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I am trying to understand the impact of unconnected wires in a cable for EMC conducted emissions tests.

My product has a shared unshielded cable for supply and communication (I know it's bad practice...). It consists of 9 untwisted wires (see picture below). The 12VDC supply is on the red and black wires. The communication (RS-485) is on the brown, pink and green wires (not optimal I confess). All the rest is unused (white, yellow, grey and blue).

cable_arrangement

The test setup is classical (see diagram below).

test_setup

When all unused wires are left unconnected on both sides, the RS-485 coupling on the supply is tremendous (>1Vpp on the LISN side for a 15m cable). I can literally read the frames directly on the supply.

However, when I tie together the blue and white wires together on both ends, the coupling vanishes (>1Vpp to something barely noticeable on the spectrum analyzer). I must insist on the fact that the two wires are connected to nothing else.

I understand the capacitive coupling when the wires are left unconnected. But I just don't get why these two wires stop the coupling when connected like that.

Did you have any similar issue or understand the physics behind?

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  • \$\begingroup\$ Why do you have 3 wires, brown, pink and green, for the RS-485? \$\endgroup\$ – SteveSh Mar 27 '20 at 14:26
  • \$\begingroup\$ We provide ground as well. I know it could be removed but any unbalance in RS-485 would result in common mode and radiation. \$\endgroup\$ – Massimo Mar 27 '20 at 15:24
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By connecting both ends of the cable, you are essentially coupling the differential signal. This can be roughly simulated:

Essentially what you see in the circuit is your setup. A differential communication bus strongly coupled to a wire and weakly coupled to the supply line. The simulation is run for two different setups (white and blue lines are open or shorted). Circuit

Open line: During the transition of the diffential signal, due to the high dV/dt, part of the signal is coupled through the parasitic capacitance and the result can be seen straight away in the supply voltage line.

Shorted line: In this case, during the transition, let's say the voltage at the pink cable starts to increase and the brown one's decrease. A strong coupling takes place between the pink and blue cables, and so does between the white and brown cable. Therefore, just a very small amount is coupled in the supply voltage line.

V(power,gnd) Green Line: Open Load Red Line: Shorted Load

I(R8) Red Line: Open Load Pink Line: Shorted Load

waveform

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with the adjacency you've described, you have exactly balanced charge injection into a differential system ( ignoring the green wire).

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  • \$\begingroup\$ Yes, but tying the blue and white together at both ends keeps the symmetry while the result changes dramatically. Do you agree ? \$\endgroup\$ – Massimo Mar 27 '20 at 15:29
  • \$\begingroup\$ OK, with the schematic of vtolentino this makes sense. \$\endgroup\$ – Massimo Mar 27 '20 at 15:50

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