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I tried to use a single directional coupler to create a reflectometer and was hoping to measure both amplitude and phase of reflected signal.

My setup is a follows:

The directional coupler is from MiniCircuits ZFDC-10-182-S+

schematic

simulate this circuit – Schematic created using CircuitLab

I have tested my design with different loads, 0R, 50R, 100R, 100R||3.3pF ect.

But only the magnitude (VMAG) measurements are correct, the phase (VPHS) measurements does not seems to be correct. And I cannot find any relation, between different loads and expected vs. measured phase.

So should it possible to measure phase information with a directional coupler?

Or am I doing it wrong ?

My measurement:

50R: VPHS=0.88V, VMAG=0.16V

Open: VPHS=1.500V, VMAG=0.76V

Short: VPHS=0.282V, VMAG=0.69V

100R: VPHS=1.455V, VMAG=0.50V

100R||10pF: VPHS=0.58V, VMAG=0.66V

100R||3.3pF: VPHS=1.18V, VMAG=0.57V

Edit: My open load now seems correct, before I measured open load as "no load" in the cable, but this resulted in a capacitance in the cable end SMA connector. I have now made an open load as an trimmed down SMA connector. I still don't get why the magnitude i higher, but that is maybe related to the load on the source being changed.

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  • \$\begingroup\$ It would be useful to show your measured data. Also to show what you're measuring when the load is a short and an open circuit. \$\endgroup\$
    – Roger C.
    Commented Jun 4, 2016 at 9:29
  • \$\begingroup\$ Ok, I have added measurements now. \$\endgroup\$
    – JakobJ
    Commented Jun 4, 2016 at 10:35
  • \$\begingroup\$ Did you consider in your calculations the effect on a phase that connection cables are giving? If possible, show how you calculated the expected phase. \$\endgroup\$
    – ivan
    Commented Jun 4, 2016 at 10:51
  • \$\begingroup\$ The phase when having an open circuit and when having 100 ohms resistor should be the same. The connection length between the IN port and those two loads (100R and the open circuit) is the same? \$\endgroup\$
    – Roger C.
    Commented Jun 4, 2016 at 11:12
  • \$\begingroup\$ I used the same 0.75m cable with SMA connectors. I assumed that the phase added by the cable would be constant, when the frequency was constant. And therefore when comparing the measurements, the phase difference between them should match the calculated. That is the phase error is just an offset, or am I wrong? \$\endgroup\$
    – JakobJ
    Commented Jun 4, 2016 at 12:14

1 Answer 1

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Doing a few calculations it seems to me that your measurements are correct. Let's take for example:

$$ 100R//10 pF = (@ 868 MHz) = 3.25-17.7j $$ Then if we calculate the reflection coefficient: $$ \rho=\frac {Z_L-Z_0} {Z_L+Z_0} =-0.69 -0.56 j$$ The phase of \$\rho\$ is -141º=219º. In comparison the phase of \$\rho\$ for a 100R load is 0º.

In your measurements with 100R//10pF you have 0.58 V -> \$\pm 58º\$ and with 100R you have 1.455 V -> \$\pm 145.5º\$. The \$\pm \$ accounts for the phase sign ambiguity of the used detector. Therefore the absolute phase difference is either \$145.5-58º=87º\$ or \$145.5+58º=203º\$, (we can't know which one to take because of the measurement phase ambiguity). We note that \$203º\$ and \$219º\$ are close enough.

If we repeat for 100R//3.3 pF we have that the phase of \$\rho\$ is \$-92º=268º\$. According to your measurements the phase difference between 100R//3.3pF and 100R is either \$145.5-118=27.5º\$ or \$145.5+118=263.5º\$ and \$263.5º\$ is quite close to \$268º\$.

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  • \$\begingroup\$ Taking into account phase sign ambiguity, everything solves perfectly. I suppose with lower frequency OP will get 99% match. \$\endgroup\$
    – ivan
    Commented Jun 4, 2016 at 17:45
  • \$\begingroup\$ I tried for many hours to make sense of my measurements, so now it finally makes sense. I think two factors was into play. 1) I used the short circuit as a base for my calculations, which was probably not an ideal short. 2) As pointed out by @ivan, the dynamic range of the AD8302 is 143 degrees, so it is not possible to span the range from short to open load. \$\endgroup\$
    – JakobJ
    Commented Jun 4, 2016 at 19:51
  • \$\begingroup\$ @JakobJ 1) Actually your short looks good: 0.28 V could be -28º degrees (minus 28º) and when the load is 100R you have 146º. The phase difference is therefore 28+146=174º (almost the 180º as it should). 2) I've checked the datasheet and the phase range of the AD8302 is 180º (but with the sign ambiguity!). 3) It seems to me that the only load that should be checked is the "open" one, it should have the same phase than the 100R one (but somehow the reference plane for both loads is not the same). \$\endgroup\$
    – Roger C.
    Commented Jun 4, 2016 at 20:04
  • \$\begingroup\$ @RogerC. you are right. If I do the calculations with the short as reference, I get same or even little better results. I think what actually helped me a lot, was your conversion to absolute degrees (0-360), the +/-180 degree is very confusing, when trying to compare results. \$\endgroup\$
    – JakobJ
    Commented Jun 4, 2016 at 20:43
  • \$\begingroup\$ @RogerC could you point to the place in the datasheet where it says that the phase range of AD8302 is 180º? \$\endgroup\$
    – ivan
    Commented Jun 4, 2016 at 20:59

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