Author Topic: Question about NanoVNA and Smith charts  (Read 631 times)

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Offline mio83

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Question about NanoVNA and Smith charts
« on: May 22, 2020, 08:22:26 pm »
Hi!

I just obtained a nanoVNA and I am very happy about it.
Until very recently, I didn't know what a VNA was, and so the nano is a very good starting point to learn about this stuff.

I have tried some measurements (S11) and it's very accurate:
- I checked the return loss (called LogMag in the nano) with a Spectrum Analyzer with TG and a directional coupler and it's spot on.
- I checked the phase of the reflection with an oscilloscope and it's also very accurate.

Very good value for 50euros :-DD :-+

I have some doubts, however, in the Smith chart trace of the nano.
For a concrete example, I have the following data point:

Code: [Select]
Frequency: 7542500 Hz
LogMag: -18.59 db
Phase: 143.61
Polar Diagram: -0.09 + 0.07j
Smith: 41.3 Ohm and 123nH

Starting from the top we have:

20*Log_10(|S11|) = -18.59

and this gives |S11| = 0.117625.

The second line gives us the phase of S11 (143.61deg or  2.50647rad).
So expressing S11 in cartesian form gives us

Code: [Select]
-0.094688 + 0.0697841 i

which is close to the reading (-0.09 + 0.07j). All good so far.

Now I can compute the input impedance Z of the DUT using the equation:

Code: [Select]
S11 = (Z-50)/(Z=50)

and we obtain

Quote
Z = 40.9805 + 5.79982 i

or, expressing it in Polar form,

|Z|= 41.3889 , with angle 0.140593 rad.


Doubt: this is where I am surprised. The Smith data reported by the nanoVNA is 41.3 Ohm and 123nH.

  • 41.3 Ohm is the magnitude of Z (rather than its real part as I would expect).
  • 123nH is the value L that satisfies:  5.79982 = 2*pi*L*f, with f = 7542500, so a value directly proportional to the imaginary parto of Z (at this specific f).

Question: is this way of presenting the Smith value correct/standard? I would expect to see the real+imaginary component of Z (modelling a pure resistor in series with an inductor) rather than what I actually get.

Can you help me understand what's going on?

Thanks!






« Last Edit: May 22, 2020, 08:25:08 pm by mio83 »
 
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Offline rf-messkopf

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Re: Question about NanoVNA and Smith charts
« Reply #1 on: May 23, 2020, 07:47:41 pm »
Question: is this way of presenting the Smith value correct/standard? I would expect to see the real+imaginary component of Z (modelling a pure resistor in series with an inductor) rather than what I actually get.

I don't have the NanoVNA so I can't comment on what's going on here. Are you referring to the marker readout function, i.e., that it reports \$|Z|\$ along with the inductance/capacitance when used with the Smith chart?

There is, of course, a variety of ways in which Smith chart data can be represented numerically, and I don't think there is a standard way of how the marker function works on various VNAs. I have a newer R&S VNA here, on which the marker style can be extensively customized. The default on the Smith chart is, however, \$\mathop{\rm Re}(Z)+ \mathord{\rm i}\cdot\mathop{\rm Im}(Z)\$, together with the corresponding inductance/capacitance.

The \$|Z|\$-together-with-inductance/capacitance style is a bit odd, but it is an exhaustive description of the underlying data. Given \$|Z|\$ and \$|\mathop{\rm Im}(Z)|\$ as well as the sign of \$\mathop{\rm Im}(z)\$ (negative for capacitance, positive for inductance) you can work out \$\mathop{\rm Re}(Z)\$ under the assumption that \$\mathop{\rm Re}(Z)\geq0\$ (formally this corresponds to \$|S_{11}|\leq1\$, i.e. a passive DUT), and hence the complex \$Z\$. This can then be converted back to \$S_{11}\$.
« Last Edit: May 23, 2020, 08:50:25 pm by rf-messkopf »
 

Offline mio83

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Re: Question about NanoVNA and Smith charts
« Reply #2 on: May 23, 2020, 10:41:25 pm »

I don't have the NanoVNA so I can't comment on what's going on here. Are you referring to the marker readout function, i.e., that it reports \$|Z|\$ along with the inductance/capacitance when used with the Smith chart?


Yes, I was referring to that.  :)

As you also say, it seems a bit odd to have displayed the pair < |Z| , inductance/capacitance >.
I agree with you that the data is sufficient to recover all the S11 information (the S11 in cartesian form is, anyway, directly available in the nanoVNA, so no problem).  :-+

I was asking because I am not yet expert/confident with the Smith Chart (just learning). And from what I understand, to (directly) draw a point in the Smith chart we use the pair <Re(Z), Im(Z)>. So I was surprised by the strange choice of what is displayed in the nanoVNA as "Smith".

Still curious to know if there is a reason why it might be handy to have the pair < |Z| , inductance/capacitance > directly available.  :-//

PS: congratulations for your new R&S VNA!!!  :-+  :clap:
« Last Edit: May 23, 2020, 11:26:25 pm by mio83 »
 

Offline virtualparticles

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Re: Question about NanoVNA and Smith charts
« Reply #3 on: June 04, 2020, 08:41:24 pm »
The accuracy of the NanoVNA is dreadful! I measured the "Load" from the supplied cal kit on a real VNA and found it to have only 22 dB Return Loss at 900 MHz.  This means that return loss measurements near 12 dB will be +/- 3.3 dB best case. Measurements below 12 dB return loss will have almost no meaning at all. I did try using a commercial grade calibration kit and it didn't help much. I suspect that the awful source and load match are the reasons for that.

After careful calibration with a commercial calibration kit I measured a 12 dB return loss standard which was 12 dB +/- 0.2 dB on a real VNA. The Nano measured 15 dB at the low end and 20 dB at 900 MHz.

This device is suitable for demonstrating VNA technology to high school kids but it doesn't make actual valid measurements.
 

Offline radiolistener

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Re: Question about NanoVNA and Smith charts
« Reply #4 on: June 04, 2020, 10:50:30 pm »
I measured the "Load" from the supplied cal kit on a real VNA and found it to have only 22 dB Return Loss at 900 MHz.

it looks like cheap Chinese 50 Ohm terminator. I bought a couple of 50 Ohm terminators on aliexpress and they also have about 25 dB Return Loss at 900 MHz. But original cal kit 50 Ohm load which I got with nanovna has Return Loss about 60 dB.

So probably you got nanovna with usual 50 Ohm terminator. NanoVNA quality also can vary from seller to seller.
 

Offline OwO

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Re: Question about NanoVNA and Smith charts
« Reply #5 on: June 05, 2020, 06:01:19 am »
There are good 50ohm terminators on taobao, it's just a lot of work to tell it apart from all the trash. I have a 50ohm standard that has -40dB |S11| at 3GHz when compared to a real professional calibration kit, and it costs less than $1. Same issue with the nanovna. On taobao there is one official seller of the original nano, and one official seller of the V2. On aliexpress and ebay you just never know, I personally knew of a seller that switched sources every week. Most sellers also don't have stock, they put up 1000 listings and then buy on-demand from who knows where. The -F at least has an official aliexpress store (but it's ex🅱️ensive), and V2 will have an official web store soon.
« Last Edit: June 05, 2020, 06:04:44 am by OwO »
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Offline virtualparticles

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Re: Question about NanoVNA and Smith charts
« Reply #6 on: June 05, 2020, 02:08:17 pm »
There are good 50ohm terminators on taobao, it's just a lot of work to tell it apart from all the trash. I have a 50ohm standard that has -40dB |S11| at 3GHz when compared to a real professional calibration kit, and it costs less than $1. Same issue with the nanovna. On taobao there is one official seller of the original nano, and one official seller of the V2. On aliexpress and ebay you just never know, I personally knew of a seller that switched sources every week. Most sellers also don't have stock, they put up 1000 listings and then buy on-demand from who knows where. The -F at least has an official aliexpress store (but it's ex🅱️ensive), and V2 will have an official web store soon.

I did use a high grade calibration kit and repeated the return loss measurement. The 12 dB reference still measured 16 dB to 20 dB from low freqs to 900 MHz. So the supplied load is awful but using a real calibration kit doesn't help because the raw source and load match are so bad. I've attached the Nano measurement of the 12 dB reference after calibration with a commercial calibration kit and a measurement of the return loss reference on a real VNA.
 

Offline OwO

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Re: Question about NanoVNA and Smith charts
« Reply #7 on: June 05, 2020, 02:22:16 pm »
10dB attenuator, open circuited, measured to 3GHz with nanovna v2:
« Last Edit: June 05, 2020, 02:24:14 pm by OwO »
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Offline radiolistener

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Re: Question about NanoVNA and Smith charts
« Reply #8 on: June 05, 2020, 03:51:47 pm »
I did use a high grade calibration kit and repeated the return loss measurement.

As I know these nanovna with lizard logo have issues with dynamic range. At low frequency at 10-100 kHz it shows just terrible performance. For comparison, original nanovna from hugen79 works pretty good at low frequency.

Someone told that he fixed that by replacing SA612 mixers and adding RF shielding.

Also there is a small issue with nanovna charger circuit, when you use it with attached power supply for charge, it may show a little difference at high frequency (600-900 MHz). When battery charge is complete (led stops blinking) influence of external power supply will be reduced, but still present. With no external power supply it shows the best result.
« Last Edit: June 05, 2020, 03:59:30 pm by radiolistener »
 


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