I am questions regarding external power for the Arduino Uno. I’ve recently made an attempt at powering the UNO with this particular lithium ion battery:
It is a 6800mAh battery with 12V potential. However, my poor UNO fried in the attempt (Area between USB and power barrel adaptor really hot, especially the crystal clock shroud). The TX/RX lights no longer blink on it with USB in, and i can’t upload sketches to it. I still am not sure what went wrong with this battery pack, as my multimeter reads out 11.45V from it. I guess that is working the regulator a bit too hard on the UNO? Or perhaps its just cheap chinese battery pack syndrome? I’m leaning towards the latter…
I’ve since bought another Arduino Uno and a Freaduino Uno as well. The Freaduino claims to support a larger input voltage. I would like to attempt to use this battery pack again, but I just don’t need another fried Arduino.
Is there a particular way to test a lithium ion pack like this?
Could someone recommend a good rechargeable battery? I would ideally like a battery pack with external power barrel plug charging, such as the above lithium ion pack, and I am straying away from 9V packs and AA packs, as ideally the project battery’s will be enclosed with the systems electronics. I am all ears though.
Thanks for any help.
Your battery probably isn't the issue. A bad wall wart could kill the Arduino, but a bad battery will almost never have any bad voltage spikes (might not even be possible) they will just die. This is more likely an application issue.
You did connect the battery to the Vin pin right? If you connected the battery to the 5V pin it had every right to fry the board up like an egg.
Assuming you are trying to power the board through Vin, what other things besides the Arduino did you have connected to the 5V supply from the Arduino. The on-board regulator is only good for a couple hundred mA before you will start overheating the board. Hooking up a big servo or another large 5V load to that supply is a very common way of overheating the Arduino. My best recommendation for this would be to use an external voltage regulator to make the 5V. Get yourself an LM7805 and a heat sink and you will quintuple your available 5V current.
You could also drop the battery voltage to something like 7.6 V, as this would reduce the power dissipation in the voltage regulator significantly.
I powered the arduino through the barrel plug adaptor, next to the usb port. This is the proper way for a Vin, correct?
Currently, the arduino is driving two IC's, and two LEDs: An RFID transceiver (SM130), and bluesmirf bluetooth modem. Could you suggest an adequate way of measuring the current draw from the uno board? Perhaps there is a monitor i could view in Windows to see how much current draw is on the USB port?
No mechanical components, servos etc on this board. Just two radio chipsets.
are you sure that the "+" is the inner part of the plug ?
The Vin will accept 12v. It is best at about 7, but not required if not to much load on the 5v power.
I would remove all the other boards, and try the system. If it works ok, then start adding the other boards slowly.
Good luck, Jack
So i connected my loads (2 chipsets for my project) to the uno I didn't fry. Used the 12V lithium battery as mentioned, and kept my finger on the regulator IC. After 30 seconds, there seemed to be a steady linear temperature rise (as tested by my finger sensor :)).... I currently dont have a useful way of measuring current, but that regulator was about to become piping hot.
I do have a LM317 regulator on hand. Would it be wise to use it, and then use the battery as my Vin without barrel plug? I'm new to regulators as a whole, thanks for all the help so far guys.
To add, the chipsets I am using have datasheets claiming to draw 180mA, and 40mA. So It doesn't seem as if they are drawing too much current (theoretically)?
Yes, IMO, it would be wise to use it : the less the difference between supply voltage and output of the arduino regulator is, the less power is dissipated in the arduino regulator. 9V is a good supply value, then you could use the lm317 and fix it to 9V output . Doing that allows the power to be "shared" by the 2 regulators :
- (12-9)x I in the LM317
- (9-5) x I in the arduino reg.
With 12V supply , the arduino reg. has to deal with (12-5) x I watts !!
As for me, I use a 7809 regulator with recommended capacitors. I soldered a jack at the output (positive inside), and connectors at the input. Then I can use my "ten-nimh-2300mAh-AA battery pack" to feed the arduino and none of my regs get too hot
I've just read your last post and if all is feeded by the arduino, the regulator will dissipate :
(12-5) x 0,220 = 1,54 W -> hot
with the lm317 at 9V output :
(12-9) x 0,220 = 0,66 W in the LM317
(9-5) x 0,220 = 0,88 W in the arduino regulator
no more heat
A long time back I powered my old arduino thru the barrel jack from a 12v battery, and I noticed the arduino on board LEDs were noticeably brighter. I measured the voltage between the arduino +5v pin and ground, and the voltage was ~8v. I subsequently powered the arduino with 12v via an 7805 regulator connected to the arduino 5vpin. I think the arduino power setup is a little "tutti-frutti".
@alnath, using the adjustable regulator, I would think 7 or 7.5 volts out would be better than 9.
Do you have a reason that 9 would be better?
7V or 7,5V would be OK yes
actually, there are 2 reasons why I said 9V :
1 - I had several 7809 , I used them and everything goes well, with all the shields I use
2 - 12-5 = 7 V then if you want to share equally between the arduino reg. and the external one, the output of the external one should be 12-3,5 = 8,5 volts
But you are right, 7volts or 7,5volts would be better, especially if you need more amperes, because it is easier to put a heatsink on the external regulator than on the arduino's
I to would set it at about 7.5 Volts. You want the two regulators to share the power load, but a through hole lm317 will naturally take much more power and can be far more easily heat sunk, so making it take the brunt of the load only makes sense.
On an side note. If this is the only thing your battery is powering and you want good battery life you should seriously consider getting a switching regulator. Linear regulators burn off the excess voltage as wasted heat that then becomes a problem. This means that only 5/12 of your battery power is being used by your electronics, the rest is lost in regulation. Switching power supplies are more complex, but they can make the same voltage change with 90% - 95% efficiency. This means they won't get nearly as hot as well.
There are plenty of standalone 5V output switching supplies that could work well if fed into the 5V input pin. If you want to keep the on-board regulator you could still use a switching regulator to get from 12 down to 7.5 and increase your efficiency significantly.
That sounds right.
Why regulate down (with heat) and regulate down (with heat) when you can regulate down only once, with little wasted power (heat). Thanks, for the idea.
Hey guys, I added the LM317 regulator in the mix for a Vin of now 8.6V. For some reason, an adjusted Vin below 7.4V would not power the arduino, and above 9.3V would heat up the arduino onboard reg really hot, and so 8.6V seemed to be a good, workable compromise.
I now face a new curious issue. My radio IC's (SM130 RFID and Bluesmirf bluetooth modem) now have an operating voltage of 3.7V. This is fine for the bluesmirf, although now my SM130 antenna is underpowered, and doesn't read RFID tags.
Why would the voltage drop to 3.6V from the 5V pin, when powered with an external 8.6V power source?
Thanks for all the help
If the onboard regulator is working, only a real heavy load, but if the onboard regulator is not getting real hot, not that.
Start by taking off all other loads from the arduino. See what the 5V line shows. Guess you do have a volt meter, so check all both voltages, Vin, and 5V.
May have oscillations.
Make sure you have the 100nF capacitor right next to the input to the regulator.