7
\$\begingroup\$

Sometimes it feels like a ground-loop is inevitable. Please follow along the following thought-experiment. It would be wonderful if you could provide solutions to the issue described.

Imagine a board where one cannot use a ground plane. There are some chips on the board:

enter image description here

A digital or analog signal flows from Chip 2 to Chip 3. Suppose the trace cannot go straight to the other chip, but needs to bend around some other circuitry on the board.

Now consider the current flow when the signal moves from Chip 2 to Chip 3:  
 

enter image description here

This looks bad. The surface enclosed by the current flow is very large. If it is a digital signal, it will surely inject lots of noise into the circuitry soldered on that surface. And if it is an analog signal, it will probably absorb noise from the circuitry on that surface.

Trying to be smart, I come up with the following solution. Let's add a return path for the signal:  
 

enter image description here

Huray, it looks like the signal is very well protected now. The enclosed surface is much smaller. It won't emit magnetic fields, nor absorb them.

But wait a minute... do you see that? Do you see the ground loop I just created? Look again:  

enter image description here

Yikes! What now? Did I just improve the situation, or did I actually make it worse?

Please help...

\$\endgroup\$
4
  • 6
    \$\begingroup\$ Stick with the original layout just reroute the signal to parallel the ground. It is always possible to create artificial situations which require bad solutions. \$\endgroup\$
    – RoyC
    Commented Sep 10, 2017 at 9:08
  • 1
    \$\begingroup\$ This situation is not just artificial, but based on real-life situations. Sometimes you cannot have a ground plane. Sometimes you cannot "reroute the signal parallel to the ground", because there are things in the way (things that perhaps themselves need to be close to the ground). In many cases, one does not have enough space on the board to make everything ideal, that's when the above described situation can become reality... \$\endgroup\$
    – K.Mulier
    Commented Sep 10, 2017 at 9:12
  • 4
    \$\begingroup\$ If a signal traverses a route, you can always make the ground path traverse the same route, unless your board does not have enough physical space. In that latter case, you need a board with more space, or more layers. The way we generally avoid accidentally getting into this situation is to place all the critical paths, out and return, first, and then clutter the board with other stuff. Running out of routing space too early is a choice. Your answer is the signal should go north from C2, then east to C3, the same way the ground goes. \$\endgroup\$
    – Neil_UK
    Commented Sep 10, 2017 at 9:45
  • 1
    \$\begingroup\$ Does an extra layer with solid ground plane actually solve all of this? \$\endgroup\$
    – K.Mulier
    Commented Sep 10, 2017 at 9:49

1 Answer 1

Reset to default
12
\$\begingroup\$

The first one uses a star grounding scheme, which works well in some circumstances: low frequencies, absence of incoming EMI/RFI... which means it is an increasingly less useful scheme in today's world...

However, before talking about the loop, I'd like to point out that your design is single-supply, thus chips draw supply current and dumps it into the ground. Using long traces for GND in this case means this current will create a voltage against the GND impedance. Since GND is used as a voltage reference, your chips will have different reference potentials.

If these are single-supply opamps, they will process this as signal. If they are logic chips, ground bounce can corrupt logic levels.

In both cases a low ground impedance is beneficial, ie. ground plane. Without a ground plane you can route horizontally on toplayer, vertically on bottom layer, and create a grid of supply/ground, which was popular back in the days of huge boards packed full of TTL chips.

Note that if your chips are +/-15V opamps, then current flows into the supplies, but they are not connected to ground (except via decoupling caps) so in this case the varying supply current will only introduce extra noise on GND if the decoupling caps are badly placed/routed.

Now, back to your question:

There is no proper solution to this. Back in the day of single-sided analog boards (think a VCR or a tape deck), they were usually placed inside a metallic enclosure which acted as a shield, and there were no cellphones. Today's hifi stuff (for example) will usually have some shielding, if the faceplate is plastic, then you can expect a bit of conductive spray paint to shield against incoming EMI.

Also, these huge TTL logic boards were designed before modern electromagnetic compatibility directives, and frequencies were low.

Anyway. If the signals are fast, then they need to be routed close to the ground return current path, or else a loop antenna is formed, which will both transmit and receive, also the extra inductance will corrupt the signal itself.

Thus, if these are logic chips, the last layout in your post would be preferable. In fact, the "ground loop" would be one cell in an oldskool "grid ground/supply" board.

Note that the ground loop is a shorted turn, so incoming RF would have to be pretty strong to induce a high enough voltage into it to corrupt logic levels. The main problem would be that it would have a resonant frequency somewhere in the spectrum...

Now, specifically about audio grounding, I can only recommend this article from Bruno "The Man" Putzeys. Also valid for other analog applications. It is also a fun read.

\$\endgroup\$
9
  • \$\begingroup\$ Very interesting! You say: "Thus, if these are logic chips, the last layout in your post would be preferable. In fact, the 'ground loop' would be one cell in an oldskool 'grid ground/supply' board." What do you mean by this? Would your conclusions change if the chips are analog instead of logic chips? \$\endgroup\$
    – K.Mulier
    Commented Sep 10, 2017 at 10:25
  • 1
    \$\begingroup\$ I would note that apart from the potential resonance, a ground loop is no problem at all. Typically you care about the voltage developed across a ground conductor, and adding more paths is hardly going to make this worse. In the limit, this observation gives rise to the ground plane. Now for analog stuff you usually want to be careful about where you reference things, which is a slightly different problem, especially when you realise that an opamp is at least a 5 terminal device not a three terminal one (Think about a class AB output stage and where the currents flow in each quadrant). \$\endgroup\$
    – Dan Mills
    Commented Sep 10, 2017 at 11:17
  • 1
    \$\begingroup\$ Yes, the last sentence is why I mentioned "properly routed decoupling caps": both caps GND pins should connect to the ground plane in the same spot, so nonlinear class-AB currents going through the caps sum nicely into a linear replica of the actual opamp output current which is then injected into GND. No need to inject distorted currents into GND when all it costs to avoid it is put both caps on the same side of the opamp and a bit of trace ;) \$\endgroup\$
    – bobflux
    Commented Sep 10, 2017 at 11:24
  • 1
    \$\begingroup\$ Yes! Make the loops small (and have lots of them) and you minimise the impedance between any two points in the mesh that you care about, and thus minimise the voltages developed between those points. The only people who sweat ground loops are audio types who should be using balanced connections or hierarchical grounds and near DC instrumentation types who again should be separating ground and reference... Note that this applies within a device, external interfaces have other issues. \$\endgroup\$
    – Dan Mills
    Commented Sep 10, 2017 at 13:54
  • 4
    \$\begingroup\$ "The only people who sweat ground loops are audio types who should be using balanced connection" Absolutely! There are always ground loops, so it is best to use a design which doesn't fall apart when there are ground loops. This is done by keeping in mind what the signal reference is. \$\endgroup\$
    – bobflux
    Commented Sep 10, 2017 at 14:04

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.