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Lab View Nano VNA system - transfer relay and negative resistance


John Galbreath 2021/08/19 20:47

1.  In the You Tube tutorial covering the transfer relay, it was mentioned that the first transfer relay was made with GaAs with disappointing results.  I was curious about the successful version and I would not attempt a relay based on PIN diode switches if I knew that others had bad outcomes.

2.  I do not know why I encounter negative resistance, but now I realize that I am not the only one.  On page 123 of the manual version r21, there in figure 106 is a resistance that is also negative.

Joe Smith 2021/08/20 04:53

>>7.  Will 2 SPDT PIN diode switches (plus SPST relay for status of relay) work for a transfer relay ?
>>Could you share your USB control details ?

>That's really up to the designer to come up with what ever circuit meets their criteria.  The manual covers the use of a transfer relay.  What specifically were you not understanding?

>>1.  In the You Tube tutorial covering the transfer relay, it was mentioned that the first transfer relay was made with GaAs with disappointing results.  I was curious about the successful version and I would not attempt a relay based on PIN diode switches if I knew that others had bad outcomes.

"I was curious ..."  Meaning it is now in the past?

>>2.  Resistance value is negative sometimes - OK to ignore ?

>I assume the user can answer that.

>>2.  I do not know why I encounter negative resistance, but now I realize that I am not the only one.  On page 123 of the manual version r21, there in figure 106 is a resistance that is also negative.

FYI, pages 118 & 144 also show examples.   There may be others.

Joe Smith 2021/08/20 05:19

For example:  Looking at S11, with nothing connected to port 1 the VNA returns  1.00083 -0.00272913 i.  Assuming a characteristic impedance of 50 ohms,  calculate resistance.  Show your work.

Hint:   While I felt it was beyond the scope of the the manual to cover the math,  section 15.2 Measuring Crystals shows a table I had pulled from a paper published by CopperMountain which contains the equations for the three types of impedance measurements.    These equations may be found in many publications.  Learn to use Google.

Jim Lux 2021/08/20 07:22

On 8/19/21 8:47 PM, John Galbreath via groups.io wrote:
> 1.  In the You Tube tutorial covering the transfer relay, it was
> mentioned that the first transfer relay was made with GaAs with
> disappointing results.  I was curious about the successful version and
> I would not attempt a relay based on PIN diode switches if I knew that
> others had bad outcomes.
>
Surplus mechanical relays are pretty inexpensive and have great
isolation and flatness. Unless you need fast switching, the diode or
GaAs probably isn't the way to go. For GaAs, you typically need a fairly
sophisticated multiple stage switch to get isolation of 80dB, and that's
the sort of out of the box performance for a mechanical relay.



> 2.  I do not know why I encounter negative resistance, but now I
> realize that I am not the only one.  On page 123 of the manual version
> r21, there in figure 106 is a resistance that is also negative.

It could be an artifact of numerical precision effects.



Jim Lux 2021/08/20 07:57

On 8/20/21 5:19 AM, Joe Smith via groups.io wrote:
> For example:  Looking at S11, with nothing connected to port 1 the VNA
> returns  1.00083 -0.00272913 i.  Assuming a characteristic impedance
> of 50 ohms,  calculate resistance.  Show your work.

On the original NanoVNA, the receiver sums 48 samples against a SIN/COS
table, using 32 bit math (16 bit ADC, 16 bit sin/cos), which is then
scaled to a single precision float.

The raw reflection measurement I and Q is probably good to about 1 part
in 100,000 (14 bit ENOB & sqrt(48) averaging).  I suspect that the drive
levels are set so that they are down a bit from saturating the ADC, so
the raw measurement is probably more like 1 part in 10,000 to 1 part in
20,000

1.00083 is ~1 part in 1000 away from 1, slightly more than the
measurement uncertainty, but I'd want to delve into the actual math
before saying "that is or isn't measurement noise"

gamma is calculated as

  // calculate reflection coeff. by samp divide by ref
  float rs = acc_ref_s;
  float rc = acc_ref_c;
  float rr = rs * rs + rc * rc;
  //rr = sqrtf(rr) * 1e8;
  float ss = acc_samp_s;
  float sc = acc_samp_c;
  gamma[0] =  (sc * rc + ss * rs) / rr;
  gamma[1] =  (ss * rc - sc * rs) / rr;


rs,rc,ss,sc all have that 1 in 10,000 measurement uncertainty.

so rr is the sum of two uncertain values squared - (x+uc)*(x+uc) = x^2 +
2x*uc + uc^2 - the uc^2 term is tiny, assuming x is 1, that makes the
uncertainty in rr roughly 4x the uncertainty (because it's the sum of
two squares).

Then the actual gamma calculation is similar, 4x uncertainty for the
numerator.

So the uncertainty in gamma (I might be off here a bit, not had coffee
yet) is going to be on the order of 8-10x uncertainty of a single
measurement, so around 1 in 1000.

One has to be careful  with divides, if rr has a small magnitude, then
its uncertainty might be magnified. It's too early to figure it out in
detail, and I'm sure there's an app note somewhere with a more rigorous
derivation.

Joe Smith 2021/08/22 09:06

I take it no luck working the math.  In the example, when looking at S11 shunt measurement, the denominator is 1-S11.   In the example, S11 was > 1.  1 - (>1) is a negative number.  And a positive divided by a negative is?  Negative.   Of course, we are interested in the magnitude, not the sign.  In this case we are looking at an uncalibrated VNA and with maximum return, we are expecting values very close to 1.  It's easy enough to look at the raw data and see what is happening.

Of course, the software has a few features that could get the random button pusher into trouble and leave them scratching their head.   But again, I am expecting the users are  BS, MS, EEs getting started with some basic RF measurement concepts.  Learning how a VNA makes these measurements would benefit most anyone interested in using one.  There are many free papers  available on this subject but it's up to you to do that research.

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