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I have two PCBs connected by a cable containing 5 sub-cables:

  • 6v power through a custom made coaxial cable (similar to that found on laptop power supplies).
  • 2x 100mbps LVDS through 100ohm impedance screened twisted pair.
  • 2x 1mbps CAN through the same 120ohm twisted pair cable.

enter image description here

Each LVDS cable is terminated at the RX end with a 100 ohm resistor. They have foils screen with drain wires.

Each CAN cable is terminated at both ends with a 120 ohm resistor. They have foils screen with drain wires.

Isolated 24v power is delivered to the left board, where it's switched down to 6v (non-isolated). Both boards contain their own 3.3v DCDC regulator (non-isolated) for the local electronics.

My questions:

At which ends should the shields be connected? I assume the LVDS shields should be connected at the source end, as shown in the diagram.

Since both ends of the CAN buses are sources, should both ends of the CAN shields be connected to GND?

Added: Both PCBs are housed in plastic cases, and there is no attachment to earth.

Passerby
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Rocketmagnet
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    This is a great graphic. – abdullah kahraman Mar 08 '13 at 16:23
  • do either side have an earth ground? – Grady Player Mar 08 '13 at 16:53
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    @Rocketmagnet: I'm interested to know what software did you use to make the graphic? – Gustavo Litovsky Mar 08 '13 at 16:54
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    @GustavoLitovsky - I used Altium's schematic editor. It's got a couple of simple line and rectangle drawing tools. It takes time, but it makes for nice documentation. – Rocketmagnet Mar 08 '13 at 16:57
  • @GradyPlayer - Nothing is connected to earth ground. The main 24v power supply is an isolated AC-DC type. – Rocketmagnet Mar 08 '13 at 16:58
  • @Rocketmagnet For consistency, I would attach shield on whichever end was closer to source of power. Not sure that even applies in your case. – Anindo Ghosh Mar 08 '13 at 17:07
  • I have two competing impulses, one is to ground every one to the data source side... but the power interconnect makes me think that is a bad idea... groundloop wise, so tie them to the left? – Grady Player Mar 08 '13 at 17:39
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    @AanindoGhosh Sometimes, it's better to ground the shield closer to the source of interference than to the source of power. **Example:** sensor on an industrial machine is connected to a DAQ by a cable. The industrial machine generates interference. In this situation, I would connect the shield to the ground somewhere near the noisy machine. On the DAQ side, I would connect the shield to the DAQ ground through a capacitor (say, 100pF). – Nick Alexeev Mar 08 '13 at 17:43
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    Think star topology when it comes to shielding. Use one single end and place to connect all shielding and don't connect the other end. You'll create ground loops (=loads of troubleshooting hours) if you don't. Shielding must not carry current. – jippie Mar 08 '13 at 19:15
  • @jippie - Connect both LVDS cables at the same end? – Rocketmagnet Mar 08 '13 at 19:41
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    @Rocketmagnet yes, that is the easiest way to keep overview and to prevent currents in the shielding. Key is however that you don't want currents in the shielding. The shielding is little more than a Faraday's cage. It works once you ground it somewhere. But if a current flows through the shield, you get crosstalk between shield and conductor. – jippie Mar 08 '13 at 20:11
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    @jippie - Hmm, just reading Ott: "Above 100kHz, it becomes necessary to ground the shield at both ends....It is common practice ... with digital circuits to ground the cable shield at both ends." – Rocketmagnet Mar 08 '13 at 20:19

3 Answers3

17

This is a difficult one to answer, mostly because RF and EMI are so incredibly non-intuitive. One might say that if someone claims to understand EMI then they most certainly do not understand EMI. I do not claim to completely understand EMI. I know a lot about it, but I have some holes in my knowledge. Consider that when reading my answer.

My main concern is that LVDS, and really any other differential signaling method that does not use isolation transformers, is not perfectly differential. There are mismatches in the differential drivers that cause common mode "noise" on the diff-pair. This common-mode noise also has a signal return path, which would be on the GND or shield in this scenario. The problem with having the shields disconnected at one end is that this signal return path would be on the power cable-- causing a huge loop area and resultantly huge EMI. While the common mode noise return current is small, the loop area his large, and so this must be accounted for in the design.

In one design of mine, I ran some 2.5 GHz signals over an 18" SATA cable. For those who don't know, a SATA cable has two diff-pairs in it and two shields. Both shields are connected together at the ends. There are no GND wires in the cable other than the shields. In my design, the shields were connected to signal GND at both ends. This design worked great, and is in volume production right now. It complies with FCC Class B, and the equivalent CE version, for electro-magnetic-compliance including radiated emissions, RF susceptibility, and ESD susceptibility.

Going on with the SATA comparison, all SATA motherboard/drives connect the shields at both ends, and they work fine at high speeds. SATA cables are available in length of about 6 inches to 2 feet-- similar to what the OP is using. Systems with SATA meet the more stringent EMC regulations. And they are shipped in the tens to hundreds of millions of units per year.

Were I designing this system, I would connect the shields at both ends. There are millions of modern systems that show this works.

7

The LVDS are differentially terminated (across phases) so there should be no net flow of current - it is balanced. The twisted pairs give you quasi TEM mode propagation so the concern of the shield here is purely electric field. terminate at one end as you have drawn to avoid introducing current loops.

Since you have implemented a differential CAN system and are going point to point rather than bussed the same arguments apply for this as for the LVDS. I would break the shield connection on the RHS but keep the one on the LHS.

Your power connection looks fine. All image current flow from the power will flow back close to the power coming in. There is no image current flow from any of the signalling because they are differential and terminated, so a signal ground return associated with power is fine.

You don't mention if there are other potential aggressor circuits/cables around. that may change this scheme.

For verification read Henry Ott's book on this subject "noise reduction techniques in electronic systems"

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  • Thanks for the answer. Why should I break the CAN shield on the RHS, and not on the LHS? – Rocketmagnet Mar 08 '13 at 18:27
  • Simply to make sure that the shields are all at the same equipotential. So if you have ground bounce on the RHS board ... it may not be a big deal. – placeholder Mar 08 '13 at 18:44
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    (may not apply to CAN) but I was taught to ground shields at the transmitting end : any noise picked up on the shield causes ground currents next to a robust transmitter instead of a sensitive receiver. –  Mar 08 '13 at 18:50
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    That's a good point, as Ott says there is no ONE solution, you have to analyze the dominant modes.. However, here which is the Tx ? I think, that breaking alternate paths for image current flow is more important. – placeholder Mar 08 '13 at 18:54
  • Yeah, it IS more difficult with transceivers... –  Mar 08 '13 at 19:02
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    For a 250mm length, 1Mbps CAN should not require any shielding except under the absolute nastiest of EMI environments. Assuming this isn't the case, why not just skip the CAN sheilding? – HikeOnPast Mar 08 '13 at 19:58
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    @HikeOnPast - Because I've already had the cable custom made. I had to have custom cables because it's apparently impossible to buy a slim twisted pair cable with 120 ohm impedance. – Rocketmagnet Mar 08 '13 at 20:04
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    @Rocketmagnet, got it. In that case, at least for the CAN buses, the shield termination doesn't really matter; do whatever is easiest from an assembly perspective. – HikeOnPast Mar 08 '13 at 20:07
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I've had problems with a similar arrangement where the R.H. 3V3 regulator circuit needed better decoupling to prevent switch-mode currents taking a partial route through data screens that were grounded at both ends. I'm not saying don't ground at both ends, just be careful about the 3V3 regulator if it is a switcher. The problem manifested itself as occasional data corruptions and I suspect that it was power current through the data screens that coupled to both wires in the twisted pair and caused the receiver common-mode "issues".

Andy aka
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