Yesterday I burned an arduino and I think today I almost burned a new one. As usually I am powering my arduino using the VIN pin supplying to it about 12v. I am using an external battery to power my arduino.
Everything works great. But sometimes I have to send a new sketch to my arduino so I connect it to USB and upload the code. But if I keep the battery connected to VIN the voltage regulator chip of my arduino gets reaaaally hot, I cannot even touch it. So I disconnect everything and wait a couple minutes.
I know I cant connect a stable 5V power supply to the 5V pin of my arduino while connecting it to USB. I know I am gonna burn my arduino. But I see no problem supplying power to arduino using VIN and USB at the same time, it should make no problem. Why?
OBS: to do this test I had no other module or component connected to my arduino, just arduino and the power suppply.
Yesterday I burned an arduino
Which Arduino. There are many. Don't know if all have Vin/USB switchover.
But if I keep the battery connected to VIN the voltage regulator chip of my arduino gets reaaaally hot, I cannot even touch it. So I disconnect everything and wait a couple minutes.
Then don't do this!
You may have damaged things already.
Does the Arduino work with just the USB connected?
Why not use the power jack?
I cant use power jack cause I am using a battery with 2 "naked" wires. It's much easier to connect them to VIN instead of using power jack. Also, as far as I know the only difference of power jack and VIN is a diode and nothing more so this shouldnt help me at all.
I am using arduino nano.
The other difference is, Vin header allows you "much easier" make mistakes, the jack won't!
You can always use something like this
Is your 12V supply regulated?, can you measure the open circuit voltage?
Here's a schematic from Gravitech FWIW.
She/he is using a battery supposably.
I had a Moteino powered at 3.3 v from a battery/regulator combination. The Moteino can handle 5 v. in, but another component in my circuit couldn't. It was also powered on the same 3.3v power rail as the Moteino. The difficulty was that if I hooked the FTDI connector from the computer up for programming, the Moteino Vin chip then put 5 v. out on my 3.3 v. power rail. Lucky I noticed this before I fried anything.
I solved that problem by building an FTDI extension cable leaving out the 5 v. line from the computer. The Moteino still programmed just fine, and I no longer had 5v. showing up on my circuit board. You could possibly build a USB cable for programming that left out the 5 v. line. Or just do a littl brain surgery on an existing one and cut the 5 v. line. Of course, it won't power an Arduino, but that's apparently what you want to avoid.
The Nano has a diode between +5volt and USB supply.
So no current should flow from battery to computer if the diode is ok.
All I can think off is an oscillating regulator. Not uncommon for some Nano clones.
See if you can "fix" it by adding a 47uF electrolytic cap (10uF-100uF) between the 5volt pin and ground.
Your idea of using a power jack instead of VIN is great, I was not aware of that connector but that wont change my problem of having a very hot temperature on the voltage regulator.
My battery provides 12v and as the time passes it decreases the voltage to as low as 10v. It's a LIPO battery, it is not regulated. What amazes me is that using only the battery everything looks fine but if I connect arduino also to USB the voltage regulator gets so hot that I cant touch it.
Guess that accursed diode is why Nano’s only have 4.75V on the 5V pin. >:(
His heat problem is still baffling though.
to the question Wawa asked “which arduino is it ?” (edit : sorry, I didn’t notice your answer )
I’d add this one : is it an Arduino, or a clone and if it is…which clone ?
Just wonder if the blocking diode is shorted and the regulator is pushing current back up the USB line? Is that possible?
Yes, a shorted blocking diode would make the onboard regulator power whatever is inside the computer.
Measure the diode (D1) with a DMM.
But I also gave a second possibility...
I have a arduino nano clone. I dont have tools to check if the diode is shorted, if so, it's probably a problem that I cant solve, right? Could any one of you power an arduino using 12v an connect it at the same time to USB and put your finger over the power regulator to check if it gets hot?
I think the question about EXACTLY what kind of 12V battery he is using should be the FIRST priority.
If he is using something like a car batery the voltage could be greater than 12V, which just for the record, is really not a good idea either because the regulator has to disipate 7V to go from 12 to 5V.
That is the reason it is getting so hot. Find a lower voltage source, like 7 to 9 V.
I dont have tools to check if the diode is shorted
How have you been measuring the voltage on your battery?
Before you make any more costly mistakes, can I suggest you obtain a regulated variable power supply, with current limit facility.
You have no way of controlling the immense current that is available from your battery, so any fault condition will cause some major damage.
Can you post a picture of how you have your projects connected with the battery please.