1
\$\begingroup\$

I have a cheap PSU of this sort, used for my 3d printer. It has a 110V input, and a 12V output with 20A maximum current. Unfortunately, it doesn't have detailed specs/data available beyond that.

Now, I would like to place a simple ON-OFF switch (like this) at a convenient physical position, and in series with the AC power line.

Of course, it's important that the switch I choose (for example a 3A-rated one) is safe enough for the input current of the PSU. By conservation of energy, can I assume that the input doesn't draw more than (12 * 20 / 110) = 2 to 3 A? Or will it be significantly more than that, since I'm underestimating losses in the form of heat?

\$\endgroup\$
2
  • 1
    \$\begingroup\$ You are correct, and the type of switch you show would be fine just as long as it is mains (120 V) rated. I'd suggest you use a 10A 120 V switch which are very common. \$\endgroup\$
    – Jack Creasey
    Dec 17, 2017 at 16:37
  • 1
    \$\begingroup\$ You're right that the nominal current will about 2 A. However many Mains power supplies do have a large inrush current when they're powered up. Then large capacitors need to charge and this results in a short peak current. So 2 A will be too little and might work at first but probably will not last long. I would opt for a 5 A or more switch, that should be robust enough. \$\endgroup\$
    – Bimpelrekkie
    Dec 17, 2017 at 16:38

3 Answers 3

Reset to default
1
\$\begingroup\$

Cheap PSU may have not power factor corrector. In such case it draws non-sinusoidal current with RMS value which is larger than P/Uin (power per input voltage). Then, input power is larger then output (because of non-unity efficiency). So, even nominal input current is estimated at about 3 A. Taking into account already mentioned large inrush cuurrents, it's a good idea to use 10 A rated switch for long reliable operation. But 3A rated switch may also work for a while :)

\$\endgroup\$
0
\$\begingroup\$

Input power supply units must have current protection. Typically a fuse. The fuse burns out during a malfunction. When the current is much higher than the current of normal operation.

The rated current of the ON-OFF switch must exceed the rated current of the protective element in the power circle.

\$\endgroup\$
0
\$\begingroup\$

If you want to allow for the power supply being less than 100% efficient, add an efficiency factor to your calculation.

If we assume an efficiency of 80%, then it comes out to (12 X 20 / 110) / 0.8 = 2.7.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.