=20
What would be the difference, Rick ?
The circuit was a simple [PSU] -> breadboard for ease of testing -> [relays= ]=20
... where each relay has a common connection to 9v, and an individual conne= ction to ground. The relays only switch at 9v (well, about 8.2v actually, b= ut let's use 9v since that's what they say on them). It seemed to make sens= e to me that that was the current required, since the measured resistance w= as 163 ohms per relay in isolation, and=20
9 =3D 163 * I =3D> I=3D 9/163 =3D ~0.055A 8 * 0.055A =3D 0.44A... and the bench PSU was reading 0.4A (it's only got one significant digit= after the decimal point).=20
So (1) there's nothing else going on in the circuit, and (2) the relays req= uire 9v (or just under, to be pedantic). I can buy that the hold-in current= would be less than the activation current, but I'm not getting why the 440= mA isn't a good approximation to the activation current.=20
For what I want, it's entirely possible that the system could come out of a= power-failure mode and want to switch on all the relays immediately, so I = have to be able to supply the maximal activation current.
Cheers Simon. (Always willing to ask stupid questions in order to learn :)