# Power Supply (9V 1A or 12V 1A versus 5V 1A)

Hello,

What would be the best choice of the three supplies if I want to get the most current (1A). My understanding is that if I go with the 9V 1A or the 12V 1A I would have to use the barrel jack and thus the voltage regulator which can cause more heating than going with the 5V 1A supply which I could use through the VIN and avoid the regulator all together).

I will be using most of the 1A…(for multiple sensors/loads)

Thank you

1A = 1A = 1A

Thanks. However design wise, would there be less heating by using the VIN?

…than going with the 5V 1A supply which I could use through the VIN and avoid the regulator all together).

No, Vin goes through the regulator.
You’ll be left with <=4volt on Arduino’s 5volt rail, which could cause problems.
5volt (regulated!) can be connected directly to the 5volt pin.
Leo…

Reduced heating of the on-board regulator is achieved by using as low a V_in as possible.
If you don't use the on-board regulator then it doesn't heat up.
Using the barrel jack gives you some protections, keeps you from increasing your ERA through carelessness, inattention.

Stay with the barrel jack; the closer to 7.5V, without going under, the better.
This will fill the bill --

7.5volt on the DC jack is the best option to power your Arduino externally, but it is not possible to draw 1Amp from the 5volt pin. The regulator is not rated for that, and it would dissipate 1.8watt if it could. I would keep it under ~700mA that way.
Leo..

If you using Arduino Uno, according to the schematic, if you run it from the 5V regulator (MC33269ST-5.0) on the board, it is 800mA, again less what the Arduino itself uses.

But only if you're running it off of less than a 9V supply. The regulator's current rating drops quickly with increasing input voltage, and at 12V it's actually only 400mA.

12V * 1A = 12W, 9V * 1A = 9W & 5V * 1A = 5W additional power will go to additional heat.

How to kill an Arduino Uno 5v pin, 650mA @ 14.8vin - Fire!

http://www.rugged-circuits.com/10-ways-to-destroy-an-arduino/

7.5volt on the DC jack (a diode-drop different from V-in) will dissipate:

7.5volt (DC jack) - 0.7volt (reverse protection diode) = 6.8volt (on V-in).
in this case 700mA will dissipate 6.8volt (V-in) - 5volt (out) = 1.8volt (across the regulator) * 0.7Amp = 1.26watt.
A hot regulator, but maybe just not hot enough to make it shut down (if the Arduino isn't in a closed box).
Leo..