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Would it make sense to connect the GND pins of an 8-pin PCIe power connector to a common point ground (which is at earth potential) and then plug that dummy connector into my new GFX card, before installing the GFX card onto the motherboard?

The setup would look like this. enter image description here

This way the motherboard and GFX card are at the same potential, before making contact, and while installing. After installing, when the GFX card is in place, i will remove the dummy connector and attach the real power connector from the PSU.

Good plan? Bad plan?

Edit: 1 answer suggests touching the bracket/faceplate to equalize the charges.

But, if the bracket is isolated from the GPU (and the surrounding circuits on the card), then touching the bracket won't do anything to the card. Which can be a good thing. Because you won't damage anything. But then a potential difference between the graphics card and the motherboard (on which it is supposed to be installed) may remain. That can result in an ESD event at the moment the graphics card slides into the PCIe x16 slot.

So, howto avoid ESD on a new graphics card? (Besides leaving it into the box.)

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  • Get an antistatic wrist band, wear it and connect it to the chassis. Don't wear fleece. – winny Jan 09 '17 at 12:49
  • @winny You mean: bonding my body to the ATX case? That way, if the bracket is isolated, then still there can be a potential difference between the motherboard and the new graphics card. –  Jan 09 '17 at 13:18
  • Yes. Start by grabbing the graphic card by the bracket and you will be fine. If you have an anti static mat, you can ground that, yourself and both the computer and graphics card by putting tha latter on the mat, professional style. – winny Jan 09 '17 at 13:23
  • Yes, this is what everyone else does. The card doesn't magically acquire a high potential, but humans and their clothing can through tribocharging (rubbing). The "antistatic" bags are usually slightly conductive, enough that simply opening the bag while wearing a wrist strap will equalise it. – pjc50 Jan 09 '17 at 13:37
  • Thanks for your answers. Yes, i will discharge myself first. But, if i would be a magician and somehow, without touching anything, i could (by thinking about it) unpack the sealed box (with the graphics card inside) and slide the graphics card into the PCIe x16 slot on the motherboard. (Which is already grounded to earth, thru the case.) Then could the potential difference be big enough to cause a minor ESD-event, in theory? –  Jan 09 '17 at 14:09
  • In theory, all sorts of things are possible. "Without touching anything" makes it a lot harder! I suppose wearing insulating ESD-prone gloves (polyester) could achieve that outcome. – pjc50 Jan 09 '17 at 15:39

1 Answers1

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Bad plan because:

no one does it that way because it is not needed

you could cause other problems by making extra connections

The best way to do this is like in your picture but without connecting anything to the graphics card. Personally I would also not connect anything to the PC case because if you place it on the ESD mat, it is sufficiently grounded already.

You're worrying too much about ESD like many people do who do not understand ESD.

A graphics card by itself is not even that sensitive to ESD, you'd have to do crazy things to damage it by ESD. The same is true for the PC's motherboard. With your wristwrap + ESD mat and proper grounding I don't believe you are even capable to destroy anything as no charge buildup can occur.

When you pick up the graphics card and it is charged, it will discharge via you and your ESD wriststrap. You can also place it on the ESD mat and any charge can flow away.

Bimpelrekkie
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  • Thanks for the reply :-) What other problems can happen by making extra connections? –  Jan 06 '17 at 13:12
  • Other problems: if you do the connection properly there should be no problem I guess. But the connection is not needed so why make it then ? You misunderstand ESD, if the card is ESD charged, the charge can escape via **any** connection on the card. ESD is not like a battery which needs a loop. ESD is a charge buildup like in a capacitor. So there is no need for GND and V+ to both touch the mat. The discharge will happen through the metal part near the connectors. – Bimpelrekkie Jan 06 '17 at 13:21
  • You might want to read the (long) 2nd post here: http://www.eevblog.com/forum/suggestions/episode-suggestion-esd-is-a-myth/ by AndyC and gain a better understanding. He summarizes the practical implications of ESD. – Bimpelrekkie Jan 06 '17 at 13:25
  • Yes! But when V+ and GND are at the same potential (i.e. connecting to the same mat) nothing happens. The charge (inside the GFX card) leaves the card when there's enough potential difference to overcome the resistance (otherwise everything would already been discharged thru the air) or when the resistance suddenly drops, for example: the moment you touch it. This discharging can go 2 ways: from the device to you or the other way around. –  Jan 06 '17 at 13:27
  • You're thinking "normal electronics" and not "static electricity". A charge buildup does not behave as you describe. **Any** conductive path (even many mega-ohms) is enough for the charge to flow away to ground. **Any** charge difference is enough to cause a discharge once there is a conductive path. Air is an insulator so the charge cannot escape through the air unless the distance is very small (for 1 cm you need about 30 kV potential difference). – Bimpelrekkie Jan 06 '17 at 13:31
  • I've been reading the 2d post in the [ESD is a myth](http://www.eevblog.com/forum/suggestions/episode-suggestion-esd-is-a-myth/) topic. Maybe, the answer, to the following question, helps me further: what happens when the ground of a device makes contact to a charged conductor, which is at a higher potential? (I understand what happens when the charged conductor is at a lower potential. Then current flows from the device to the conductor until the ground of the device is at the same potential as the conductor.) –  Jan 06 '17 at 14:07
  • Higher or lower potential, it does not really matter as we're dealing with **charges**. A charge has a certain value (large or small) and a sign: positive (too few electrons) or negative (too many electrons). When a conductive path is there the charges **equalize**. The electrons fill the holes left by the positive charges. A + 1 Coulomb charge exacly neutralizes a -1 Coulomb charge. Any charge left will neutralise to ground (the earth which is a large object) when a path is there. – Bimpelrekkie Jan 06 '17 at 14:17
  • When GND makes contact with some positive charge, this charge wants to take the path the other way around: to V+. Because, at that moment, V+ is charged more negative than GND. But, does this really happen? –  Jan 06 '17 at 14:48
  • *Because, at that moment, V+ is charged more negative than GND* No that is not true, the GND and V+ nodes are on the same device and they are connected via some conductive path (the components on the device). Then both GND and V+ are **the same** from that charge's point of view. Even if the device was a 9 V battery it would make no difference, the 9 V is much-much smaller than the potential energy difference between separate charges. An ESD discharge can easily be 1000 Volts. A few volts makes no difference there. – Bimpelrekkie Jan 06 '17 at 14:55
  • Thanks :-) Okay, when i create a good positive charge on some conductor somehow, such that there is a few kV potential difference between the conductor and the device, and i connect the conductor to the GND of the device, **then the conductor won't discharge?** (Let's assume i'm a magician and i can connect the conductor to the device without discharging to my body.) –  Jan 06 '17 at 15:20
  • *then the conductor won't discharge?* No, there will be an exchange of charges (one discharges bit the other charges) because conductor and device have different charges and when connected they will **equalize the charges**. The total resulting charge will be the average of all charges. – Bimpelrekkie Jan 06 '17 at 15:35