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I have a custom PCB designed to interface an Arduino Nano's ADC pin that had 12v from a bench power supply inadvertently connected to a header pin meant for 5v input. The ATMEGA328P now gets extremely hot even removing the Nano module from the PCB. It seems to program, though it will occasionally spit out an error during verification that can be overcome by programming it again. The ADC can still be calibrated and matches within reason (<1% error) the displayed bench DMM. The Nano module normally reads approximately 4.5V, which now reads 3.5V on the 5v rail.

I thought I had designed robust protection into the PCB after doing research into diode clamps, crowbar circuits, etc. I settled on the following circuit that takes the incoming raw signal, passes it "through" an ESD array (DRTR5V0U2SR-7) and then into the circuit below.

schematic

simulate this circuit – Schematic created using CircuitLab

My calculations say that 1.2mA should get through the 10k resistor and simulations in LTSpice (with modeled internal clamp diodes of the microcontroller) show at most 726uA (assuming a 4.5V rail.) Most of it is going through the bat54 diode and approximately 400nA through the internal diodes.

Removing the Nano module and powering the bare board draws very little current, so I do not believe anything shorted on the adapter/support PCB. Testing the BAT54S array shows a voltage drop of 0.25V from input to 5V rail and 1.25V from rail to ground. Reversing the leads reverses the readings. Comparing the values to the other BAT54S in circuit show the same values.

I thought originally that maybe I backfeed the regulator on the Nano due to the clamping diodes, but at 1mA and various power hungry modules, like a connected 20x4 line LCD, I think the rail would absorb that amount of current. My next thought was the TVS conducted a large amount of current that overwhelmed the rail, but due to the low current when the bare PCB is powered I do not believe the TVS is damaged.

I'd like to improve my understanding and hope to learn and prevent this in the future. What am I missing?

Edit: I've updated the schematic with the DRTR5V0U2SR-7, to more clearly show the internal diodes from the DRTR5V0U2SR-7 datasheet. The 12V was applied to the node labeled RAW INPUT which is the pin header on the support board. The 12V from the supply was connected while the USB was powering both the Nano, and the adapter board (the Nano was in the socket on the PCB.) The voltage was applied for approximately 5+ seconds before I realized I had moved the leads to a pin expecting 5v not 12v. Later, I measured the short circuit current to be current limited at 1 amp on the bench supply.

The BAT54S diode array was measured while still mounted to the PCB. However, there are about 7 of the BAT54S spread around the PCB and they all measure identical while measured in circuit.

When I removed the Nano, the fault still exists, and it is definitely the microcontroller getting scorching hot as that part heats up to finger burning levels while the heat slowly transfers through the board to the regulator on the opposite side.

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    \$\begingroup\$ Please confirm exactly which node in your diagram the 12 volt was inadvertently connected to. Please also state where the TVS is connected. It doesn't appear to feature in your diagram. \$\endgroup\$
    – Andy aka
    Commented May 13 at 17:40
  • \$\begingroup\$ How was the TVS connected, what was on TVS VCC and GND pins and was the 12V directly applied to TVS IO? \$\endgroup\$
    – Justme
    Commented May 13 at 17:42
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    \$\begingroup\$ Please note that these diodes are only rated for 30V reverse voltage, so they are kind of frail for protection purposes. \$\endgroup\$
    – Lundin
    Commented May 14 at 9:29
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    \$\begingroup\$ It sounds like you have blown a protection diode in some IC, perhaps the MCU or voltage regulator. If you can locate which part that's getting hot then that might help (avoid touching it directly with your fingers though, it may be >100°C). It might matter if the board was otherwise powered when you slipped with the 12V - in case a voltage regulator gets a reverse voltage on the output pin, when no supply is present on the input side, they tend to go poof pretty much instantly. \$\endgroup\$
    – Lundin
    Commented May 14 at 9:38
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    \$\begingroup\$ @RobertSusmilch - Hi, Please note that a clarification comment from Fabio Barone which you probably didn't yet see (as it was previously part of an answer), has now appeared in the comments above here. Please respond in a comment here back to Fabio about that, as you have kindly been doing to other requests for info. TY \$\endgroup\$
    – SamGibson
    Commented May 15 at 15:31

3 Answers 3

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Testing the BAT54S array shows a voltage drop of 0.25V from input to 5V rail and 1.25V from rail to ground.

If that 1.25V is the forward voltage drop of a single BAT54 when tested with a DMM, then I'm afraid there's something wrong with that diode.

Also: The 5V clamp will only work if the 5V can safely absorb the excess current.

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    \$\begingroup\$ A suggestion; if most of your [would-be] answer is questions, consider raising those questions as comments instead -- and voting to close the question for lack of information. \$\endgroup\$ Commented May 13 at 23:34
  • \$\begingroup\$ Rail to ground implies that's the other diode, measured from cathode to anode. Hard to measure that reliably when the diode is mounted. \$\endgroup\$
    – Lundin
    Commented May 14 at 9:33
  • \$\begingroup\$ @TimWilliams Understood, but seeing as the OP was a "new contributor" I figured an answer may be more encouraging. Cheers. \$\endgroup\$ Commented May 14 at 21:25
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    \$\begingroup\$ @TimWilliams - Thanks for the comment. I'm sure Fabio is trying to help too. || Since the first point in the answer, although written in the form of a question does also give an answer, that's allowed. Effectively it's being written as an assumption - it isn't a bare question i.e. one where this answer is waiting for a response. However there was also a bare question to the OP. I have moved that to be a comment requesting clarification, where such questions to the OP should be. \$\endgroup\$
    – SamGibson
    Commented May 15 at 15:26
  • \$\begingroup\$ There are several of these BAT54S diode array on the PCB. I tested all of them to see how they behaved while in circuit. They all showed the same 0.25v and 1.25v behavior. Also, removing the Nano from the socket shows negligible power draw while bench powered through the 5V power pins (it is powered by the Nano's onboard regulator.) The PCB functions completely normally when I plug in a new Nano board. In fact, the ADC seems to still function as well on the damaged Nano since it tracks the input voltage with my DMM (within 0.3%) even though the VCC rail is at 3.6v. \$\endgroup\$ Commented May 15 at 15:30
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The AVR was damaged through overvoltage 5V supply pin, not through ADC input.

Your TVS was connected from the TVS supply pin to 5V, therefore the 12V from IO pin went through the TVS to 5V, maybe a bit clamped by the TVS.

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When you connect 12V to the "raw input" node, current will flow through the "upper" diode into the 5V rail.

The diode will have a voltage drop of a bit less than 1V.

So basically, you short the 12V rail to the 5V rail with a diode (ie 1V drop).

So there are 3 possibilities :

  • the 12V source is current limited (or has high resistance), and drops to about 5V+1V=6V
  • the 5V voltage rail can't sink enough current, and it's voltage rises to about 12V-1V=11V
  • both rails are very stable, and you get a HUGE current through the diode, probably burning it

My guess is the 2nd option.

If so, your 5V rail is now at 11V (it might destroy several components on the 5V rail). It also means that your input clamping circuit (D1) now clamps to about 11V+0.5V=11.5V, instead of clamping at about 5.5V. So you got potentially far more current than you expected (nb : it all depends on how the atmega itself was powered at this instant, it it's supply was still 5V or 11V).

In the future, if you want to protect from input voltages above your voltage rail, I would advise you put a resistor before the ESD diode. If you know the highest possible voltage (and you have this voltage rail available), you can also consider diverting the ESD to the highest voltage rail (but do the clamping with D1 towards your 5V rail)

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