Powering an Arduino Without Wires




Wanna ever thought of powering an Arduino without wires? Check it out.

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Step 1: Get These Stuff

You need these for this.

2.Battery Holder (6v)
3.4 x AA batteries

Anyone try it with a 9v battery. If works, please post a comment.

Step 2: Powering

Put in the batteries in the holder and hold it according to the pic.

Do not hold it the other way. Arduino wont light up then.

Step 3: Place

Place it accordingly.

Remember, it must touch the both ends of the ICSP headers.

Step 4: Ta-Da

Ta-Da , the Arduino lights up.

So you have learnt a silly process to power up an arduino without wires.

For projects, tape it to avoid power failure.

Comment? Suggestion? Problem? Please post in comments.
Feel free to send me a message.

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    25 Discussions


    5 years ago on Introduction

    I don't care if the words were included as part of urbanization. I refer you to proper English. Period. Learnt is NOT a word. Period.


    6 years ago on Step 4

    My comment is that the word is LEARNED, not learnt.
    Otherwise, neat trick!
    I would also do one other thing.
    The Arduino processor can heat up a lot when it has more than 5V going into it.
    It will run on just about anything between 3 and 5 volts though, so take out one battery, and replace it with a dummy cell made out of rolled up tinfoil. This will reduce the voltage to 4.5V and save possible damage to your ATmega328 chip.
    Remember, the power that goes into the ICSP header does NOT go through the 5V regulator on the board the way that power going into the power input jack does!

    2 replies

    Reply 6 years ago on Step 4

    Not a problem. The last thing you need is to have people coming back here screaming that your "trick" burned out their ATMega processor.

    I actually print warnings right on my circuit boards that have the "Regulator Bypass" function that there is a risk of damage to the processor if using more than 5V on an unregulated board.
    The reason I have a regulator bypass is for people that do want to operate their circuit on batteries. The regulator is a great hinderance when running on less than 5.5V input.

    AVR Warning.jpg

    Reply 6 years ago on Introduction

    The comments needed to be removed because the conversation was not constructive, somewhat rude, and there was no value added. It was very close to violating the "be nice" comment policy below the comment box.


    6 years ago on Introduction

    This is the worst method of powering arduino. 4 batteries like these give 6V which is too much for ICSP header. 3 would be better. 9V baterry will surely destroy your arduino. But only if you're too poor to buy 9V baterry socket. If not, you can connect it to Vin pin. It gets regulated...

    4 replies

    Reply 6 years ago on Introduction

    agreed 9V would kill it.. But if you use rechargeable AA's, that would be 1.2V each for 4.8V, which the 328P will run on just fine, as long as you don't have a ton of sensors. You could also use slightly used alkalines to have less than 6V for more safety, which is likely what this instructable is doing.

    I've never actually worked with an Arduino, but from what I understand, the Arduino wants 9-12v @ min250mA.

    This solution, while clever, would not deliver enough power to actually do anything.

    But it is a good concept, good job, expand on it!
    Keep it up! :)

    4 replies

    You are very much wrong. I refer you to the schematics: http://arduino.cc/en/uploads/Main/arduino-uno-schematic.pdf This particular arduino runs off 5V, the actual current consumption varies depending on what you're driving and the status of the microcontroller. I don't know where you got the 9-12V 250mA figures from, but that's entirely incorrect.

    Notice that in the schematic. the barrel jack connector feeds into two MC33269E-5.0T3 chips, these are linear regulators with a dropout of about 1V and a maximum supply of 20V. It also goes through a diode, I don't know what the value of this diode is but it's going to be either a silicon diode or schottky, so the input voltage ratings should be something like 6.2V (Schottky) or 6.7V (silicon) up to 20V or perhaps a more reasonable maximum voltage based on thermal characteristics. This is actually in-spec only, you could actually run at a lower voltage successfully, it would just be out-of-spec (which frankly is fine for most DIY projects)

    The ICSP header bypasses the linear converters, and are wired to the microcontroller directly. I refer you to the ATmega8U2 datasheet: http://www.atmel.com/Images/doc7799.pdf which states an operating voltage of 2.7-5.5V.

    Incidentally this means that with 4AA cells, you are risking overvolting the microcontroller. Four full alkaline cells produce 6V total, which risks blowing out the microcontroller (in reality, probably not much will happen, but you're pushing the limits a bit). Four NiMH cells would be safer, as they quickly drop to 4.8V total.


    Reply 6 years ago on Introduction

    if you did over voltage the uP, this would result in a thermo overload and there are builtin protections in the chip to cover this. However the initial spike my cause an issue if Murphy shows up and your chip is marginal...

    As I said, I havent used Arduino myself, I got the information from a site, I did a more thorough search now and came up with this: http://arduino.cc/en/Main/arduinoBoardUno wich states that the input voltage limits are 6-20V.

    As you mention, this is on the barrel jack, I havent looked at the schematics, so I'm sure you're right about the regulators.

    Thank you! :)


    6 years ago

    just solder a barrel connector to the baytery pack.then you won't risk shorting out your arduino by touching the wrong spot.

    1 reply

    Reply 6 years ago on Introduction

    This. It could be that the power regulator is not in between the icsp header and the chip. With a barrel connector you are always good (5v - 12v) as the power regulator is then in between.