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I am trying to find relays for my application and I read a data sheet which looks fine but specifies

Minimum applicable load: 10mV 10µA

In my circuit, it is expected that the relay closes but no voltage and current is applied. You can think of it as 2 relays in series where one is open and one is closed, so there is no current. This sounds like something I'd do since I was at school.

Why would a relay require a minimum voltage and current on the load side? Is it allowed to operate that relay under my conditions or not? What could possibly break in a relay if I don't respect this requirement? What does "minimum applicable load" mean? When and how do I need to consider this value?

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    \$\begingroup\$ A link to that datasheet would help. Probably it's not a limitation of the relay itself but the smallest voltage / current at which the manufacturer has evaluated this model. So they guarantee that it will work with this load. It does not mean it will not work with less load. It is just not guaranteed. \$\endgroup\$ – Bimpelrekkie Jan 29 '16 at 12:17
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    \$\begingroup\$ The note says that it is a "reference load", and that you have to evaluate it with your own load. I think this confirms what I stated above. \$\endgroup\$ – Bimpelrekkie Jan 29 '16 at 12:22
  • \$\begingroup\$ If you never run any current through your relay what is it then good for? \$\endgroup\$ – PlasmaHH Jan 29 '16 at 12:23
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    \$\begingroup\$ @PlasmaHH: not "never", but it may happen. If there is current, it will be much higher than that. \$\endgroup\$ – Thomas Weller Jan 29 '16 at 12:37
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The primary reason that almost all relays have a minimum load requirement is that the mechanical action of closing coupled with an actual current flow are required to 'whet' the contact and break through a layer of oxidation that invariably builds up.

That is one reason that small signal relays generally use expensive contact alloys which resist oxidation, but as the phone company found out decades ago, even pure gold contacts can have issues in a high humidity environment. While oxidation doesn't affect the gold contacts, repeated cycles of moist/dry air would deposit an insulating layer.

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    \$\begingroup\$ Thanks. Given I have many close cycles without current and just a few with current (ratio ~ 1:100) but my current is significantly higher than the minimum (ratio 1:1000, say 10 mA instead of 10µA). Would that be sufficient to revert the oxidation effect? \$\endgroup\$ – Thomas Weller Jan 29 '16 at 14:25
  • \$\begingroup\$ Generally, I would say yes, but it depends on the actual composition of the relays and how much wiping action the contacts generate when closing/opening. Small signal rated reed relays might be the best bet for overall reliability. \$\endgroup\$ – R Drast Jan 29 '16 at 14:30
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    \$\begingroup\$ is it "whet" (ie sharpen), or "wet", as in "mercury wetted.." below? \$\endgroup\$ – James K Jan 30 '16 at 20:18
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    \$\begingroup\$ It should be "Whet". The action that results from the contacts wiping. \$\endgroup\$ – R Drast Mar 27 '17 at 9:33
  • \$\begingroup\$ I think you would find more sources referring to wetting current. - en.wikipedia.org/wiki/Wetting_current \$\endgroup\$ – KalleMP Apr 4 '17 at 8:32
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Mercury wetted (reed) relays are used where you cannot guarantee minimum load currents. These have a film of mercury on the contacts that will make contact even with no current as it forms a liquid junction. Generally the contact resistance will be slightly higher than the best gold contact resistance. The switching lifetimes may be much higher that other types. The load currents are generally lower as they are only usually needed for small signals.

In your case where the live loads are of a respectable value almost any dry relay type will work reliably.

The longevity of a dry contact will not appreciably change even if no current flows but age and lack of load current will slowly increase the low current contact resistance as mentioned by others. This means that marginal low current signals will suffer higher than specified contact resistance until such time that an adequate current has been switched.

EDIT:
It is also possible for the thin oxide layer that forms on the relay contacts to cause the low current resistance to increase and remain high until reasonable load currents have been switched that will 'burn' it away.

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  • \$\begingroup\$ Wouldn't an ordinary reed relay be sufficient? \$\endgroup\$ – Peter Mortensen Feb 5 '16 at 8:20
  • \$\begingroup\$ We do not know. The mercury wetted type does not have a degenerative contact resistance and at zero load may have near unlimited contact cycling rating. \$\endgroup\$ – KalleMP Feb 9 '16 at 14:09
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Relay will work at any load (below maximum of course). However if you use less than minimum permissible load, it may fail before it should (less switching operations). The reason for this being after time some corrosion can get on the contact. Minimal current is measured/calculated to be able to break through contact corrosion and it also slows down the corrosion process itself.

You can use the relay with less current, but if you are using it in a device that needs high reliability or you can't change the relay if it fails, it's not a good idea.

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The relay may not work as expected below the minimum load. If the current you're switching is below 10uA, you may find that the relay still conducts a significant current even in open state, due to internal capacitance. In a similar way, an open relay can easily pick up 10mV from EMI, so if your actual signal is that small, the relay may still seem to pass it through.

Of course, such phenomena depend on the frequency of the signals you're switching and your environment, which the datasheet stresses out:

Minimum applicable load is reference value. Please perform the confirmation with actual load before production, since reference value may change according to switching frequencies, environmental conditions and expected contact resistance...

So your relay will work just fine even if the signal you're switching is not actually present. Your signal must simply be above the threshold when it's there.

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  • \$\begingroup\$ Your answer is much about the open state. I wanted to know about the closed state. \$\endgroup\$ – Thomas Weller Jan 29 '16 at 12:43
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    \$\begingroup\$ In closed state you're getting a contact resistance below 50mOhm, maybe more if you're below 10mV. \$\endgroup\$ – Dmitry Grigoryev Jan 29 '16 at 13:04
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Either your application or explanation don't make sense. If there is no current flow, then you have a permanent open, this means you can remove the relay and save a component, time, and money. However, if you are controlling a 10mA load (per your comment), then you can use the relay and not "worry" about the "minimum applicable load."

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    \$\begingroup\$ As I said, consider 2 relays in sequence. If relay 1 is open, I can close relay 2 and there is no current flow. If I then close relay 1, the current flows. I cannot remove one of the relays in that case. \$\endgroup\$ – Thomas Weller Feb 10 '16 at 7:42

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