1

I am not too familiar in electronics, but I am currently using them for a Computer Science project.

I am connecting an electromagnetic door lock to a SPDT relay, so that the door can stay on NC and then be opened when I power the coil.

The problem is that I don't know how to to power the common pin as the electromagnet requires \$12\ V, 110\ mA\$. The best that I could find is a \$12\ V, 1\ A\$ plug that I don't think will work due to the power differences.

The link to the electromagnet I plan to use: https://uk.banggood.com/DC-12V-60kg-Visible-Installation-Door-Cabinet-Magnetic-Lock-Access-Control-System-p-1241735.html?rmmds=search&cur_warehouse=CN

Any help to do this would be much appreciated.

SamGibson
  • 17,870
  • 5
  • 40
  • 59
Marcel
  • 13
  • 2

1 Answers1

3

The fact that the power supply is rated \$12\ V\$ at \$1\ A\$ only means that the power supply can output up to \$1\ A\$. The actual current output will be based on the requirements of the specific circuit.

If I put to that supply a circuit that draws \$110\ mA\$, as in your case, that supply will output \$110\ mA\$ at \$12\ V\$, which is a little over a watt of power. If I connect a circuit that requires more than \$1\ A\$ to function, that specific supply won't be able to give enough current to the circuit since the maximum current the supply can provide is \$1\ A\$.

So if I want to power a circuit that requires a \$12\ V\$ rail at \$2\ A\$, for example, I'll need a supply which can output more than \$2\ A\$.

In your specific case a \$1\ A\$ supply for a \$110\ mA\$ circuit is well over enough. You could even do it with a much smaller \$0.5\ A\$ supply if all you need is \$110\ mA\$.

SamGibson
  • 17,870
  • 5
  • 40
  • 59
Davide Morgante
  • 148
  • 3
  • Your answer assumes that "power supply" means regulated, constant voltage supply. Of course most of the power supplies that you can by these days are regulated, constant voltage supplies, but even so, your assumption hides information that might help the OP to better understand what's going on. – Solomon Slow Jun 16 '20 at 18:43
  • @SolomonSlow Yeah..I'm not really an expert. This answer was migrated from physics SE, there my answer was, I hope, adequate. Here is not so! – Davide Morgante Jun 16 '20 at 19:26
  • Thank you, I did not realise that the supply would drop the amps to what it needed to be. Thank you – Marcel Jun 16 '20 at 19:47