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So if my arduino will draw around 800mA @5V and the dc-dc buck converter around 266mA @5V plus solenoid 1.2A @12V, is it a good idea to get 12V 4A (48W) power supply? Since Filip Franik told me that it is a good practice to get 2 times the current I need just in case. Then what will happen to the current that are no in use?

  • Good practice is to calculate your needs and multiply them by 2 "just in case" So if you need 1.5A find a power supply that has a 3A max.a So you will need one that's around 36W (3A * 12V) – Filip Franik Nov 2 at 20:08
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    do not use the arduino as a power supply ... it tends to become an expensive fuse – jsotola Nov 2 at 22:20
  • I am using arduino for relay module, lcd, buzzer and keypad. Not using it as a power supply, I am trying to make the 12V the power supply for solenoid and arduino. – Fiery Kenichi Nov 2 at 23:26
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So you want to use the 5V voltage regulator on the Arduino serve as the supply for your 5V components, and feed the 5V regulator from the 12V supply that powers your solenoid?

The problem with that plan is heat. The voltage regulator on the Arduino is a poorly heat-sinked (heat-sunk?) "linear" regulator. A linear voltage regulator is basically a solid state circuit that converts excess voltage to heat.

The higher the supply voltage you provide to the voltage regulator, the more heat it generates for a given current load.

If you feed 12V into the Arduino's voltage regulator, it needs to reduce the voltage from 12V to 5V. 12V - 5V is 7V. So it will convert 7V to heat. You're only drawing ≈220 mA. Total power that will be converted to heat is 7V*0.22A, or around 1.54 watts of power. You should budget for at least 50% more current than you think you'll need, so lets use 330 mA. 7.0*0.33 is about 2.3 watts.

From a little research I did, it sounds like a typical Arduino voltage regulator can dissipate around 1 watt of heat. You're well over that limit even at your low ≈220mA number, so you need a solution that won't overheat your Arduino's voltage regulator.

If you were to instead supply the Arduino's voltage regulator with 7.5V, it would have a lot less work to do. That's a voltage drop of 2.5V. So for a 0.33 max current draw, that would be 2.5*0.33, or about 0.83 watts. That would be under the 1 watt ceiling. However, it sounds like your project is based on a 12V supply.

You could instead use an external switching voltage regulator to reduce 12V to 5V. (search for "boost voltage regulator".) Those are much more efficient, and don't generate nearly as much heat. You could then feed regulated 5V directly into the USB input on your Arduino, as well as supplying it to your other 5V components.

  • I have decided to choose the 12V 1.5A (18W) supply connected to 12V 1.2A solenoid and a DC-DC Step Down Buck Converter, stepping down from 12V to 7V, that will cause the current to increase from 1.5A to 2.5A for the Arduino. AMS1117 5.0V SMD device (0.8A Adjustable/Fixed Low Dropout Linear Regulator) Max Input Voltage is 18V, Current limit is around 1.5A - 1.6A, My current is around 2.5A minus off 1.5A, what will happen to the 1A? – Fiery Kenichi Nov 3 at 17:48
  • I read a article, it says since there is no maximum current rating for the power supply connected to Arduino, the Arduino will draw as much current as it needs (and the voltage regulator can cope with). So its fine? – Fiery Kenichi Nov 3 at 17:48
  • I don't understand what you are describing. – Duncan C Nov 3 at 18:11
  • I totally forgot about DC-DC Step Down Buck Converter Output Current: Rated current 2A, maximum 3A (Additional heat sink is required). So from what I currently know buck converter will not create 2.5A just like that. – Fiery Kenichi Nov 3 at 18:12
  • I am describing about the the ampere since buck converter steps down voltage (while stepping up current) from its input (supply) to its output (load). – Fiery Kenichi Nov 3 at 18:15
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If my supply increase from 12V 1.5A to 4A. My arduino will draw around 800mA @5V (571mA @7V) plus solenoid 1.2A @12V, is it a good idea to get 12V 4A (48W) power supply? Since Filip Franik told me that it is a good practice to get 2 times the current I need just in case.

I know there is the efficiency of the converter is calculated by dividing the output power (Pout) by its input power (Pin) and then times the input current to get output current.

So for example 7V divide 12V is around 58.3% x 4A = 2.33A. Even if the solenoid takes 1.2A @12V, the current left is 2.8A, so the current is 1.63A, @ 7V. If arduino current draw what it needs and as long as it doesn't draw more than 2A, I think it is sufficient and safe.

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