My gate opener uses a AC 240V to AC 16V transformer. I want to connect the transformer to the power supply on the wall and use an extension cable to transfer the output AC 16V of the transformer to the gate opener. But the distance between the power supply on the wall and the Gate is about 30m.

The Gate opener has a 12V battery inside. The specification is:

  1. Motor: 120 Watt 12v DC
  2. Transformer: 16v AC 1 Amp
  3. Battery change voltage: 13.8v DC

What kind of Cable I can use as the extension cable? Can I use a 2.5mm heavy duty garden ectrical cable?

  • 1
    \$\begingroup\$ Do you know how much current the opener needs at 16 VAC? This would help to calculate the minimum diameter for the cable. Will the cable be inside some type of conduit to protect it from earth contact? \$\endgroup\$
    – user57037
    Commented Nov 3, 2021 at 2:02
  • \$\begingroup\$ Low V conducts higher current . Use AWG tables for 1% loss to allow for 8x start surge \$\endgroup\$ Commented Nov 3, 2021 at 2:41
  • \$\begingroup\$ @mkeith Thanks. The door opener uses Motor: 120 Watt 12v DC, Transformer: 16v AC 1 Amp. Cable will not inside conduit. \$\endgroup\$
    – lei lei
    Commented Nov 3, 2021 at 2:49
  • \$\begingroup\$ It sounds like the transformer is too small. 1 Amp 16 V is only 16 Watts. The motor is 120 Watts (10 amps). Maybe you can put a 12 V battery at the gate, and use the power from the transformer to charge the battery. \$\endgroup\$
    – user57037
    Commented Nov 3, 2021 at 2:57
  • \$\begingroup\$ @mkeith Yes. The opener has a 12V battery inside. \$\endgroup\$
    – lei lei
    Commented Nov 3, 2021 at 2:59

1 Answer 1


You possibly need a heavier gauge wire than what the motor uses due to the distance.(if it wasn't for battery) .

At 16V a 250W, a 12Vdc motor uses 10A.

1A AC is just the battery charger.

2.5mm or AWG10 is 3.277 mohm/m x30m => 100mohm x 2wire x 1A = 0.2V loss on 16V or 1.2% loss.

You can use any wire for a float charger.

  • \$\begingroup\$ Thanks Tony. I do not understand the voltage loss you calculated. Why multiply by 10A but not 1A? \$\endgroup\$
    – lei lei
    Commented Nov 3, 2021 at 4:19
  • \$\begingroup\$ Loss at 1A would be insignificant (within normal line-voltage variation). That assumes the battery operates the motor, not just a (remote-control?) bit of ancillary equipment. \$\endgroup\$
    – Whit3rd
    Commented Nov 3, 2021 at 4:45
  • \$\begingroup\$ I started answer before you commented about battery . \$\endgroup\$ Commented Nov 3, 2021 at 6:47

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