I'm powering a Nano from 12v (it needs to be 12v because that's need to drive some LED ribbon elsewhere), and it gets hot enough that I can't hold my finger on the voltage regulator. Is this a problem (not holding my finger, just the heat!)? Will it damage it when I run it for 4hrs like this, and should I use an external regulator instead? I hope that answer to that is "no", because I've already built the pcbs, but I'm worried it might damage it
markvr:
Is this a problem (not holding my finger, just the heat!)?
Yes.
markvr:
Will it damage it when I run it for 4hrs like this
That's hard to say...but it's definitely not good.
markvr:
I hope that answer to that is "no", because I've already built the pcbs, but I'm worried it might damage it
It's a design error. The on-board regulator is tiny, designed for powering the board and not much else.
If you could drop the 12V down to 7V before you feed it into the Nano that might fix it and you can use the same PCB. Get an LM317...
(7V is ideal but 9V might work too...basically you're reducing the excess volts it has to dissipate)
These may be of some use:
OR
Yup too much power being disipated via the linear regulator, if it fails there is a chance your atmega will also get a full 12v
The risk increases with load (sensors, led's)
Maybe you are running too many LEDs off the onboard regulator. This is probably drawing too much current, causing it to heat up.
cjdelphi:
Yup too much power being disipated via the linear regulator, if it fails there is a chance your atmega will also get a full 12vThe risk increases with load (sensors, led's)
The Nano uses a 78M05 regulator, which include an internal overload protection and a thermal shutdown protection.
This is one of the best feature these regulators have.
Then IMO, there is no risk for the atmega to get a full 12V . If it gets too hot, which is likely to happen in these conditions, the 7805 will put itself in protection mode -> nano is not powered anymore as long as it stays too hot or the overload exists.
I agree with Fungus, it would be a good idea to drop down the power input : an 7809 or a 7808 with a little heatsink if needed, and it's done.
It reduces the power to dissipate in the internal regulator :
with a 7809 -> new power to dissipate = previous power to dissipate - 3xI
with a 7808 -> new power to dissipate = previous power to dissipate - 4xI
All the power to drive the LED ribbons goes through darlington arrays (they are 12v), so there shouldn't be much power going through the nano. All it's doing is driving some TLC5940 LED multiplexor chips, which is why I'm surprised it gets so hot.
I'll investigate adding a seperate LM317. Just don't have much time left to make new PCBs!
markvr,
try to add simple resistor 1K between +12v and Nano's Vin pin.
I'm novice in electronics.
tock:
markvr,
try to add simple resistor 1K between +12v and Nano's Vin pin.
I'm novice in electronics.
sure it won't get too hot XD , but it won't work either 
remember U = R x I
Why not?
If 100 ohms resistor?
Another thing in question is the 12 volt power supply. Many power supplies that state 12 volts, will actually produce more, maybe 13.8 or 14.8 volts. So check the voltage with a meter. Either way, I would cut the voltage down before putting it onto Vin (as previously stated, 7 volts would be optimal). You can step it down several ways, use a resistor, use several diodes, or use a voltage regulator. Best way is a voltage regulator.
If you have a spare power transistor 1amp or more, you could buy a 7v zener and use that to drop the voltagedown...
tock:
Why not?
If 100 ohms resistor?
because the nano needs some current to work, for itself, and for some outputs if needed .
Say we need 7V on Vin ... then there must be no more than 5V on the resistor.
With 1k , the limit is 5/1000 = 5mA
with 100 ohms , 50mA
etc....
I wouldn't consider a resistor as a solution : Vin is not regulated, it is high when you don't drive any output,
and drops down as the current increases, until it is too low to power the nano.
It is just a (bad) way to limit the current .
The best solution is an external regulator (linear or switch), and a good design that stays under the absolute maximum ratings
@alnath, I would not say using a resistor in series with Vin is the best practice, but:
It can work. If you know how much current through Vin, the nano is using (test with amp meter), You can calculate the resistance needed assuming you know the voltage (test with volt meter), of the 12 volt supply.
There's more than one way to skin a cat.
If I was to guess, I would guess a resistor between 300 and 700 ohms would do the trick (but don't use my guess, use the VOM).
@Jackwp, sure, it could work, but one day, months after, you decide that you need one more led, you make the changes in the sketch, try it with the USB only plugged in, everything works well. You power it with your "power supply" .... oups !
And it takes you 5 hours (and possibly a "Plz help me" post here 
) to find out why it has stopped working
More than one way to skin a cat, sure, but one have to make it "dans les règles de l'art" 
Yes alnath, you have a good point. Looking at ohms law, anytime you change one, the other two can change.
I think since you are sending the Vin to the onboard regulator, why not count on it to do its job, and just help it a bit by not over driving it.
I tested my leonardo, with power in to Vin. It ran until Vin dropped below 4.3 volts, and it started running again when voltage got above 4.8 volts. So, I think the voltage on Vin is really not to critical, unless it is over 10Volts, or below 5 volts. Just my thoughts. Maybe we will just have to agree, to disagree.
Sure it will work under 7V , but then, the power is not regulated anymore.
I think we are a little off topic here (my fault 
) , because OP is, as far as I can guess, aware of all of this.
I posted because Tock's advice "try an 1k series resistor" couldn't work and, as he admitted being a novice in electronics, I wanted to explain why
Sure it will work under 7V , but then, the power is not regulated anymore.
Actually the power is still regulated by the onboard 5V regulator.
alnath , I give up, you are right, or maybe we both are. Lets have a beer, and let this thread go.
no, 78xx regulators need the input to be at least 2V higher than the output to regulate.
with input under 7V, the output is just the input minus a value I've never calculated
Edit (I don't want to add more "off topic" posts ) : Jackwp, it is not a matter of being right or wrong, and I'm sorry if it seemed to be, my english is not as good as I wish it was
If you happen to come here (south of France), I'll be happy to pay you a beer ( just a perrier for me )
alnath , we can discuss this further, I am sure I can learn from it. But should we not start a new thread? Something like "Vin voltage range".
If you want to, start a new thread. I do still have a couple thoughts, questions.
Thanks, Jack