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enter image description here

A certain quantity of these traces is connected straight to the ground-plane.

What happens when an unpowered board (and completely disconnected from any peripheral) is on a grounded antistatic mat?

Will the ground-plane reach equipotential with the mat, because the GND traces make contact with the mat?

Or nothing happens, because both the V+ traces AND the GND traces touch the mat?

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  • \$\begingroup\$ Is the board powered or unpowered? \$\endgroup\$
    – user39382
    Commented Apr 12, 2017 at 20:07
  • \$\begingroup\$ @duskwuff Good question. Thanks. Answer: unpowered. (Just edited it.) But also curious what happens if the board is powered on. For example: testing outside the case. \$\endgroup\$
    – user127725
    Commented Apr 12, 2017 at 20:09
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    \$\begingroup\$ I suggest an option C: the sharp bits on the board slightly damage the surface of the antistatic mat but nothing else happens. \$\endgroup\$ Commented Apr 12, 2017 at 20:57
  • \$\begingroup\$ Nothing to see here. Move on. You can even power the board up while it is resting on the anti-static mat, and it will function normally. Or at least, any malfunction will not be due to contact with the anti-static mat. \$\endgroup\$
    – user57037
    Commented Apr 13, 2017 at 6:12
  • \$\begingroup\$ @mkeith "Nothing to see here" is too simple. When you apply the same voltage to V+ and GND, nothing happens. So, when ALL pins touch the mat, you're applying the same voltage to ALL pins. Then a hypothetical static charge won't flow from the motherboard, via the mat, to earth. At least, that's my guess. (Yeah, that hypothetical static charge on the motherboard is very low. Maybe even near zero.) \$\endgroup\$
    – user127725
    Commented Apr 17, 2017 at 12:14

3 Answers 3

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What happens when an unpowered board (and completely disconnected from any peripheral) is on a grounded antistatic mat?

All potentials on the board settle to zero (ok maybe not the battery and RTC circuit) but they would float to whatever ground is (which is close to 0V) It would be almost like attaching a 1MΩ resistor to every point on the board and then to ground. If there was air in between the board and the mat it would be even higher.

schematic

simulate this circuit – Schematic created using CircuitLab

Will the ground-plane reach equipotential with the mat, because the GND traces make contact with the mat?

Eventually it will, depends on how much charge exists on the motherboard. which should be really low.

The mat is 1MΩ (or more, I just measured mine and across the whole mat is 10MΩ to the button, yes I realize its not super accurate, but its close) to ground, most mats are made of dissipative material so this also won't affect the high speed signaling (noticeably anyway) if you run the motherboard right on the mat.

We do have ESD tables and they are 1MΩ also but underneath that thin 1MΩ layer there is a conductive metal layer. I would not run a motherboard on this because of a risk of capactivly coupling the signals to ground or to each other. This probably wouldn't affect the operation of the board but it could so I would not do it.

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  • \$\begingroup\$ Yes, "All potentials [...] would float to whatever ground is" and i also agree with: "It would be almost like attaching a 1MΩ resistor to every point on the board and then to ground." But, connecting every trace to the same ground (which is the conductive bottom layer underneath the dissipative top layer) also means: 0V potential difference between every point = 0V between V+ points and GND points = no flow of charge. So, can there be a transfer of charge between the board and the mat in order to reach equipotential? \$\endgroup\$
    – user127725
    Commented Apr 13, 2017 at 8:48
  • \$\begingroup\$ (Equipotential between the ground-plane of the board and the conductive bottom layer of the mat.) \$\endgroup\$
    – user127725
    Commented Apr 13, 2017 at 8:49
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Most anti-static mats are static dissipative. The mat will not short out any traces, and unless you have very, very high impedance signals it should be fine.

If you have a conductive surface, things change. Some bags that are delivered with electronics are coated in a thin layer of conductor. Some people suggest putting the motherboard on this bag during testing, but this is not advisable, as it can short out connections.

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Remember that the mat has a pretty high resistance to earth, many megaohms. Earthed wrist straps are typically connected to earth through 1 Mohm.

This is to leak static away rather than short it hard to earth. Shorted static is likely to damage something its charged up on, such as when a board with a static charge on is dropped down on the mat. It also makes people jump. So the mat is not a sheet of earthed tinfoil. This is also for safety, as the consequences of an electric shock from the mains or wherever would be much worse if people were connected straight to earth by their earth strap or the mat they're leaning on.

Therefore the tracks on your board that touch the mat are touching a large resistor to earth. I would say that the impedance between V+ and earth on your board there is pretty low because of a lot of decoupling capacitors across the rail and a lot of conduction paths, some resistive. So the effect of touching either rail to the mat produces a series circuit with the impedance to earth much higher than the impedance between the tracks. Nothing dramatic happens.

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  • \$\begingroup\$ I'm gonna assume you meant "many, many megaohms"? \$\endgroup\$
    – Joren Vaes
    Commented Apr 12, 2017 at 20:25
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    \$\begingroup\$ Err...dunno what you mean? Ahem... (shuffles from foot to foot, avoids eye contact with 'edit'...) Thanks, @JorenVaes :-) \$\endgroup\$
    – TonyM
    Commented Apr 12, 2017 at 20:28
  • \$\begingroup\$ @TonyM Indeed, you want the static charge to leak away instead of a short. But, will it leak away? ALL pins touch the same mat = ALL pins touch the same ground = zero potential difference between V+ pins and GND pins. So, no flow of charge inside the board. Then how can the hypothetical static charge leak away (from the board, via the mat, to earth) if it can't move? \$\endgroup\$
    – user127725
    Commented Apr 17, 2017 at 12:19
  • \$\begingroup\$ @marty, a static charge creates a potential difference with respect to earth, not the other rail. I'm not sure why that can't move when you connect a resistor between said statically-charged surface and said earth. \$\endgroup\$
    – TonyM
    Commented Apr 17, 2017 at 12:25
  • \$\begingroup\$ @TonyM Thanks. Yes, if there is a static charge on the board, then there is a potential difference between board and earth. If you provide a dissipative path (with a resistor) the static charge can leak away. But now there is not just 1 path. It's like there is an individual path between every trace (on the backside of the board) and earth. This means that also the V+ pins are touching the mat. When V+ and GND touch the same thing, there is no potential difference between V+ and GND. So, any surplus electron will stay between V+ and GND. It can't flow to earth. At least, that's my guess :-) \$\endgroup\$
    – user127725
    Commented Apr 17, 2017 at 12:50

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