Many OMRON's relay specify the contact resistance with this:

The contact resistance was measured with 10 mA at 1 VDC with a voltage drop method.

Such as http://www.omron.com/ecb/products/pdf/en-g6a.pdf.

Then, how the test circuit are arranged? It's apprently 1VDC/10mA is not equal the contact resistance they stated. If we already know the current flow through the contact, then why they specify the voltage '1VDC'? Apparently it's not the voltage drop on the contact. Any special reason?

  • \$\begingroup\$ Why do you bother, probably they used a DC source of 1V and have put a series trimmer resistor to adjust current to 10mA, at that point they have measured the voltage drop across contact. Or they could use a current source 10mA and they measured the voltage. But for sure the specified resistance is correct, right, so what's the point. \$\endgroup\$ – Marko Buršič Feb 20 '16 at 10:11
  • \$\begingroup\$ From ohm's law, we only need know the current flow through the contact, then we can measure the voltage drop on the contact. Why they specify their source voltage? Apparently, if they use a series resistance, and given the current flow through the contact, the source voltage has no direct relation the voltage drop on the contact. \$\endgroup\$ – diverger Feb 20 '16 at 10:17
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    \$\begingroup\$ My guess is that the test equipment is set to a voltage limit of 1V and a current limit of 10 mA. When the contacts close the equipment measures the voltage drop at 10 mA, but the 1 V open circuit voltage also effects relay switching performance. That potential difference might help at breaking trough any oxide layer on the contacts and significantly alter wear mechanics. As this is just a wild guess I won't post it as an answer. \$\endgroup\$ – jms Feb 20 '16 at 10:21
  • \$\begingroup\$ Who knows, maybe it's a typo or bad translation, or whatever. I think ii is a waste of time thinking the relation between 1VDC and 10mA, as they state that resistance is 50mohm, there is no relation in between. \$\endgroup\$ – Marko Buršič Feb 20 '16 at 10:22

Relay contacts are often not reliable for very small voltages and currents. 1V @10mA means that a voltage of 1V appear across the relay contact when open and 10mA flows when it is closed. In practice, this could be done more than well enough with a simple 100\$\Omega\$ resistor and a 1.00V source, since the failure level of 50m\$\Omega\$ would represent only 0.05% error (measure the voltage across the closed contact).

10mA and 1V is a relatively easy current and voltage for a relay to switch, at least a relay of this particular type.

The failure rate at low currents is specified as 'P' = 60% confidence of 0.1 failure in 10^6 operations at 10uA/10mV, and that's with the relay clacking away at once per second. Caution is called for if you are operating infrequently at low level currents. This kind of relay is sometimes used in precision instrumentation- and it may make sense to 'exercise' the relays before taking a reading.

  • \$\begingroup\$ Thanks for the explain of "failure rate". I find when OMRON give this parameter, they sometimes say: "This value was measured at a switching frequency of 120 operations/min and the criterion of contact resistance is 50 Ω." Such as omron.com/ecb/products/pdf/en-g6e.pdf. This mean, when they do the test, and when find the contact resistance achieve 50Ω, then they think it's a failure, right? \$\endgroup\$ – diverger Feb 21 '16 at 1:58
  • \$\begingroup\$ Yes, that is correct. 10uA doesn't give you much voltage drop with 50 ohms (0.5mV). \$\endgroup\$ – Spehro Pefhany Feb 21 '16 at 6:37

I'm with @jms on this. You should be able to simulate the Omron test using a bench power supply. Set the \$V_{MAX}\$ to 1 V and \$I_{MAX}\$ to 10 mA and try testing some switches or relays that haven't been switched for a while and see what voltage drop you get across the contacts. Repeat the test with \$V_{MAX}\$ = 24 V and see if it changes. I would expect to see a lower resistance when the oxide / grime layer is hit with 24 V. Edit: See Spehro's caution below and his answer to prevent high current from any PSU output capacitors.

enter image description here

Figure 1. Allen-Bradley pentafurcated contact.

I know that for industrial control station modular push-buttons there are often special low-voltage contacts available. For example explanation see the Allen-Bradley - Bulletin 800E Two-Across PenTUFF™ Contact Blocks. This type of contact is designed to give reliable switching on low-voltage, low-current devices such as PLC / micro inputs.

Word of the day


verb, past tense: bifurcated; past participle: bifurcated


divide into two branches or forks. "just below Cairo the river bifurcates"

I think Allen-Bradley's "pentafurcation" is a trademark.

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    \$\begingroup\$ I would strongly advise against doing this with a bench supply - most have a least some capacitance across the output, which makes it into a capacitive discharge (contact) welder. \$\endgroup\$ – Spehro Pefhany Feb 20 '16 at 11:09
  • \$\begingroup\$ Point taken. I have always considered them as having direct drive output with little capacitance effect. You are correct that this can't be assumed and your 100 Ω series resistor solves that. \$\endgroup\$ – Transistor Feb 20 '16 at 11:20

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