I was playing with my VNA yesterday and noticed in the Smith chart view of a surface mount device that the trace went from an open at 30 KHz, through inductive to some capacitance at 6 GHz. Nothing new.

As I placed the marker on the spiraling trace and moved it along, I found two interesting spots along the way.

At one spot, the real and imaginary values became the same positive number and that yielded a value for the part of about 120 nH which is what I think this part was.

As I kept moving the marker along into higher frequencies, the real and imaginary numbers became the same number but the imaginary number became negative and the device was a capacitor.

I'm very used to all of this behavior but my question is about the point where the real and imaginary values were equal and the parts value seemed rational. I ran into this early in my career when trying to figure out why a SMD didn't act as I expected in an RF circuit. Not only were the PCB parasitics unexpected but the value of the SMD I was trying to use to tune the circuit wasn't working for me. I struggled for weeks on the problem until I figured out how far the value of a SMD varies over frequency. I'm curious about the real and imaginary numbers becoming equal and the value of the part becoming that listed on the bag.

Any comments?

  • 5
    \$\begingroup\$ Telling us what was the actual part you measured, and sharing the Smith chart would help us give you more useful answers. Also, please ask a more specific question than "any comments?". We prefer questions that have right and wrong answers, not just invitations to general discussion. \$\endgroup\$
    – The Photon
    Nov 6, 2022 at 15:53
  • \$\begingroup\$ What do you mean by "open". In electronics we say open when we mean open circuit i.e. infinite impedance. If you mean "short" then say so. Show the impedance graph because words can be so ambiguous. \$\endgroup\$
    – Andy aka
    Nov 6, 2022 at 16:26

1 Answer 1


The VNA measures the impedance of the load using a resistance bridge with R = 50 ohms. The measurement is most accurate when the load is 50 ohms. The VNA has a source impedance of 50 ohms.

There are many explanations about the design/operation of the VNA bridge if you need more detail.

The bridge is "balanced" when R = X = 2PiF*L = 50.

The phase Phi = tan-1(X/R) ... 45 degrees = tan-1(1)

So, I think that's a simple explanation... assuming the inductor is the only load on the VNA.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.