I went through Sparkfun's website on how to power a project because I wanted to calculate how much battery capacity I needed for my project. I saw an ATmega 328 micro-controller that is used as the brains for the circuit: it draws about 20mA under normal conditions.

So that got me thinking: how much current will an Arduino Mega draw from a LiPo battery of 7.4V with 5,000 mAh 2s 25C under normal condition? Please anyone do help me out and pardon me, I am new to this.

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    Do you have a reasonably accurate multimeter? You can put it in series with the power circuitry and measure the power draw in your specific situation. – Paul Mar 10 '17 at 12:48
  • It should be a clamp ammeter otherwise you risk smoking your multimeter with the large currents. – Avamander Sep 7 '17 at 20:06
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    A clam amp meter for measuring milli amps??? – MatsK Sep 7 '17 at 22:17
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    I once measured the current draw of my Uno in normal conditions (active mode), and it was around 40 – 50 mA. A bare ATmega328P in POWER_SAVE mode is about 2 µA. Never measured on a Mega. – Edgar Bonet Sep 8 '17 at 8:02
  • ya, i dont think you need to worry about a clamp style meter for measuring the current draw of a microcontroller...... – Chad G Apr 6 '18 at 16:27

The important thing to realize here is that the capacity of the battery does not affect the amount of current drawn. Your Mega will draw the same whether it is connected to a 1000 mAh battery or a 5000 mAh battery. The larger battery will just last longer.

Tests on my Mega2560 Arduino show that (with no peripherals connected) it draws about 70 mA from a 7V supply, and slightly more (72 mA) from a 12V supply (the voltage regulator has to throw away some of the extra voltage as heat). (However, see below)

The exact amount it would draw in practice would be strongly correlated to what things you have connected. For example, LEDs, motors, shields, other devices.

I have a page about power saving that you might find useful if your objective is to minimize power usage.

Edited to add more about power consumption

It appears that my initial claim that the current consumption goes up with an increase in voltage was wrong. With Majenko's help I made up a detailed set of automated tests, increasing the Mega2560 power input by increments of 200 mV, taking a reading from both the power supply and my multimeter, averaging each one over ten readings, and plotting the results. This is what I got:

Current vs Voltage plot

The two measurements seem to be consistently different by 0.2 mA. What this shows is that at 7V, the current consumption is 65.3 mA, and at 12 it is 64.1 mA, about a milliamp less.

I do, however, stick by my original assertion that the capacity of the battery will not affect consumption, and a larger battery will just last longer.

  • Sorry, but I just read this now (since it was bumped to the homepage) and I'd like you to further explain the "the voltage regulator has to throw away some of the extra voltage as heat" sentence. The only thing that should increase the current consumption is the RN5C-D couple (20k, which draw 0.35mA at 7V and 0.6mA at 12V, for an increase of 0.25mA); the fact that you are experiencing a 8 times larger increase is puzzling on my side. Even thermal increase in the 1117 has a much lower impact (50°C change about 5% the quiescent current - which is 6mA, so 0.3mA). What am I missing? – frarugi87 Apr 6 '18 at 13:14
  • I'm seeing about 100mA from my Mega2560. It's static within about 1mA or so all the way from about 6.6V upwards. – Majenko Apr 7 '18 at 15:01
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    @frarugi87 - upon re-measuring I am still finding between 1 and 1.8 mA more consumed at 12V. I can't fully explain that but am working on it. Which revision Mega2560 board do you have? Mine doesn't seem to have a revision number, so it is probably the first one released. – Nick Gammon Apr 7 '18 at 22:24
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    @frarugi87 It appears that I was wrong. See amended answer. Much more extensive testing shows that the current consumption decreases slightly with an increase in voltage. – Nick Gammon Apr 9 '18 at 6:44
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    @frarugi87 You are right I am getting off the topic, but just for your info, running the loop backwards (12V down to 7V) gave a very similar plot, so it seems the part is heating up, and that is causing the difference. – Nick Gammon Apr 9 '18 at 21:59

I saw an ATmega 328 microcontroller that is used as the brains for the circuit: it draws about 20mA under normal conditions.

the datasheet provides the best guidance on that. typically 10ma - 20ma is fairly standard for 8-bit mcus. other components on the board, like opamps, or regulators, leds, may consume a little bit more so I would say 20ma - 30ma total is a good estimate for the board.

So that got me thinking: how much current will an Arduino Mega draw from a LiPo battery of 7.4V with 5,000 mAh 2s 25C under normal condition? Please anyone do help me out and pardon me I am a newbie to this.

the datasheet showed 20ma typical so I would put the board to 30ma total. 5000mah/30ma = ...

  • Note that one of those “other components on the board” is itself a microcontroller that may draw its own 20 mA. – Edgar Bonet Sep 8 '17 at 7:55
  • And looking at the measurements in another answer, the current is 65-70mA – frarugi87 Apr 10 '18 at 8:06

You may be right that AtMega chip will draw relatively small amount of energy. But there are other components in the Arduino that will draw much more. (Voltage regulator as a best example.) In the end, the current drawn can (and will) be 10x larger.

And there is another problem, you have to know how much power will connected sensors draw.

What you can do is build your own 'less greedy' Arduino and use sleep abilities of AtMega chips.

Here are some links to go through:




  • Why do you say tht the voltage regulator is an example of what draws much more current? usually the voltage regulator by itself consumes moderate current (6mA in the worst conditions), and only when using the Vin power supply. What drains the most are the two microcontrollers. and there are also some LEDs.... – frarugi87 Apr 10 '18 at 9:29
  • How did you get 6mA? – Divisadero Apr 10 '18 at 10:05
  • Sorry, I remembered wrongly. It's 6mA typically, 10mA worst case. From the schematic you see that the voltage regulator is an NCP1117; its datasheet has a parameter called "Quiescent Current", which has that value – frarugi87 Apr 10 '18 at 12:41
  • "Quiescent Current" is the current the regulator disipates when there is no load... POWER dissipation of the voltage regulator will be much bigger. Power, not current. For sure using 9V battery and regulator is much worse (energy-wise) than providing directly 5V. (Just guess, but i think 9V battery (as it is compound of cells) has less capacity than the same size 5V battery...) – Divisadero Apr 10 '18 at 13:34
  • BUT I do agree, that voltage regulator was not the best example to start with! – Divisadero Apr 10 '18 at 13:36

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