I'm in a pursuit of finding the fastest PCB relay there is and there is a Chinese manufacturer that produces DPDT PCB relays that are rated 5V, 12V and 24V. Which of these would have the fastest switching time? Can you conclude that without doing a measurement? (cause, you know, the shipping takes like 3 weeks or so...)
I'd argue that higher voltage rated relays would contain more windings, therefore would take more time to magnetize up to the switching point.

P.S. knowing the "problem-solving mentality" of the SO folk, I do not want to switch to a SS-Relay or transistor, I am building a highly useless thing called a relay CPU :)

Edit1: thank you for all the comments. Usually, cheap chinese manufacturers don't provide useful datasheets at all. And buying more expensive relays can get really expensive quite fast. But still, thank you all for infos :)

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    \$\begingroup\$ the voltage of the relay doesn't directly say anything about its speed. However, anyone who produces relays and is worthy of your money will have characterized the relays they sell and will give you a datasheet. No info from the manufacturer, no business. It's really as simple as that. \$\endgroup\$ – Marcus Müller Oct 6 '17 at 21:31
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    \$\begingroup\$ For some reason I have a feeling the switching time is not going to be your bottleneck... \$\endgroup\$ – Eugene Sh. Oct 6 '17 at 21:31
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    \$\begingroup\$ @Trevor but it's current that creates the magnetic force that actuates the contact. If the coil resistance is scaled to the voltage that just ends up being a wash. \$\endgroup\$ – ratchet freak Oct 6 '17 at 21:33
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    \$\begingroup\$ Never hurts to get this documented on a question like this. web.cecs.pdx.edu/~harry/Relay/index.html \$\endgroup\$ – jonk Oct 6 '17 at 21:38
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    \$\begingroup\$ "I'd argue that higher voltage rated relays would contain more windings" ... more likely the same number of windings and thinner wire ... \$\endgroup\$ – Trevor_G Oct 6 '17 at 21:39

We know that ampere-turns is constant for the same force, so the number of turns n must be proportional to the reciprocal of the nominal coil current. The inductance is proportional to n^2.

We can also see that coil power is generally constant for a given design of relay, across coil voltage ratings. So the coil resistance must increase with the nominal voltage squared.

The effect of the inductance is a time constant L/R. Since L and R both increase with the square of nominal coil voltage rating, there is no first order effect on the inductance-delayed rise of force. Of course there is inertia in the moving parts and the delay due to the mechanical parts moving will not change with coil voltage rating.

If you look at data sheets from reputable makers such as Panasonic you will see the same max/nominal operate times stated across coil voltages.

For those who want to get an intuitive feel for this, the wire is thinner, there are more turns, and it fills about the same volume on the bobbin.

The release time is generally stated to be lower than the operate time, but take care that this is usually specified without a flyback diode- which will greatly increase the release time.

  • \$\begingroup\$ If one really wanted to activate fast, perhaps a trick from the stepper motor world could be borrowed: choose a low voltage coil, but implement a high voltage driver to overcome the inductance, with chopping current control to keep from then burning out the coil. But really fast applications might consider solid state switching, if possible. \$\endgroup\$ – Chris Stratton Oct 7 '17 at 16:36
  • \$\begingroup\$ +1 for complete answer. Pretty sure the mechanical inertia will be orders of magnitude longer then inductive delays anyway. \$\endgroup\$ – Trevor_G Oct 7 '17 at 16:36
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    \$\begingroup\$ @Trevor It's a little complicated- the inductance which I referred to (for simplicity) above as if it was a constant is not really constant- it varies as the relay clapper moves and closes or opens the magnetic circuit. During drop-out the inductance drops rapidly as the clapper leaves the pole piece, causing the coil current to increase just as the spring in the contacts is being released, assuming a flyback diode, further slowing the drop-out. I think you're right in that the inductance is much less of an effect on the operate (pull-in) time than the mechanical inertia. \$\endgroup\$ – Spehro Pefhany Oct 7 '17 at 19:36

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