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Even though this is already a well-debated topic, I couldn't find an extensive discussion about what really is drawing so much power in Arduino's board.

Here is the preamble: I have an Arduino Mega 2560, I measured its power consumption in a steady state with just a red LED turned on with a 560Ohm series resistor. The power supply (Vcc) was given to Arduino from the Vin pin and the measurement were as follows (below the schematic set-up):

  • 30mA @Vcc = 5.2V;
  • 65mA @Vcc = 8.3V.

Test condition

What I would like to understand is:

  1. From a hardware perspective, I don't see what is drawing all of this current in Arduino's board;
  2. Why the current drawn from the power supply changes so much with the power supply's value (i.e. what part of the board changes its consumption with the power supply value);
  3. It also seems that turning on an LED through a digital output of Arduino draws a higher current with a higher power supply (Vcc). Why is that?

Note that, by looking at the datasheet of the ATmega2560, the uC mounted on my Arduino, I saw that the current Icc drawn from the power supply is max 14mA (I suppose when no output is sourcing current, I couldn't find the test conditions, unfortunately). I thought the microcontroller was the major cause of power consumption, but it doesn't seem to be so!

Another thing I considered is the onboard LDO, but the input voltage should change the power dissipated by the LDO, not the current drawn from its output.

It's even more strange if you think that regardless of the Vin, all the electronics after the LDO is working at the same voltage.

EDIT1

After Majenko pointed out a big flaw in my set-up with his answer, I re-measured with Vcc = 12.1V and I measured approximately the same Icc observed at Vcc = 8.1V. I completely agree with Majenko's answer and I would just like to go a bit more in details about what happens while the board is working under voltage.

Here is my analysis:

The minimum operating condition when supplied from Vin is 7V, and that's the "knee" point in Majenko's answer.

However, the maximum quiescent current of the LDO is 10mA, and the Icc of the ATmega2560 is 15mA, so even if we suppose they don't draw any current at Vcc=5.2V, and then draw 25mA together at the minimum operating voltage at 7V, this doesn't add up to the actual increase of 40 or 50 mA that Majenko calculated I observed from 5.2 to 8.3 (which should be the same as from 5.2 to 7V).

And we're not even taking into account that at 5.2V the micro is perfectly able to turn on and keep on the LED, so it's somewhat working and already drawing some current.

Other things that we can take into account, but don't change the facts, are:

  • a second LDO (LP2985) to generate the 3V3, whose quiescent current should be around 100uA since the circuit is not drawing any current from the 3V3.
  • a second microcontroller related to the USB programming, whose Icc may be up to 20mA but this one is OFF when supplying from Vin. --> Edgar Bonet correctly pointed out that this is not true!
  • Some voltage dividers and comparators.

So what is the cause of this increase in current consumption from 5 to 7V?

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    There is one major flaw in your measurements: 5.2V is below the minimum input voltage for the board, so the regulator won't be regulating to 5V. Create more data points over the voltage range and draw yourself a graph. I'll bet it won't be a straight line. – Majenko Jun 7 at 12:10
  • It's not clear from your question how you are measuring the board's current. If you are measuring current draw at the Vin pin, that is still on the input or power-supply side of the LDO, not on the regulated side. If this is the case, increasing current draw (disspiation!) with increasing supply voltage is to be expected. A diagram of how you are measuring the board's current draw would help greatly. – JRobert Jun 7 at 12:11
  • There is a second MCU on the board, used as an USB–serial bridge, which draws its share of current. – Edgar Bonet Jun 7 at 12:18
  • Re “this [second microcontroller] is OFF when supplying from Vin”: No, it's ON. See the schematic. – Edgar Bonet Jun 7 at 15:21
  • Ah you're right! I checked the schematic but the UVCC pin tricked me :-) – Elia Jun 8 at 19:45
5

There is a basic error in your methodology which leads to incorrect conclusions.

You cannot determine what is going on from just two data points. Instead you need far more.

Here is a voltage sweep from 5V to 12V with a 0.05V resolution (overkill, but the default setting for the script I use) connected to a Mega:

enter image description here

As you can see the current steadily climes in a pretty linear fashion up to a certain point, then it flattens out.

Everything up to the point where it flattens is below the minimum input voltage for the board. During this time the voltage regulator isn't getting enough voltage, so can't regulate its output voltage to 5V. So the board is running under voltage at this point. Once it reaches that "knee" point the voltage regulator can regulate the voltage properly - so most of the components on the board start seeing a steady 5V and their current consumption flattens out.

There are a handful of components on the high-voltage side of the regulator - most notably the voltage divider for detecting the input voltage and feeding it to the comparator that drives the P-channel FET to turn the USB power on and off - which will consume slightly more current when the voltage increases, but that is only very small amounts.

So, in short, by comparing a voltage that is below the minimum input voltage for the board with one that is above the minimum input voltage, you are comparing an invalid result with a valid result and inferring that everything in between is valid.

| improve this answer | |
  • You're right about the big flaw in my set-up! I updated the question with new results at Vcc=12.1V and also extended the question thanks to your considerations. I hope you can share your thoughts about my analysis because something is still not in the right place. Also, how did you calculate the current consumption? I didn't know about any "mathematical" model. – Elia Jun 7 at 13:27
  • @Elia I didn't calculate anything. It's what my bench power supply reported as being consumed. It's all scripted through a USB connection to control the power supply. – Majenko Jun 7 at 13:30
  • @Elia As for the under voltage scenario, just think about Ohm's Law and what it means. – Majenko Jun 7 at 13:31
  • I thought it was a simulation, it's even better then! :-) I thought about it in terms of Ohm's law, also because the increase is quite linear, but if you consider 40mA at 5V and roughly 90mA at 7V, you get a differential resistance of roughly 40Ohm in that range, and I don't see anything that may show this resistance. Also, there are many semiconductors components that don't strictly follow ohm's law. – Elia Jun 7 at 13:37
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    @Ella Think about the simplest thing: the power LED. You have an LED with a forward voltage that is fixed. You have a resistor in series with it with a resistance that is fixed. You increase the supply voltage - what happens to that bit of the circuit? – Majenko Jun 7 at 13:38

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