From my own experience, burning microcontrollers is quite easy. Put the 5V at ground, GND at VCC and in an instant your chip is burned.

What exactly goes on internally that causes it to stop functioning entirely? For instance, if I were magically able to open a chip and rearrange all its semiconductor connections and fix it, where exactly would I need to look, and what would I need to do?

If this is chip-specific, please choose any that could answer my question or give me an idea at least.

  • \$\begingroup\$ You look for a square of metal or oxide brake down \$\endgroup\$
    – GR Tech
    Commented Jan 26, 2014 at 21:31
  • 5
    \$\begingroup\$ In addition to that I agree with Spehro Pefhany's explanation; Many IC's now have Diodes that allow them to survive inverse power supply. Although this is something not to rely on \$\endgroup\$
    – Mark
    Commented Jan 27, 2014 at 7:11
  • \$\begingroup\$ @GRTech Gate oxide breakdown is an unlikely failure mechanism for a reversed power supply. \$\endgroup\$
    – W5VO
    Commented Jan 27, 2014 at 8:41

5 Answers 5


Most commercial IC circuits are isolated from the substrate material by a reverse-biased P-N junction (including CMOS parts). The substrate is usually tied to the voltage expected to be most negative.

If it isn't, then that junction becomes forward biased and can conduct a great deal of current, melting metal or heating the junction to the point where it no longer acts as a diode. That is typically at a voltage of about 0.6V, but the IC makers play it safe usually by telling you not to go lower than -0.3V.

(referring to the below diagram, but not shown, the substrate would be tied to pin 5)

enter image description here

Most CMOS parts have another twist that if part of the chip has a normal Vdd and another part sees a big negative current it will trigger a big parasitic SCR that is a side effect of the structure, then the device's power supply draws a large current which causes overheating, melting etc. if the current is not externally limited. That is called latch-up.

enter image description here

  • 2
    \$\begingroup\$ Nice answer, lots of upvotes, unfortunately it is incorrect. Latchup is a different phenomenon. In IC design it can be avoided by having enough substrate contacts, this is also checked during design with automated tests. \$\endgroup\$ Commented Jun 16, 2015 at 15:40
  • 1
    \$\begingroup\$ @Rimpelbekkie Nope. The trigger current can be increased, but the effect cannot be eliminated totally except by going to an insulating substrate such as sapphire because the four layers of a thyristor are still here. The current is not limited in the situation under discussion here. \$\endgroup\$ Commented Jun 16, 2015 at 15:51
  • 1
    \$\begingroup\$ Nope what ? Latchup is a real phenomenon, no doubt there. Is it the reason that a lot of current flows when the supply is reversed ? NO ! If you disagree, explain to me how the thyristor equivalent circuit pictured above can conduct when VDD is negative with respect to ground. To trigger the thyristor, VDD must be positive and enough voltage must exist across Rwell and/or Rsub. This can only be caused by having too few and too distant substrate contacts. I have been designing ICs for 25 years, have yet to see one with a latchup issue. \$\endgroup\$ Commented Jun 16, 2015 at 17:26

What releases the magic blue smoke when you exceed working voltages or reverse the supply voltage?

Applied to any 'chip'

Excessive current producing excessive power dissipation (\$I^2 R\$) and/or excess voltage causing insulation breakdown due to high internal field strengths coupled with the lack of thermal conduction from the devices inside the chip.

Consider the non-linear, asymmetric (polarity sensitive), physically small nature of the internal devices and their small heat conduction paths. Couple this with low voltage destruction of very fine insulating layers (high field V/m) producing bi-directional low resistance conduction pathways.

The internal individual device temperature rises very quickly and destroys its semiconductor/insulating properties. Once destroyed this produces other low resistance pathways causing multiple cascading failures across other devices on the chip.

All this happens very quickly and its very much a one way event. (Think Humpty Dumpty - Putting all the pieces back together won't get you back to where you started from - Humpty has left the building)

How could you repair it?

Basically you can't cause magic doesn't exist. There would be so many interacting faults in the circuit that it would be nigh impossible to localize any fault. (Remember even in a 'simple' IC you are dealing with hundreds of thousands of devices.) All faulty devices would have to be identified and replaced at the same time (assuming you had the ability to reconstruct all faulty devices at an atomic level) - miss only one and you have to start again when you power up.

Simple solution (and most cost effective in time and money) throw the dead bug away,learn by the experience, replace it with a brand new full spec chip and next time be more careful with the power supply.

  • 3
    \$\begingroup\$ Not true for all simple ICs. Something like the 555 or typical opamps or motor drivers are fairly simple, tens of transistors, not hundreds. \$\endgroup\$
    – Passerby
    Commented Jan 27, 2014 at 1:13
  • 2
    \$\begingroup\$ @Passerby well the OQ did start with microcontrollers and I was basing my answer on this. Whether its 5 or 5 million devices inside the chip it still holds true. Simple devices such as the 555 may be more robust but once you start to destroy internal structures one fault will lead to another. \$\endgroup\$ Commented Jan 27, 2014 at 22:36

A late answer, I came here via another question but noticed that actually none of these answers address the real reason why almost any IC / Chip can be fried by applying a reversed supply voltage.

The real reason is that all chips need to have ESD protection on all pins that are not supply pins with a circuit like this:

on-chip ESD protection circuit

So almost every pin has this ! That is a lot of diodes in parallel. You can easily destroy all these diodes by reversing the supply. And that actually destroys your chip.

Latch-up as mentioned above is an effect that occurs when the supply has the correct polarity but a current is sinked or sourced on an input or output causing a malfunction as explained above. It has nothing to do with reversing the supply ! If you think I'm talking nonsense then please look up how a Latch-up test is performed. There is specialized measurement equipment to do such test.

Please read this excellent article explaining latchup and note that the supply is "normal" so not reversed ! When still in doubt read the EIA/JEDEC STANDARD IC Latch-Up Test EIA/JESD78.

  • \$\begingroup\$ Hi, sorry I am late to the party. If you apply reverse voltage and burn the ESD diodes, why would the chip be destroyed? It would still work, albeit have no more ESD protection, right? \$\endgroup\$
    – tobalt
    Commented Dec 17, 2021 at 7:25
  • 1
    \$\begingroup\$ @tobalt You're assuming that all the ESD diodes fail open, then you're right. But what if one or more fail short, which is often what happens. Also, even if the ESD diodes survive, there are still MANY PN-junctions (diodes) inside the circuit itself that you will destroy by applying a reversed supply, as then, these PN junctions go in forward mode. Note how reversing the supply violates the absolute maximum ratings of almost any IC, that means there are no guarantees that the IC will still work. It could but all bets are off. \$\endgroup\$ Commented Dec 17, 2021 at 9:06
  • \$\begingroup\$ Yes right, I was assuming the open-fault, which is quite a stretch. Thanks. \$\endgroup\$
    – tobalt
    Commented Dec 17, 2021 at 9:10

What exactly goes on internally that causes it to stop functioning?

An excess of current, junctions can resist current only in one direction, when polarity is inversed they become short-circuits. Heat is generated, junctions burn as well as other overheated elements.

If I were magically able to open a chip and rearrange all its semiconductor connections and fix it...

You cannot fix it (in practical) because many junctions are now broken/evaporated, as well as their immediate environment.

Protection against polarity inversion is quite easy (a diode), however it generates a voltage drop and additional heat, the manufacturer doesn't embed it on the chip, the IC user may add an external diode if necessary.


As the semiconductor structures are very small, it's indeed quite an easy task to burn them down.

  1. Clearance distance - if you apply a large enough electric field between two conductors, there will be a breakdown. This, being on a chip, causes terminal malfunction. This mainly occurs on the Gate of a FET structure.
  2. Semiconductors in principle are non-linear, polarity-sensitive devices. This in turn renders the entire device very non-linear and polarity-sensitive.
  3. Million other reasons that I can't think of right now...

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.