An OSL that can go up to the GHZ can costs several hundreds if not thousands of dollars. What makes them so special?

My theory is that a load involves some sort of special resistor and mantaining a 50ohm load into the GHz might not be so easy, same thing with the short, a piece of wire is a transmission line at those frequencies, inductance/capacitance can have a big effect. Does that justify the huge cost?

More importantly, what about an open? What makes an open from the calibrator so different from just leaving the line unterminated? An open is an open is an open, so why do we need it?

What are the technicalities behind a good OSL calibrator that an amateur with a soldering iron, some N type connectors and some RF SMD resistors can't do?

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    \$\begingroup\$ What's so difficult about making your own, like this youtube.com/watch?v=9QRpWjLFApA \$\endgroup\$
    – BobT
    Commented Aug 26, 2020 at 3:03
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    \$\begingroup\$ At 1 GHz, the wavelength of a wave in free-space is around 30cm. It is not THAT exotic. It is certainly possible to terminate a 50 Ohm line with a 49.9 Ohm SMT resistor at 1 GHz, and get very low reflections from it. As for the rest (opens and shorts) I don't know with certainty. Good question. \$\endgroup\$
    – user57037
    Commented Aug 26, 2020 at 3:54
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    \$\begingroup\$ @mkeith You would need at least two resistors, a single resistor makes a 90degree bend, hence an inductor,; but two resistors cancel out to a first approximation, four would be better, but then you are making a flat capacitive plate. I've used pairs of 100ohm resistors at 2.4gHz, and not noticed any artifacts on my cheap spectrum analyser. If you can afford equipment good enough to spot the discrepancy, you can afford a decent 50.000 ohm terminator. You can always make your own, then take it to a University lab measure, it's performance and stick a label on your terminator \$\endgroup\$
    – BobT
    Commented Aug 26, 2020 at 4:34
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    \$\begingroup\$ @BobT some chinese OSL kits seem very promising up to around 6 GHz, and cost a fraction of the price. They are, in my opinion, a better alternative to DIY. \$\endgroup\$
    – S.s.
    Commented Aug 26, 2020 at 4:44
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    \$\begingroup\$ Another thing that makes really good ones expensive: I once called up (then) Agilent for help with coefficients for a 15-year-old 85053? 8 GHz kit, and was put straight through to a PhD engineer who could sort me out. So for your $10k you get more than a fancy wooden box. \$\endgroup\$
    – tomnexus
    Commented Aug 26, 2020 at 10:01

2 Answers 2


Metrology grade calibration standards are machined to very tight tolerances to provide precise and repeatable results when performing an SOLT calibration. At lower frequency (ones of GHz or below), this is less of an issue, but as the desired measurement moves to higher frequency, the more important this becomes (in particular, for the phase accuracy of the measurement).

Consider what an SOLT calibration is fundamentally doing. If you're not familiar with it, research the 12-term error model for vector network measurements. The SOLT calibration method allows for the mathematical derivation of these 12 error terms, assuming it is possible to measure a perfect Open, Short, Load and Thru at the reference plane of your DUT. For the math to work out, each standard is measured, and the raw measurement of the standard is compared to the given characteristic of the standard (for the open/short, this is typically described as capacitance/inductance coefficients to a polynomial). A perfect physical standard will have exactly the response described by the polynomial, and since this is known, then the raw measurement can be compared to the known value, and the difference is used to determine the error term of the test setup.

Assuming everything is perfect, your reference plane will be well defined once the error terms are applied to the raw measurement, and the measurement you acquire will end up at the DUT reference plane. Of course, nothing is perfect, and there are tolerances in calibration standards. For an SOLT calibration, a "perfect" set of open and short standards will have exactly the same offset length. Any imprecision in the distance between the cable reference plane and the actual open/short standard will introduce uncertainty in the standard measurement, which will result in a poorly-defined reference plane. Similarly, if the load standard isn't close to the specified system impedance (i.e. it has a poor match), uncertainty in the error terms will be introduced. All of these uncertainties occur if the manufactured connector strays from its designed values.

Especially for high frequency measurements, an "open is an open is an open" is not at all a true statement. Quick example: consider an air-dielectric coaxial RF connector with a velocity factor of \$0.9\$, and we want to measure the response of our DUT at \$f_0 = 40\mathrm{GHz}\$.

Our wavelength ends up being:

$$ \lambda = \frac{c}{f_0} = \frac{0.9\cdot3\mathrm{e}8\ \frac{\mathrm{m}}{\mathrm{s}}}{40\mathrm{e}9\ \mathrm{s}^{-1}} = 6.75\ \mathrm{mm}$$

Divide that by 360 and you get that each \$18.75\mathrm{\mu m}\$ equates to 1 degree of phase error. So for every 20 microns your open standard's connector is off from it's specified offset length, you're adding over a degree of phase error to the measurement of the calibration standard at the frequency of interest. This isn't even considering the loss of the cable and standard connection interface (which will add uncertainty to the magnitude of the measurement). Now consider that the short and the load standards may also have uncertainty, and all of these uncertainties will compound.

In addition to calibration accuracy, it is important for a calibration standard to have repeatable results. If a VNA is calibrated with the same standards day-in and day-out, and each calibration results in wildly different calculated error terms, the measurements that come out of that instrument will not be very useful if accurate characterization of the DUT is the goal. Metrology-grade standards, in addition to being accurate to their specified values, are also typically built with tight tolerances to ensure repeatable characteristics after many connect/disconnect cycles (assuming they are used properly, i.e. torqued correctly, kept clean, etc.).

In short, all of the extra machining precision required to provide accurate and repeatable calibration standards combined with the relatively small market (i.e. low economy of scale) for metrology-grade components equates to an expensive device that is seemingly very "simple."

Some decent further reading is the following presentation from Maury Microwave that describes some of the important metrics for metrology-grade connectors (pin depth, concentricity, repeatability, etc.): https://www.maurymw.com/Support/pres/connectors-precision_or_not.pdf

As @mkeith alluded to in the comments to your question, it's certainly possible to make one's own standards (the load is the easiest since if it's got a somewhat decent return loss, it won't affect the calibration very much). It really boils down to a trade-off of how much accuracy is required need. Ballpark area on a smith chart for a low-frequency passive device? Probably fine. Trying to characterize e.g. the impedance of a high-frequency transistor for end-use in a multi-stage power-amplifier design? Not so fine.

  • \$\begingroup\$ Thank you for your answer, what I did not find in your answer is, how exactly is an open from an OSL calibrator? what is so special about it? how is it constructed? \$\endgroup\$
    – S.s.
    Commented Aug 26, 2020 at 4:41
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    \$\begingroup\$ @S.s. I think the value of the precision "open" is not how well it represents infinity ohms, But how far along that "infinity" begins, so it's a question of effective length, rather than effective value, there will be some fringing field at the end of the conductor. So if phase is important to you, you need to have the reference short and reference open at the same equivalent electrical length, so physical distance on one will be shorter than the other. On the other hand having a 50ohm termination in the wrong place doesn't matter, as 50ohms looks like an infinite length. \$\endgroup\$
    – BobT
    Commented Aug 26, 2020 at 5:03
  • \$\begingroup\$ @S.s. I'm not entirely certain what you mean by "OSL calibrator." Can you provide a link to an example? If you're just talking about a passive calibration standard (just like a typical coaxial SOLT calibration kit with male/female short/open/load standards), then my answer describes what is important for an open standard (namely an open circuit with a well-defined electrical length). If you're talking about something like an eCal module, those use a different error model and have little to do with an SOLT calibration model at all. \$\endgroup\$
    – Shamtam
    Commented Aug 26, 2020 at 11:29
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    \$\begingroup\$ @S.s. The answer your question of "what is so special about it?" is that it's length is well-defined, stable, and repeatable. Beyond precision machining to ensure consistent physical dimensions (which provide consistent electrical length, minimal discontinuity, etc.), I suspect good calibration standards also attempt to compensate for environmental conditions like temperature and humidity. \$\endgroup\$
    – Shamtam
    Commented Aug 26, 2020 at 11:33
  • \$\begingroup\$ @Shamtam yes I am talking about a passive calibration standard that includes a short, an open and a load. \$\endgroup\$
    – S.s.
    Commented Aug 26, 2020 at 12:18

This is a great question. As a passive RF designer, it was difficult to justify the high prices of an OSL. So I designed my own. This is what I found along the way. Starting with the Open and Short, my first product. The secret is to have consistent 50.0 Ohm impedance for the same lengths. This means matching or equivalent length of teflon and air sections. In addition, blunt end of the open has parasitic and fringe capacitance. This has to be minimized without affecting the electrical length. Then the polynomial coefficients can be determined to match the capacitance at any frequency.

Then I moved onto a precision load. By precision load, I mean a return loss of over 42dB (VSWR of less than 1.016:1) This corresponds to a resistance of 50.8 Ohms. Not too difficult, but thus is not a dc problem. Maintaining this resistance above much greater than 500 MHz is not easy. Considering that the resistor is changing electrical length versus resistance. The resistor is 50 Ohms at full length; but at half of the length, the resistance to the grounded end is 25 Ohms. So the mid point of the resistor would be non reflective or matched in a 25 Ohm transmission line section. The optimized transmission line geometry around the resistor are very complicated. The best calibration kits have loads that are precision (>42dBRL to 4GHz or 6GHz, and are functional (~26dBRL) all the way to 18GHz for standard accuracy use.

In the end, the NMOSPL series of cal kits provides an economical alternative to expensive cal kits, without compromising quality.

Disclosure: My company produces the NMOSPL series of cal kits.

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    \$\begingroup\$ To avoid the whole answer being deleted as spam (i.e. insufficiently disclosed promotion) I have added a clear statement that you are mentioning products from your own company. (While that connection could be inferred, a clear statement of the affiliation between you and the mentioned product is needed.) \$\endgroup\$
    – SamGibson
    Commented Sep 9, 2023 at 22:15

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