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I'm using a 12 V, 3 A wall charger to supply a project:

How I plan on setting it up

The project includes a motor (which needs 12 V, 3 A), an Arduino, a servo, and some other components. Since I don't want to fry my electronics, I originally thought of using a linear voltage regulator like the 7805 but I don't want it to overheat.

I got a buck converter to have a steady 5 V supply instead of the linear voltage regulator.

My question is: How much current would pass to the electronics after the buck converter, and could it damage something by being 3 A?

I'm still pretty new to electronics and I'm not completely sure if this the best way of doing this project. Any help would be appreciated!

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    \$\begingroup\$ It's the arduino, servo, sensors etc (powered by the buck converter) that determine the current at the output of the converter. What is the maximum total power requirement for these modules? \$\endgroup\$
    – devnull
    Commented Jul 18, 2022 at 20:48
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    \$\begingroup\$ The power supply does not push amps. The load draws as many amps as it needs and the supply either is able to provide it or not. Likely a duplicate question. \$\endgroup\$
    – Justme
    Commented Jul 18, 2022 at 20:59

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How much current would pass to the electronics after the buck converter and could it damage something by being 3 amps?

If your motor uses 3 amps then, the 3 amps isn't like a rampant tsunami of electrons trying to invade any "low-lying" electronics that happens to be in the way. Current is used where it is needed as per ohm's law.

Of course you still need to make sure your wiring is good or you will get odd effects and possibly damage; look up star-pointing of cables.

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Power supplies do not deliver 100% of their rated current at all times. Given a 12V 3A supply, this specifies that the supply can deliver up to 3A at 12V, if the components being powered require it. An Arduino will never "request" (draw) more current than it needs (unless the Arduino is broken). The UNO, when powered from a 7V supply will draw 42 mA with no other sensors connected.

Take the current draw from all components being powered by the supply and add them up. Then subtract that from the supply's rated current. For example, given your 12V 3A supply powering an Arduino with sensors (~200mA), motor (~100mA), and a random few LEDs (~500mA), that's a total current draw of ~800mA. So your power supply will have 3000mA - 800mA = 2200mA left over to power other components you might add later.

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  • \$\begingroup\$ The current at the output of the buck converter is not the same as in the input. What is in the answer would be valid for a linear regulator. \$\endgroup\$
    – devnull
    Commented Jul 18, 2022 at 21:30

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