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As I understand it inrush current is the current when the contact closes. Resistance is not minimum yet, and still inrush current can be several times the nominal current, like 80A on a 10A relay. How come that the inrush current doesn't weld the contacts?

edit
case in point: this relay can take 800A (!) inrush for 200\$\mu\$s

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  • \$\begingroup\$ The (W) in spec possibly means Wolfram (Tungsten). This metal is hardest one to weld. +AgSn etc, is the filler and coating for tungsten sponge. The choice of material is the reason that contacts can survive high currents. \$\endgroup\$
    – user924
    May 30, 2012 at 2:25

7 Answers 7

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Relays aren't perfect switches, and will have a certain contact resistance, which may be several tens of milliohms. In power applications this has to be taken into account. A relay with a contact resistance of 10m\$\Omega\$ carrying 16A will dissipate 2.5W in the contact!

It has been suggested that contacts tend to weld more on opening than on closing. I don't think that's correct. Firstly, in most relays release time is much faster than operate time. Secondly, yes there is often a nasty arc when opening, but that arc is actually a sign that contact anode and cathode are in fact separated, and then they can't weld anymore. That doesn't mean that arcs are harmless. They're a powerful HF transmitter and cause much EMI. And they burn the contact's coating. In AC switching they will extinguish on the zero-crossing, after maximum 1/100 or 1/120 second (doesn't count for very high voltage switching), but in DC this may take longer. That's why DC ratings for a relay will be significantly lower than AC ratings.

So contacts tend to weld upon closing, and you rightly mention that the contact resistance isn't minimum yet during the inrush, so it looks strange that exactly then a higher current is allowed. It's all to do with time. Closing a contact usually takes several ms, but most of that time is used to build the magnetic field in the coil and the travel of the contact's anode also takes some time. The actual time between first contact and final closure is very short. Add to that that the current isn't 80A yet at the first contact; current can't go from 0 to 80A in a nanosecond. So while the current builds up, the resistance decreases. All in a very short time, so that the total dissipated energy in general won't be too high.
For situations where this isn't good enough there are relays with a separate faster tungsten contact to improve closing performance. (In Dutch it's called "voorloopcontact", I don't know the name in English.)

edit (in reference to comments to different answers)

There was some discussion whether the relay would weld upon opening. Steve quoted Siemens: "[welding occurs upon] opening and immediate reclosing of contacts". If you want to weld the contact this is definitely the way to do it. Opening will probably draw an arc, and reclosing it during this arc means that there's current during the high resistance phase at the first contact. Contact + high energy = high risk of welding. Even if the arc already extinguished the air in the gap may still be ionized, meaning that it may breakdown during reclosing before contact is made, and that there's already current when closing. So basically the same situation.

Further reading:
Tyco application note: Relay Contact Life

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    \$\begingroup\$ Has anyone ever seen a graph plotting contact resistance vs time during closing? \$\endgroup\$
    – stevenvh
    Jul 10, 2011 at 7:04
  • \$\begingroup\$ 1up. Also, it depends on the load. A purely resistive load has an inrush current equal to its steady-state current (if it weren't for things like incandescent bulbs that have a lower resistance while still cold) and won't cause as much trouble upon switching on as a capacitive load. Really bad inrush events happen when empty capacitors are connected to a supplying line and charge very rapidly while a contact may not yet be fully closed. \$\endgroup\$
    – zebonaut
    Jul 10, 2011 at 12:25
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    \$\begingroup\$ Although relays could only actually weld upon closing, an arc generated on opening could certainly cause damage to the contacts in such a way as to increase their initial resistance the next time they close. \$\endgroup\$
    – supercat
    Jul 10, 2011 at 18:23
  • \$\begingroup\$ @stevenvh - I don't think we have a specific term for a pair of tungsten contacts that close first and open last. They are quite common on power relays - a set of tungsten contacts to take the pain, and set of lower resistance contacts to carry the current. "voorloopcontact" just means fore contact or in-the-front contact, which makes sense. \$\endgroup\$
    – Cybergibbons
    May 29, 2012 at 13:45
  • \$\begingroup\$ @Cybergibbons - most literally "pre-run". Odd that there wouldn't be a name for it. What do they write in the datasheet then? Anyway, thanks for your reply. \$\endgroup\$
    – stevenvh
    May 29, 2012 at 17:24
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The effect of high currents is to heat up the metal. The resistance of the metal converts electrical energy into heat energy. If the current spike is very short, the thermal energy is not as great, and it can be dissipated before it melts the metal.

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"Contact welding in contactors occurs only during closing of contacts or during opening and immediate reclosing of contacts"

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  • \$\begingroup\$ This seems to confirm what I said: it's upon closing that they weld. If it would have welded upon opening there would be no need to re-close. \$\endgroup\$
    – stevenvh
    Jul 10, 2011 at 17:56
  • \$\begingroup\$ Yeah, but it needs to first open to draw the arc then close to weld. I think we both had a piece of it, but the wording here makes it clear:) \$\endgroup\$
    – SteveR
    Jul 10, 2011 at 20:34
  • \$\begingroup\$ Getting back to the original question I still say It is somewhat safe to exceed the rated relay current while the relay remains closed for short durations. If the relay opens, then re closes during that high current period (whether it be contact bounce, control coil circuit problems, or whatever) than its very possible the contacts will weld. If the actual weld occurs during the arc or after when the relay drops (gotta drop sometime!) is irrelevant. \$\endgroup\$
    – SteveR
    Jul 10, 2011 at 21:43
  • \$\begingroup\$ This is a phenomenon well known to electricians. In fact, it can be simulated by a double pulse applied to an electrical system. The first pulse, when it ends, leaves a non-zero residual "condition". The second pulse starts with non-zero initial conditions and generally "reinforces" the effect of the first pulse. \$\endgroup\$
    – user288518
    Jun 11, 2021 at 15:36
  • \$\begingroup\$ This phenomenon also appears when lightning strikes a high voltage line. The wave propagation makes it possible to double the voltage of the wave arriving at an electrical substation. The only remedy is to use a piece of line of adequate length at very low impedance. \$\endgroup\$
    – user288518
    Jun 11, 2021 at 15:46
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It is the current that the relay has to interrupt that you need be concerned with. That is when the contacts tend to weld.

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  • \$\begingroup\$ I don't think that's correct. I've known contacts welded which were switched off with no current. \$\endgroup\$
    – stevenvh
    Jul 10, 2011 at 2:03
  • \$\begingroup\$ @ stevenvh - it's True that if you put enough current through a closed contact it could weld together. However when a contact opens with high current it has a tendency to arc. This arc will be greater at DC potentials rather than AC. AC will not be a problem usually if it opens close to or at the zero crossing interval. It is this arc that often welds the contacts. I've seen it many times. Think of how an arc welder works by drawing an arc along its electrodes. \$\endgroup\$
    – SteveR
    Jul 10, 2011 at 10:24
  • \$\begingroup\$ The arc welder is not a good comparison; there you deposit material from the electrode, you're not welding the electrode to steel. The electrode may stick to the steel if you don't strike it properly, that's welded, but you can never have the electrode welded to the steel as long as you have an arc. Never! I also mention the arc, and the opening of the contacts in my answer \$\endgroup\$
    – stevenvh
    Jul 10, 2011 at 10:48
  • \$\begingroup\$ When the contact opens under high current and begins to arc, the arc will sustain if there is a sufficient current remaining. This arc tends to "Hang Fire" resulting in a welded bond of the two contacts. \$\endgroup\$
    – SteveR
    Jul 10, 2011 at 10:53
  • \$\begingroup\$ I still don't agree. Even if the arc persists, the contact opens wider and wider, there's nothing that brings the contacts closer again. And that's needed for welding, because there's no material to fill the gap, like your electrode would provide in arc welding. You can only weld contacts together if they're in contact already. Once you have the arc while opening it's too late for that. \$\endgroup\$
    – stevenvh
    Jul 10, 2011 at 11:00
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Well, in general it welds pretty easy :-D

If it does not weld for you yet, it will, or you have your 80A peak for too few microseconds.

Alternatively, if it's AC current, you was lucky too many times to switch at 0-crossing.

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One reason missed in the answers so far is that the current is limited by contact heating. Over a short time frame the heat transfer from the contacts is higher than that of 'steady-state' dissipation because the thermal gradient leads to more heat transferred through conduction. More heat transfer means a cooler 'hot-spot' and hence higher current handling.

It should also be noted that this will work better when the relay is cold - once the relay has warmed to operating temperature, the tolerance for such a current 'surge' will be less. This can cause problems in equipment with NTC inrush limiting - If the equipment was switched off for a short time, then switched back on, the caps might have discharged, but the relay and NTC may still be warm. This would create an above-normal inrush when the relay is least able to handle it.

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  • \$\begingroup\$ The integral of i2t.dt precisely makes it possible to calculate the global energy of the impulse and to predict (by the global knowledge of the "physical system") the rise in temperature which results from it (complicated calculation, but tests allow everything. likewise to assess the effects). \$\endgroup\$
    – user288518
    Jun 19, 2021 at 9:40
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The inrush current can be specified for I2t pulses and not just for permanent RMS currents. DHR circuit breakers are also, as relay, specified in the main train lines for overloads. See the triac specifications for information on the curves giving the peak current pulse for a given pulse duration. The relay does not handle the inrush current, but it can support it ... if well sized. You need to know how inrush current can be.

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