I have a question regarding some conducted RF immunity (IEC 61000-4-6) tests. We inject noise starting at 150kHz until 80MHz. The signal injected is AM at 80%. The level is 20Vrms.

Here is the setup of the test:

enter image description here

In our setup, both unit and auxiliary equipment are floating, there is no direct path to ground. The cable being tested consists of a power conductor, a ground conductor, and an RS-232 line. Our goal is to assess the margin of our one-wire communication (RS-232.)

The power supply is designed to generate a voltage drop across the negative terminals of both the equipment under test (EUT) and the auxiliary equipment (AE.) This setup allows for a circulating current. Additionally, the RS-232 data line is connected to a CMOS that shares the same ground as previously mentioned.

If any additional current is introduced to the existing circulating current, it will impact the input of the CMOS. This outcome will result in a failure.

We did the test four times at 20Vrms and we got a fail (temporary loss of data) at 3.5MHz and 3.65MHz. When I convert this to a wavelength the cable's length is exactly a quarter wavelength. (3.5MHz -> 80m).

Here is a graph of a current probe monitoring the current 10cm away from the EUT. enter image description here

I can deduce that from 100kHz to 4MHz the system behaves like an inductor. Another observation is that at the critical 3.65MHz the system sees close to an open circuit, which would mean that the standing waves are the highest and hence some resonance.

I would like to get some help maybe visualizing a bounce diagram or some other way in order to understand how is the energy flowing in the system, and what causes these fails at the above mentioned frequencies. Just to be clear. I do not need suggestions to solve the problem of “fails”. Thank you Here is the setup in the lab:

enter image description here

  • 1
    \$\begingroup\$ My hunch is that with a 1/4-wavelength antenna there is maximum voltage difference between the two ends of the 20m cable and this is corrupting the data. Ideally, the RF should result in equal (common-mode) voltages on all the conductors of the cable, but this depends heavily on the common-mode impedance balance of the conductors. It would be helpful if you could provide a schematic for the cable and its terminations. \$\endgroup\$ May 31, 2023 at 18:46
  • \$\begingroup\$ The cable is not shielded? \$\endgroup\$ May 31, 2023 at 20:15
  • \$\begingroup\$ @Just_visiting, I contacted the Mechanical team to get the specs for the cable \$\endgroup\$
    – DRF
    May 31, 2023 at 20:23
  • 1
    \$\begingroup\$ Galvanic grounding is not necessary (indeed even undesirable in some cases) to have shielded signals. \$\endgroup\$ May 31, 2023 at 22:54
  • 1
    \$\begingroup\$ Something you might want to consider is to repeat the test with ferrites at each end of the 20m cable, one close to the EUT and the other close to the AE, and see if this helps. If so, you can work with electrical and mechanical engineers to move the ferrites into the enclosures, either on or off of the PCBs. \$\endgroup\$ Jun 1, 2023 at 14:08

1 Answer 1


This is a little complicated and at the "bleeding" edge of my knowledge, but let me offer an explanation. The basic path for current is from the RF source for the test, which is referenced to the reference/ground plane in the chamber, to the coupling network that puts energy into the cable, then via the parasitic capacitance from the cable-connected PCBs back to the chamber's reference plane and RF source. (Neglecting other paths for simplicity.) You've pointed out that there is a failure where the 20m cable is a quarter wavelength, and that likely means that the cable is a quarter-wave monopole antenna. Below this quarter-wavelength frequency (3.5/3.65 MHz), a monopole antenna looks capacitive. So from 150 kHz to the fail frequency, it's capacitive and getting less capacitive as frequency increases, and at quarter-wave resonance it's resistive. There is maximum voltage across the antenna at resonance, and that voltage difference is the likely cause of the failure. This signal will get rectified/detected by any semiconductor junction (just like in a crystal radio receiver) and the 1-kHz demodulated signal can now find its way into the circuitry to cause problems.

In his book, "Electromagnetic Compatibility Engineering," the late Henry Ott had "Appendix D Dipoles for Dummies (as well as for all the rest of us without a Ph.D.in electromagnetics)," and much of what I described can be found there. The publisher (Wiley) has made some of the book freely available and this is the case for this appendix and it is available here: Appendix D

Hope this helps.


Your Answer

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

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