0
\$\begingroup\$

I need some advices on a design and not values in datasheet.

I know that the carry current is the current that can be supplied continuously to contacts without exceeding the maximum temperature when the contact are not switching.

  • how can I be sur that contact parts of the relay won't melt or stick together?
  • how can I calculate the real temperature at the contact?
  • Is the maximum current in a material only determined by Joule losses?
  • How can I calculate the maximum switching current possible.

Can someone please gives me some explanations?

thx

\$\endgroup\$
1
\$\begingroup\$

You don't normally calculate this stuff yourself, because the relay manufacturer doesn't give you all the information you would need. Instead they give you a whole range of information about currents, voltages and load types.

The ability of a switch to open a circuit reliably over a large number of operations depends on the circuit voltage (how many volts will be across the contacts once they're open), the load current, the contact material, the speed of operation, the gap in the contacts, any other arc quenching measures (blowout magnets,etc).

You pretty much have to rely on the relay manufacturer to tell you this stuff, or at least the consequences of it. If they don't then you can assume it's a low-quality or unsuitable relay which will fail, thereby saving you needing to do any further thinking.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Thanks for your response. However, I m more interested in the manufacturer part in order to really understand how things work and how they calculate it. \$\endgroup\$ – John Oct 13 '15 at 9:26
  • \$\begingroup\$ My limited exposure to people who make switches leads me to think there is very little calculation, and an awful lot of doing-what-was-done-before and testing things. The ohmic losses in the components of the switch are probably not a big deal in most situations, it's interruption which is the hardest bit. Contacts don't (generally) stick because of the ohmic heating of their metal masses, it's because of the very high temperatures and metal transfer in the arc. Of course, small relays tend to have lots of thermoplastic parts too, which can also be a problem. \$\endgroup\$ – user1844 Oct 13 '15 at 10:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.