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I have a schematic for a current sensing circuit using a dedicated TI differential amplifier part for current sensing applications. The schematic also includes an "impedance matching" part for connecting the voltage output of the differential amplifier to a BNC coaxial cable. This coaxial cable then can be attached to the PCB and then plugged into an oscilloscope.

The impedance matching bit consists of a potential divider of resistance 5-MOhms and 75-Ohms. I don't understand how or why this would work (my understanding of impedance matching is a bit iffy I'm not going to lie). I was told this bit of the circuit was taken from the internet and I've tried googling around but couldn't find the original source. Could anyone help explain how this potential divider impedance matches to a 50 Ohm BNC coax?

Secondary question:

I tried simulating it in LTSpice to maybe get some insight into how it would work but I don't think my model of a coaxial cable was correct. I found some typical values of inductance and capacitance from here and then measured the resistance of the cable and included that too. I also looked up the input impedance of the oscilloscope which said it was 1-MOhm and an input capacitance of 13pF. I don't think I'm simulating it correctly however, is this circuit model a good enough approximation? From the article they linked it says there's also a parallel conductance, but I have no way of measuring this so have omitted it. enter image description here

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Firstly let's look at the coax in your question: -

enter image description here

With lumped parameters of 36.8 nH and 31.2 pF, the actual characteristic impedance can be calculated by: -

\$Z_0 = \sqrt{\dfrac{L}{C}}\$ = 34.34 ohms.

So, clearly it is not intended to match the signal because there is a significant mismatch.

Next the amplifier (INA240A). The data sheet tells us that it doesn't provide any useful signal outputs above 1 MHz and for a transmission line to be "significant" at 1 MHz, it would be longer than about one-tenth of the wavelength of 1 MHz (300 metres) at 30 m or more.

So, unless you have a very long piece of coax there is nothing about the attenuator that is related to impedance matching.

Let's say your cable was 300 metres plus. Because we are talking about fairly low frequencies (less than 1 MHz), the coax's characteristic impedance would not look like 50 ohm or 75 ohm - it would have a complex impedance and this is also not-matched by the 75 ohm resistor.

In short, my conclusion is that R2 and R3 are just a simple attenuator and that anything told to you about why it is there is a red-herring.

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  • \$\begingroup\$ Sorry I mistyped the impedance its actually meant to be 76.8n, I did check the characteristic impedance calculation to make sure the values in that example and it did roughly match 50-Ohms. So when impedance matching is it the reactances that need to be matched specifically and the resistive elements can be mismatched? \$\endgroup\$
    – genericpurpleturtle
    Apr 26, 2018 at 12:10
  • \$\begingroup\$ For high frequencies it is the ratio of the reactances that produce a resistive characteristic impedance and, for good matching, this can only be matched by resistors (or other transmission lines aka coax etc.). \$\endgroup\$
    – Andy aka
    Apr 26, 2018 at 12:20
  • \$\begingroup\$ Why can it only be matched by resistors or other tranmission lines? Why could one not use an inductor and capacitor combination? Coul you please know a good source which explains transmission lines and impedance matching be it book or internet. The stuff I've found on the internet hasn't been very comprehensive. Finally just to clarify, I can get away without the impedance matching and just directly connecting the output of the amplifer to the BNC coax and then to the scope? Also thank you for your help :) \$\endgroup\$
    – genericpurpleturtle
    Apr 26, 2018 at 17:08
  • \$\begingroup\$ @genericpurpleturtle go searching for the words "characteristic impedance" and look for this formula: \$Z_0 = \sqrt{\dfrac{R+jwL}{G+jwC}}\$. As frequency gets high it becomes the formula in my answer. Yes you can match with inductors and capacitors but it becomes accurate at only a small range of frequencies. If your cable is a few metres long I don't think you'll have a problem matching or not. I don't know of a good source because I rarely need to look stuff like this up LOL!! \$\endgroup\$
    – Andy aka
    Apr 26, 2018 at 18:18

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