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I have a 500W 24V power supply for running 24V LED strips. The power supply says it's rated for ~21 amps (500W / 24V). Assume input is US 120V household power.

I want to control the 120V supply with a relay driven by 5V Arduino logic signal.

I'll be driving 4x 24V 96W LED strips.

From A = W / V:

A = (4 * 96W) / 24V = 16

Does this mean I need a relay rated for 16 Amps or 21 amps listed on the power supply specs?

OR

Do I work back to Watts from the relay's specs?

Example relay is rated at 5A / 220V AC:

W = 5A * 220V = 1100

Though I've read choosing a 16A or above relay is a good idea since it shouldn't be the first thing that fails in a power surge.

More info: Here's the power supply I have: https://www.amazon.com/gp/product/B01IU8QBCO/ref=oh_aui_detailpage_o07_s00?ie=UTF8&psc=1

And will be purchasing one of these for 12V landscape lighting: https://www.amazon.com/gp/product/B06XR3ZLSG/ref=ox_sc_saved_title_8?smid=A3S4SVMWTDK1H2&psc=1

Looking at a relay like this to be on the safe side: https://www.amazon.com/AC100V-250V-2-Channel-High-low-Trigger-Arduino/dp/B077W1NVLM/ref=pd_ybh_a_15?_encoding=UTF8&psc=1&refRID=RAE8FR66XX0GN9YQANXR

EDIT: Suggested alternative: Though I'm confident wiring mains I like to avoid it. I'd suggest anyone not confident to go buy one of these (I'll be getting one for my project): https://www.adafruit.com/product/2935

https://www.amazon.com/Iot-Relay-Enclosed-High-Power-Raspberry/dp/B00WV7GMA2/ref=sr_1_1?s=wireless&ie=UTF8&qid=1542226953&sr=1-1&keywords=Controllable+Four+Outlet+Power+Relay#customerReviews enter image description here

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  • \$\begingroup\$ check the power supply ..... it may have a remote shutdown signal \$\endgroup\$ – jsotola Nov 12 '18 at 19:24
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It depends on which side you are switching.

If you're switching the secondary then you got it right.

A = (4 * 96W) / 24V = 16

Choosing a 16 A relay here is fine but it might be a good idea to choose a higher rating to be on the safe side.

If you are switching the mains voltage, I suggest the following:

Looking at the information given we can see that the power supply has an efficiency of 82% and a maximum power output of 500W. With this information we can calculate the maximum power input by taking:

500 W / 0.82 = ~610 W

With this information we can now calculate the current on the primary side:

610 W / 120 VAC = ~5,1 A

So a relay rated for atleast 5 A would be enough in your case since you won't be using the full 500W.

Be careful if you choose to control the primary side!

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  • \$\begingroup\$ I am indeed switching the mains voltage. This power supply gets really good reviews but also doesn't seem to have any smarts to switch itself off when not under load (or at least control the fan). As this will be part of a home automation system it will be trivial to add logic to turn the power supply on/off when needed. But as you stated messing with mains power isn't to be taken lightly. Secondary side is fine since any relay will only be handling a small fraction of the total load (LED strips are cut into 11 different zones). \$\endgroup\$ – Geordie Nov 13 '18 at 0:49

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