratory teacher was equating VSWR with, what she called, S-parameters as a way of obtaining power. She seemed to be suggesting that the db scale on the VSWR meter are S-parameters, which, by definition, are the logarithmic ratio of reflected & incident power, and, hence, may be employed to obtain power.

I see from your response that her suggestion is incorrect.

In the following post I will attach my entire lab notebook for your reference and hope I will stand corrected with your guidance.wirefree

 

[wirefree]

Lab notes :

Page 1/3: https://ibb.co/gHoOTL

Page 2/3: https://ibb.co/d3ca8L

Page 3/3: https://ibb.co/fTOtv0

 

[DaveE]

The Decibel scale is normally defined as:
dB = 10*Log(P) where P = Power in Watts.
dBm = 10*Log(P) where P = Power in milliWatts.
The factor of 20 is used when people use the dB scale to express voltages (or currents), because the power associated with it is the voltage squared. e.g. 10*Log(V^2) = 20*Log(V)
I don't think it really matters in any fundamental sense, it's just the common convention. So watch out for errors by a factor of 2.

 

[tech99]

Thank you for sending the lab notes. I cannot read the writing very well. I think the "VSWR Meter" is just a voltmeter. Maybe it has a power or decibel calibration?
By the way, I am very interested in your study location.

 

[davenn]

Thank you for sending the lab notes. I cannot read the writing very well.

you do know you can click on the image in the link and it is magnified and is very easy to read ?

 

[wirefree]

By the way, I am very interested in your study location.

Appreciate your interest. Thanks very much!

It is quite lovely here in the foothills of the outer ranges of the Himalayas where I am now settled. I studied abroad mostly - America & Australia - but have decided to settle here in a small town, well, a lil' city now sadly, where we enjoy sub-tropical weather.

People here bring down with them from the hills great Alpine folk cultures, songs and music. The flute characterise their sweet outlook towards life. And Goddess worship is a great tradition in these hills.

For India, with its vast & age-old culture, 330 millions Gods, and much metaphysics, it is testament to the practical outlook of everyone in this country that the scientific temperament has taken hold. But it doesn't help if my teacher is reading S-parameters on a VSWR meter.

I think the "VSWR Meter" is just a voltmeter. Maybe it has a power or decibel calibration.

We've been using this equipment for determining VSWR and, thereafter, setting the lab bench to VSWR=1 before starting our experiments.

YouTube just taught me that a Vector Network Analyzer (VNA) helps measure S-parameters. But our equipment is certainly not one. Here's what it's scale looks like:

https://ibb.co/b0MVYLHowever, slide 2 of this Berkeley presentation points towards what you tried to explain in your initial post - that SWR may indeed yield a way forward.

http://rfic.eecs.berkeley.edu/~niknejad/ee242/pdf/eecs242_lect5_sparam.pdfJust a gentle reminder to end: my question is really about arriving at power on a given port at microwave frequencies.

With warm regards,
wirefree

 

[davenn]

Just a gentle reminder to end: my question is really about arriving at power on a given port at microwave frequencies.

I will keep this as practical as possible ( non-maths ... I am bad is complex math )

OK, so below, I have designated the standard port numbers on the actual unit (upper part of image) and you can
see in the lower part how that corresponds schematically.
Now, NOTE, it doesn't matter if it is waveguide or coaxial, it's the same process

( Assuming an ideal coupler and terminations) ...

Firstly ... in the lower section above NOTE the arrow directions of the main line and the coupled line ... They are reversed

So, let's say we input 10mW (10dBm) of RF into port 1 with output port 2 and coupled port 4 terminated.
Port 3 will then directly show a RF level 10 dB ( it's a 10dB coupling) down on the on the input ... so you will read 1mW at Port 3
using a RF power meter. In this configuration, you are measuring forward power. The RF level on port 3 is directly proportional to
the RF level on port 1.
Increasing the input power on port 1 to 100mW will result in the port 3 RF level increasing to 10mW

( knowing this ... you could calibrate/scale your meter dial so that the scale shows the actual port 1 input power

(Ok so far ? ... you may already understand this ... I am just making sure)

Now if you remove that termination from port 4 and place it on port 3 and put your power meter on port 4, your reading will be 0mW.
In this configuration, you are measuring reflected power.
This is because the termination ( load) on port 2 is absorbing ALL forward power and there is no reflected power
getting coupled to port 4.
This is your ideal VSWR of 1:1

OK so far ? Now let's look at what happens if you change the port 2 termination/load impedance. Because there is now an imperfect match, not all
RF from the port 1 input is going to be coupled into the load on port 2 and some of it is going to be reflected back along the line towards
port 1. The result being that a portion of that is going to be coupled into the "coupled line" and will appear at port 4 with a
level 10 dB down on the amount of reflected power in the "main line"

this is going to result in your meter displaying a VSWR that isn't 1:1 and that is going to directly relate to the amount of reflected power.
For a given mismatch/VSWR, increasing the input power on port 1 is going to proportionately increase the level of reflected RF
power from the P2 load and therefore the power level detected at port 4

OK ... now the orientation of your directional coupler as shown in your notes image...

shows that it is set for measuring VSWR ( return loss).

P1 input, P2 terminated output, P3 terminated couple, P4 measurement port.

And because no real world system is ideal, there will be some reflected power and that can be used to measure the input power
using the reflection coefficient that tech99 referred to in post #2Hopefully that was of some help as a visualisation of the process Dave

 

[Tom.G]

I have a problem with slide 3 on the link that @wirefree posted.

http://rfic.eecs.berkeley.edu/~niknejad/ee242/pdf/eecs242_lect5_sparam.pdf

The second equation on that slide is:

Pavs = Vs²/8Z_o
Pavs = Power available from source
Vs = Source Voltage
Z_o = Real Resistance of source

If you take for example Vs=10V and Z_o=10Ω, you get 100/80 = 1.25W; whereas conventionally W = E²/R = 100/10 = 10W.

What am I missing?

EDIT: Oh never mind! After sleeping on it I realized the key phrase was "Power available" from the source. The factor of 8 is 2³, the square of half the numerator voltage times half the current.

 

[wirefree]

Greatest regards to @tech99 for the succinct answers. It often helps to hear it straight on.

Many thanks also to @davenn - your descriptive explanation has merits too obvious to state.For the benefit of those who will stumble upon this post in the future, I conclude this thread by listing key points from all the above helpful posts as well as my readings on the topic since starting this thread:

1. VSWR is the ratio of the amplitude of antinode to node of a standing voltage microwave in a waveguide.
2. A Power Meter, not VSWR Meter, is best suited for measuring power of a given signal at microwave frequencies.
3. If a VSWR Meter is to be employed for the purposes of measuring power, one must proceed by first determining the Reflection Coefficient and proceeding as suggested in @tech99’s post above.
4. A VSWR Meter does not measure s-parameters; a Vector Network Analyzer does.

@tech99 and @davenn : Please do correct me where I am incorrect or incomplete.

My gratitude to all those who stopped by to read this post and, of course, to those who contributed.