MY ARDUINOS KEEPS DYING!! :(

Using Arduino Project Guidance
1

Hi,

Now is the fifth arduino that I burned in the last week so it's time to take a breath and try to discover what is going on. My setup is exactly like this

Regularly my arduino is powered using a 12V power supply (which can handle up to 2 amps). At the same + line I feed a really powerful LED and I feed my arduino nano in parallel.

All GNDs are connected together and it all works fine. The problems arises when I disconnect my 12V power supply from the circuit and connect my arduino nano to my computer to upload a sketch. If I forget the arduino nano connected to the USB (computer) for more than 3 or 4 minutes my arduino dies, it smells like smoked and it does not work anymore.

So I am wondering: when my power supply is disconnected, will the LED draw current from my VIN pin (instead of the power suppply) and this high current (300ma) is burning it? Should I use a diode in my VIN pin in order to only allow current to go inside my arduino and not allow current to get out?

2

They don't usually just die, they're KILLED. :frowning:

From the sound of your description, when you disconnect the power supply, you're drawing current backwards through the regulator's internal diode to supply the LED with (just under) 5V via the Vin pin. That's a big no-no. It will kill the regulator.

You could disconnect the LED before connecting to the computer if the power supply isn't connected, or you could add a diode as you say, or you could supply power via the DC input jack, (with the LED connected before the jack), so that the onboard diode stops reverse current flow. Edit: It's a Nano, not an UNO. I missed that. :frowning:

3

Yeah man, it didnt die, I killed... :frowning: :frowning: :frowning: and it's also killing my pocket ::frowning:

Anyway, I have evertying soldered to my arduino, it's not easy and practical to unsolder everything everytime I upload a sketch. So removing the LED is not a good option.

If you are sure about what you are saying, so the energy is getting out of the VIN pin (instead of getting in) and I expected this may be causing my arduino death.

Your idea about using the DC input JACK is amazing but only if you are sure that energy cant get out of it. Does the DC POWER JACK only accepts energy in? Are you are it does not allow energy to get out?

Also, are you sure that when arduino is being powered by USB, it can allow energy to go out in the VIN pin? I took a great look at google and didnt find anything to support this.

4

batata004:
If you are sure about what you are saying, so the energy is getting out of the VIN pin (instead of getting in) and I expected this may be causing my arduino death.

As I say, the regulator has a built-in diode that allows current to flow from the output to the input, and it's not intended for continuous current, so that's my guess as to what's going on.
Having noticed that you have a Nano, not an UNO, I'm assuming that they have the same or a similar regulator, so this would still apply. I can't read the regulator numbers from the Nano schematic I'm looking at now.

Your idea about using the DC input JACK is amazing but only if you are sure that energy cant get out of it. Does the DC POWER JACK only accepts energy in? Are you are it does not allow energy to get out?

There's a diode between the DC input jack and the Vin pin, intended to avoid damage ifthe power is connected in reverse-polarity, but it would also serve to stop current flowing back from 5V, through the regulator the the DC input jack.
I just went back and re-read your opening post, noticing that this is a Nano, not an UNO. A Nano doesn't have a DC input jack, does it, so that fix won't work. Adding a diode as you suggested is the answer then.
Forget everything I said about the DC input jack. It only applies to boards that actually have one. (I wasn't referring to the USB jack.)

Also, are you sure that when arduino is being powered by USB, it can allow energy to go out in the VIN pin? I took a great look at google and didnt find anything to support this.

As I mentioned, the diode that's built into the regulator would allow 5V to flow to the Vin pin.

Edit: This is from the NCP1117 datasheet:-

Protection Diodes
The NCP1117 family has two internal low impedance diode paths that normally do not require protection when used in the typical regulator applications. The first path connects between Vout and Vin

That's the regulator used in an UNO, so assuming the same or a similar low dropout regulator is used for the Nano, that will be where the current conducts back to the LED.

Edit2: As an afterthought: You could always solder an inline connector into the wire leading to the LED, so you can disconnect it for programming. You could just leave a short pigtail that you can plug the LED into.

5

Your idea of an inline connector is good, I never thought about it!

Indeed my nano does not have a DC power jack, I thought that when you said that I could solder one in my arduino but after looking in the internet it's no possible. So the diode will be the only good alternative which is safe and easy to use.

But just a final question: you said that when I power my arduino using USB it provides 5v in th VIN pin? Shouldnt it be a little less cause there are some circuitry between the 5v pin and the VIN pin which should decrease the 5v correct? And also, if when I power my arduino using USB the vin has only 5v, my LED should not power up right? But it did!

6

batata004:
But just a final question: you said that when I power my arduino using USB it provides 5v in th VIN pin? Shouldnt it be a little less cause there are some circuitry between the 5v pin and the VIN pin which should decrease the 5v correct?

Yes, that's the regulator that I mentioned, which drops the Vin voltage to 5V. When conducting backwards, the voltage will drop slightly due to the internal diode, but only by about 0.6V.

And also, if when I power my arduino using USB the vin has only 5v, my LED should not power up right? But it did!

It won't be as bright. Look at it this way. If the 12V power supply is disconnected, and the board is connected to USB, the only possible source of power is 5V - the USB voltage.
The LED will not draw as much current as it did from the 12V supply, but it will still draw too much continuous current for that diode.

The only possible current path from 5V to Vin is through the regulator, so this must be what's happening.
Make sense?

Here's the schematic for a Nano. Note the regulator near the bottom, and the Vin and 5V tags. That must be the current path, if the LED lit while only 5V was applied via the USB port. Nothing but the regulator connects to Vin.:-
(Right-click and "View Image" or similar for a full-sized view.)

Arduino Nano Schematic.jpg
Arduino Nano Schematic.jpg1163×802 133 KB

7

What led are you using that is taking 12v 300mA?

8

What value resistor was in series with the LED? about 40 Ohms ?
Are you using a Chinese Nano clone ?
If so, the schematic could be like this http://actrl.cz/blog/wp-content/uploads/nano_ch340_schematics.pdf where there is a schottky diode between the USB and Vin. This could have failed under the stress, but I guess the Nano would still work if powered via Vin.
Anyway, the regulator is easy to change (I've done this) and so should the diode be easily changeable.

9

Great man, your diagram made it very easy! Now I understood it, the currenct from USB has to pass thgourht 5v pin and if its too much current it may destroy arduino.

Thank you! This problem of mine is not "rare", I am amazed that there are no more idiots like me making this same mistake.

10

6v6gt:
What value resistor was in series with the LED? about 40 Ohms ?
Are you using a Chinese Nano clone ?
If so, the schematic could be like this http://actrl.cz/blog/wp-content/uploads/nano_ch340_schematics.pdf where there is a schottky diode between the USB and Vin. This could have failed under the stress, but I guess the Nano would still work if powered via Vin.
Anyway, the regulator is easy to change (I've done this) and so should the diode be easily changeable.

That Schottky diode actually appears to be between USB +5V and the regulator's output, not Vin. Current can only flow from +5V to Vin as I described earlier. Vin only connects to the regulator's input. (I think you misread the line from the Schottky anode as joining Vin, but it doesn't. There's no blob where they cross. And that would cause a whole raft of other problems if they were joined at that point.)
Edit:

a schottky diode between the USB and Vin

Looking again, I guess you meant "between the USB and +5V". A simple typo makes more sense.
Still, I doubt that the Schottky diode has been damaged. Just the regulator. They don't like conducting backwards. That internal diode can handle a very short surge to 15A, but not very much continuous current at all. Certainly not enough to power a high-current LED.

11

batata004:
Great man, your diagram made it very easy! Now I understood it, the currenct from USB has to pass thgourht 5v pin and if its too much current it may destroy arduino.

Thank you! This problem of mine is not "rare", I am amazed that there are no more idiots like me making this same mistake.

The voltage from USB connects to +5V, (via a Schottky diode), then flows back through the regulator to your LED which is connected to Vin. The whole Arduino isn't destroyed at that stage, just the regulator.

12

I've killed one Nano (clone) when a GND wire got loose and bumped a 5v lead. The rectifier diode on the opposite side of the pcb from the USB socket was visibly popped.
Frankensteined on a replacement 1N4XXX diode and all is well.

13

@OldSteve so my 5 arduinos that I supposed killed are still alive? I just need to make a surgery and remove the regulator? I see a small chip that has 3 pins is it the regulator? Can I buy spare ones and solder it back? Are you sure in this process I didnt damage anything else? My nano is a clone, not original.

14

OldSteve:
That Schottky diode actually appears to be between USB +5V and the regulator's output, not Vin. Current can only flow from +5V to Vin as I described earlier. Vin only connects to the regulator's input. (I think you misread the line from the Schottky anode as joining Vin, but it doesn't. There's no blob where they cross. And that would cause a whole raft of other problems if they were joined at that point.)

Oops. You're right. I didn't look carefully enough.
But if the regulator has failed, but not shorted, the Nano should still work if powered by the USB. If the Schottky diode has failed the Nano should still work if powered via Vin. (If I read the thing correctly now, that is).

15

6v6gt:
Oops. You're right. I didn't look carefully enough.
But if the regulator has failed, but not shorted, the Nano should still work if powered by the USB. If the Schottky diode has failed the Nano should still work if powered via Vin. (If I read the thing correctly now, that is).

All depends on what happened after the initial failure. If the regulator died short-circuit Vin to Vout, and 12V was later applied to Vin, all sorts of damage could have happened. That's why I said earlier: "The whole Arduino isn't destroyed at that stage, just the regulator."
If the regulator failed short-circuit to ground, connecting USB would short the 5V to ground.
I think they'd need a careful examination to see exactly what's failed.
If the regulator failed open-circuit, then you're right, the board should work when powered from USB, but the LED would no longer light since there's no longer a current path back to Vin.

There are a few possibilities.

16

What I'd do is remove the regulator from one, then try powering the board with a regulated 5V to the Vcc/+5V pin, (not from USB, just in case), and see if the board still works. Chances are that as 6v6gt says, the regulator failed 'open-circuit', in which case there probably isn't any other damage..

17

Yeap, it's too complicated to me.. I am not too good to solder small pieces so I think I am gonna buy new arduinos and dispose the dead ones. Even if is failed "open circuit", as you said, my soldering skills are not too good to solder that small regulator in place correctly.

Thank you anyway, at least you taught me how not to destroy my other arduinos :slight_smile:

18

his problem of mine is not "rare", I am amazed that there are no more idiots like me making this same mistake.

There ARE more people out there with the same issue - most of them keep it as their secret and keep burning their devices until the purse is empty :slight_smile:

19

@rpt007 at least now those unfortunate people can find this thread and maybe stop killing their arduinos. :slight_smile: Thank you all, you were awesome here.