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Typically ESD diodes are placed at the input, output or external contact points for product to prevent product being damaged by ESD.

What I don't understand is where the return path is for this ESD shock on the battery operated device. Since battery positive and negative are relative terminals, battery negative connected to the ground net is just relatively the lowest potential in that circuit.

Where does the current go since it has no way to actually go to earth?

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    \$\begingroup\$ You do not always need a return path when charges move. Consider a charged baloon, it has charge, you can move it from one corner of the room to the other, you acting as its electromotive force, and no charge will have to flow the opposite way (even though, some of the charges on other objects in the room might redistribute them differently due to the change in the perceived electric field). Return path is a thing for circuit theory, where voltage and current are inherently intertwined. \$\endgroup\$ – Sredni Vashtar Oct 11 '20 at 15:42
  • \$\begingroup\$ It flows through the battery-powered-product's self-capacitance to the ground plane (in the case of formal testing). \$\endgroup\$ – Andy aka Oct 11 '20 at 16:33
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    \$\begingroup\$ @SredniVashtar return path is required for circuit analysis. Without return path, Kirchhoff's current laws don't work. I think it is better to use a model based on capacitance to understand and explain ESD rather than invoke the idea that charge is transferred without a return path. \$\endgroup\$ – mkeith Oct 11 '20 at 17:26
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    \$\begingroup\$ Charge simply cannot flow through a capacitance, so models which assume it does are fundamentally wrong. The only way the charge gets through is if there is a leakage path or another discharge. Rather, in the absence of an actual return path, charge flows into the self capacitance of the target, and stays there. If you measured the voltage of an isolated target with respect to something electrically neutral, you'd see a step change as each ESD event delivered a bolus of charge into that capacitance. Absent a leakage path, it would then remain for all time. \$\endgroup\$ – Chris Stratton Oct 12 '20 at 0:06
  • \$\begingroup\$ @mkeith that's exactly my point. Kirchhoff's laws are not generally valid. The fact that engineers keep trying pushing them even when they are not applicable (like KVL in Lewin's experiment) is due to the fact they are more comfortable using them (it's like that drunkard who looked for his keys under a lampost because... that's where the light was). A capacitor models charge displacement via induction - a discharge, where the charge is actually transferred from a surface to another is more akin to a failing capacitor, where there is a dielectric breakdown, than to a working cap. \$\endgroup\$ – Sredni Vashtar Oct 13 '20 at 8:32
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It's very tempting to casually say that the discharge is the return path, but it would be more correct to say that it is a path of resolution which dilutes an imbalance of charge created by some other means.

We tend to think of current flow circuits but actually, circuits are just one particular behavior of charge.

Another is the accumulation of charge (say, triboelectric, ie, feet on carpet charge). This is a "static" charge - absent a pathway to neutralization, the excess or shortage of electrons "just sits there" on a charged object.

When the charged object comes close enough to an object connected to a large reservoir of more neutral charge (either via wiring back to the earth itself, or simply a conductive object having its own charge capacity such as a doorknob) then the electric field resulting from the difference in charge may in exceed the breakdown voltage of air for a specific geometry, and a spark may result.

In the case of diodes intended to present a shunt path for ESD, that reservoir of more neutral charge could anything from an actual grounding system, to the bulk of the devices's own supply rails/network. Ultimately an ESD event is a brief current spike to equalize charge, and since the possibility of damage from over-voltage is ultimately about over-voltage relative to other semiconductor segments equalizing charge with the powered or unpowered supply rails is sufficient.

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  • \$\begingroup\$ It might be worth considering a Van de graaf generator here as a contained example of static electricity. The belt forms one side of the circuit, whilst the discharge path forms the return path. \$\endgroup\$ – user1937198 Oct 11 '20 at 17:53
  • \$\begingroup\$ You can imagine parasitic capacitors, and then the return path is the parasitic capacitance \$\endgroup\$ – user253751 Oct 12 '20 at 14:42
  • \$\begingroup\$ @user253751 not really, no. \$\endgroup\$ – Chris Stratton Oct 12 '20 at 14:46
  • \$\begingroup\$ @user1937198 How not? \$\endgroup\$ – user253751 Oct 12 '20 at 14:47
  • \$\begingroup\$ See the explanation above, and remember that charge can flow into a capacitor as a destination (eg, the self capacity of an isolated object) but not through one, so there is no actual "path back" absent an actual one. \$\endgroup\$ – Chris Stratton Oct 12 '20 at 14:48
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There is none unless you count the free space capacitance every object has with its surroundings (which itself is just a way to model how a body can hold a charge). The ESD transient is like the initial charge equalization and mediation that happens when you first hook up something for the continuous closed loop current flow you are familiar with when there are a bunch of transmission line effects as the two ends of the circuit "communicate" with each other to sort out what steady state flow is going to happen.

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  • \$\begingroup\$ You hit on a key point there, it is the capacity to hold charge of the electrically interconnected discharge target(s) that matters; their capacitance with respect to ground is just a result of that when there happens to be a planet nearby. \$\endgroup\$ – Chris Stratton Oct 11 '20 at 17:43
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In a real ESD event from a human, there isn't exactly a return path. The human somehow acquires a net charge, and then, when the human touches some other object, some of that charge migrates into the object. Generally speaking, very little charge transfer occurs from a person to a battery powered device unless the battery powered device is somehow earth grounded. This is based on my personal experience in handling battery powered devices.

However, during ESD testing the situation is a bit different. In testing the table and the ESD "gun" are grounded together and to earth ground. There is capacitance between the device under test (DUT) and the table, which is metallic. That capacitance may provide the return path for the ESD gun. For devices which plug into mains power, the device is also earth grounded via the power outlet, and that may be an additional return path for such devices.

Whether an object is earthed or not is somewhat material in this discussion simply because the ESD event is more energetic when the object is grounded to earth.

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    \$\begingroup\$ While that may be descriptive of a setup, it's actually the movement of charge between objects of differing charge that is at play. No planet is necessary. \$\endgroup\$ – Chris Stratton Oct 11 '20 at 17:42
  • \$\begingroup\$ I mention the earth ground because it is a common node between the ESD gun, the table, and, if it is mains powered, the device under test. \$\endgroup\$ – mkeith Oct 11 '20 at 17:59
  • \$\begingroup\$ But while present, it's not actually important to the mechanism. If you find yourself with a surplus or deficit of electrons, then when you get near something with a different balance, a discharge can result. The only need for a third object (planet or cat) is to create the initial imbalance of charge on the source object. \$\endgroup\$ – Chris Stratton Oct 11 '20 at 18:01
  • \$\begingroup\$ The ESD gun is discharging a capacitor through the DUT to ground. So the common ground node is pretty important. \$\endgroup\$ – mkeith Oct 11 '20 at 18:05
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    \$\begingroup\$ You are hung up on the word earth. My point is just that the table and the ESD gun are connected together at their ground nodes. If the DUT is mains powered it is ALSO connected to the gun and table by way of its ground node. For battery powered devices, the metallic table is an electrode and the DUT is another electrode in a capacitor. \$\endgroup\$ – mkeith Oct 11 '20 at 18:09
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Thinking of a return path is not helpful in case of ESD. In general, charges flow from high to low electrostatic potential. In case of an ESD event, e.g. your hand is on a high potential with respect to the floor/earth and a discharge happens from your hand to earth through the device. Charges do not return to your hand. The point is that the potential difference between you and the device is not defined and therefore can be in the order of kilovolts. If you now make contact with the device, charges accumulated on your body move to the device due to this potential difference. This event can raise the potential of the whole device with respect to the environment, which is not problematic. All you want to avoid with esd protection measures is the potential difference appearing between any two points whithin the device's circuitry.

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  • \$\begingroup\$ But battery operated products are not connected to earth Ground or atleast the GND net of pcbs is not connected to earth ground. \$\endgroup\$ – Curious KP Oct 11 '20 at 15:38
  • \$\begingroup\$ @CuriousKP it doesn't really matter if gnd and earth are connected, I added further explanation. Hope it makes it clearer \$\endgroup\$ – Sim Son Oct 11 '20 at 15:56
  • \$\begingroup\$ @CuriousKP - they are through the capacitance to ground. About 4pF for a cell-phone. Larger equipment will have more, a human body about 100pF. \$\endgroup\$ – Kevin White Oct 11 '20 at 17:11
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Everything has capacitance to earth with larger objects having more.

When I was working on touch sensing in phones we would use the figure of 4pF for something the size of a cell-phone. A human body has about 100pF.

The return path is through those capacitances if there is no direct conductive path.

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    \$\begingroup\$ It's not really the capacitance with respect to earth that is at play, it is in effect the capacity to hold charge. Capacitance relative to earth is in part a result of that, but it is not the cause. If you were floating in interstellar space with a silk scarf and a glass rod and after creating a net charge on each of those by rubbing them brought one near enough a neutrally charged metal sphere, you'd get an electric field and eventually a discharge in vacum... even with no planet handy. \$\endgroup\$ – Chris Stratton Oct 11 '20 at 17:36
  • \$\begingroup\$ @ChrisStratton - agreed, but since the Earth (with capital E) is the largest object around it acts a good proxy for the universe. It's capacitance to the universe dwarfs eveything else and the mutual capacitance to the Earth is much larger than to the universe. \$\endgroup\$ – Kevin White Oct 11 '20 at 17:44
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    \$\begingroup\$ That is precisely the mistake in this thinking. In most cases it is not the capacitance to earth which typically dominates, but the self capacity to hold charge. \$\endgroup\$ – Chris Stratton Oct 11 '20 at 17:57
  • \$\begingroup\$ @KevinWhite The capacitance of the earth, considered as an isolated sphere, is only around 710 microfarads. Its "capacitance to the universe" does not "dwarf everything else". A man-made capacitor 1000 times bigger than the earth's capacitance will easily fit in your pocket. \$\endgroup\$ – alephzero Oct 12 '20 at 0:43
  • \$\begingroup\$ The mutual capacitance of a human or device to the earth will be about 100pF or less. That is many orders of magnitudes lower than the capacitance of the isolated sphere the size of the earth. \$\endgroup\$ – Kevin White Oct 12 '20 at 14:18
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When some charge enters an IC pin, the typical ESD diodes shunt it to Vee, Vss, Vcc or Vdd.

So the charge ends up being dumped to the power supply.

This will be the same for battery or AC line operated equipment. There's no need for a GROUND at all.

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    \$\begingroup\$ It's perhaps instructive to think about what would happen if you were to drag your feet across a carpet in a tall insulating building, and then get your finger near a D cell sitting on an insulating platform. If you get a discharge to the negative casing, you are discharging to the capacity of that metal can to hold charge. If you get a discharge to the positive terminal, you are probably discharging through the electrochemical cell to the capacity of the negative terminal/can. Key in this is that discharge doesn't just happen, surface charges in the nearing objects re-arrange first.. \$\endgroup\$ – Chris Stratton Oct 11 '20 at 18:22
  • \$\begingroup\$ Very true. How electrostatic charges operate is astonishing. Very many years ago I was operating some AV equipment in a hotel. Recently built, it had air conditioning and low humidity. It also had a lot of 'nylon' carpeting. You could hardly walk any distance without discharging some charge to door handles etc. Most worrying for me, the kit I was using also suffered ESD strikes. They were so huge that not only could they 'blow' the 13A fuses in the UK mains plugs but you could hear them blow as well ! \$\endgroup\$ – Graham Stevenson Oct 11 '20 at 18:45
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Current isn't only the flow of charge. Displacement current, due to changing electric field, flows through insulators and vacuum. It's what allows current to flow through a capacitor. It's the sort of current that induces the magnetic field in an electromagnetic wave.

If you use Kirchoff's Current Law, you should realize that it involves all current, including displacement current.

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    \$\begingroup\$ Displacement current is a thing, yes. But displacement current will not let the charge delivered to one plate of a capacitor escape to the other. The capacitor still ends up charged, and stays so eternally unless there is some actual mechanism for the excess or defecit of charge to escape. \$\endgroup\$ – Chris Stratton Oct 12 '20 at 15:01
  • \$\begingroup\$ @ChrisStratton The question was what the return path for the ESD current is. The flow of current through the protection diodes is physical charge, but the return current is displacement current. \$\endgroup\$ – John Doty Oct 12 '20 at 15:08
  • \$\begingroup\$ You're missing that an isolated target ends up with a net charge. And you're assuming that there is something in proximity which can be a target of the displacement current, when in actuality nothing is needed. Yes, a spark discharge radiates energy as an EM wave. Yes, that can induce current flow if it ever reaches something. But the actual surplus or deficit of charge remains trapped, because charge is conserved. \$\endgroup\$ – Chris Stratton Oct 12 '20 at 15:11
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    \$\begingroup\$ No. You cannot apply KCL to a non-circuit. Remember these spheres are isolated - there's no path for any current take except for the actual discharge current which cannot be its own return or there would be no net transfer of charge. There will be redistribution of the surface charges in each as they near, and after the discharge, but that current can't "go anywhere". You simply cannot complete KCL for any circuit between the two, because there isn't a complete path, and KCL fundamentally requires one. \$\endgroup\$ – Chris Stratton Oct 12 '20 at 15:46
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    \$\begingroup\$ Displacement current is not current of moving charge carriers. It has the unit of current, and as such it is also a source of the magnetic field, but no charge is actually flowing. The concept of displacement current was introduced by Maxwell to satisfy charge conservation in Ampere's circuital law. \$\endgroup\$ – Bart Oct 13 '20 at 10:56

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