In a fan motor I have, the written specifications are as follow :

"dual power: 110/230v 50/60Hz 22/20w".

I would like to power the motor directly, without electronic parts between any 220V wall power socket and the motor ( only a standard power plug unit). To select the appropriate 220v power plug I need to know the amperage.

Using the formula A=w(20)/V(230), I get A=0.0869

  1. Are my calculations correct ?
  2. If they are, then I am lost. What kind of power plug could I use to get such a low amperage?

The plug only needs to support an amperage equal to or greater than your device requires. A plug that supports excessive amperage will not cause any issues.

  • \$\begingroup\$ Is is true for any other case for the selection of a power plug ? \$\endgroup\$ – Sam Jun 6 '19 at 19:51
  • \$\begingroup\$ I would add that even though @Sam will be powering the fan at 220 V, he should consider the worst case that someone may switch the fan to 110 V and expect the plug to be adequate (should make no difference in this case since it'll still be a small 0.2 A current draw). \$\endgroup\$ – RaphaelP Jun 6 '19 at 21:37
  • \$\begingroup\$ And I partially disagree with you last sentence since using a plug that supports excessive amperage could make @Sam end up with a heavy duty high power plug that expects thick wires. So, the possibly thin fan wires could make for a bad connection with the plug's terminals and get little support from the plug's casing. \$\endgroup\$ – RaphaelP Jun 6 '19 at 21:41
  • \$\begingroup\$ So to clarify, more amperage isn't a problem but a too large up-difference would require new considerations. Thank you guys! \$\endgroup\$ – Sam Jun 7 '19 at 17:40
  • \$\begingroup\$ The plug may be bigger than necessary, and that could be inconvenient but there won't be any safety issues. \$\endgroup\$ – Drew Jun 9 '19 at 2:15

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.