Sorry for the newbie question (just started learning electronics).

I just saw a tutorial that says that the 5V output pin on the Arduino delivers more amps (if needed) when connecting more voltage to the board (connecting to the 9 - 12V instead of 5V). Does that mean that the 5V pin then delivers more than 5V as well?


Time for a crash course in electronics and power:

The 5V pin only ever provides 5V. That's why it's the "5V" pin. (assuming, of course, that you give the board more than 5V).

In simple linear electronics you have three values:

  • The voltage
  • The current
  • The load

Generally two of the three are either "fixed" or "known" and the third changes depending on the other two. For simple resistive circuits (where the load is a simple resistor) this relationship is Ohm's Law.

  • If the voltage is fixed, and
  • If the load changes, then
  • The current will also change to keep the relationship stable.

A "power supply" (any device which provides power to some other device) generally has two parameters: Voltage and Current. One of those is always fixed. The other is always variable, and quoted as a maximum that the supply can give.

For the Arduino's 5V pin the parameters are:

  • Voltage: 5V
  • Current: 800mA Maximum

That means that the 5V pin has a fixed voltage of 5V, but the current is free to change depending on the load up to a maximum of 800mA (or a lower limit if you are providing the input power from a source with a lower current. There is also heat to consider with the Arduino - see below).

The two values, voltage and current, combine together to give the "power" of the power supply, through simple multiplication. So the Arduino's 5V pin is 5V × 0.8A = 4 watts maximum.

When you provide power to the Arduino through the barrel jack it first goes through a linear voltage regulator. This is the device which takes the incoming voltage and reduces it to 5V. Since energy can neither be created nor destroyed it does that reduction in energy by converting some of it to heat.

Say you connect something up that needs 100mA and you plug in a 9V power supply to the board. Let's also assume that the Arduino itself needs 50mA (I like nice round numbers).

  • That's 150mA in total for the "load".
  • The incoming voltage is 9V
  • The outgoing voltage is 5V

The power of the input is thus 9V × 0.15A = 1.35W The power of the output is then 5V × 0.15A = 0.75W

The power difference is 0.6W

That 0.6W is converted from electrical energy to heat energy by the voltage regulator and "thrown away".

So the "heavier" your load (the more current you demand from the 5V pin) the greater the heat that is generated by the voltage regulator.

When the regulator gets too hot one of two things happen:

  1. A good quality regulator will shut down and you will lose power to the board
  2. A cheap Chinese regulator will likely "fail short" and feed your 9V direct into the Arduino causing components to release their Magic Smoke™ and stop working
  • 1
    Thank you so much, made things clear! :) – Araw Aug 4 '19 at 11:20
  • Don't you mean to say "A cheap Chinese regulator will likely "fail short" and feed your 9V supply voltage directly into the Arduino causing components to release their Magic Smoke™ and stop working?" (I've always loved the idea that electronics contains magic smoke.) – Duncan C Aug 6 '19 at 0:12
  • @DuncanC Yep - typo. – Majenko Aug 6 '19 at 0:19

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