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With higher voltages (more than about 5V), microcontrollers may be damaged. Is this because the voltage actually physically damages them, or because it allows excessive current to flow? - and thus increases power dissipation beyond safe limits. How does this apply for other devices?

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    \$\begingroup\$ could be any or all of the above, very dependent not only on the device but on which pins of said devices the voltage is applied to. \$\endgroup\$ – Mark Nov 14 '10 at 0:41
  • \$\begingroup\$ It is almost always power that damages things, by overvoltaging enough current flows to allow power to transfer to damage things. \$\endgroup\$ – Kortuk Nov 14 '10 at 1:00
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    \$\begingroup\$ I want to know the answer to this question to. Another way of rephrasing it would be "Suppose I have a 20V power supply but one end of it has a 100kOhm resistor on it, so it can't supply much current. Is it possible to damage a microcontroller board with this power supply? \$\endgroup\$ – DavidEGrayson Nov 14 '10 at 2:42
  • \$\begingroup\$ @davidEGrayson, it is not simple to broadly define that. \$\endgroup\$ – Kortuk Nov 14 '10 at 23:29
  • \$\begingroup\$ @David - Consider asking that as a separate question if you want to know the answer to it. When you do so, add some more specifics. \$\endgroup\$ – Kevin Vermeer Nov 15 '10 at 20:17
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It's primarily because the insulating layers in the device can only withstand a certain voltage. The insulation breaks down with excessive voltage and causes internal shorts.

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    \$\begingroup\$ How will internal shorts damage a microcontroller permanently? What exactly about the silicon will change, and what causes it to change? \$\endgroup\$ – DavidEGrayson Nov 14 '10 at 2:47
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    \$\begingroup\$ The shorts allow large currents to flow, which, because nothing is a perfect conductor, causes a temperature increase, which eventually melts things. \$\endgroup\$ – endolith Nov 15 '10 at 2:07
  • \$\begingroup\$ The thinnest insulating layer (and thus the most susceptible to a voltage breakdown) is the gate oxide in a field effect transistor. I'm not sure what the failure mode is when the gate oxide is punctured (open or short), but the transistor will not function as intended. This failure would not be thermally induced, since the gate current (and thus power dissipation) should be negligable. As Endolith mentions, any short allows uncontrolled current flow, and open circuits or unactivated transistors can let internal circuitry float or get into states that can be self destructive. \$\endgroup\$ – W5VO Nov 15 '10 at 6:29
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P,I,V are well known, but rarely designers and users pay attention to dV/dt.

In power electronics the damage is caused by dV/dt, say about 5000V/microsecond. At this speed the multiple layers of semiconductors (which very often have parasitic thyristor somewhere) open wide and cause avalanche of destructive events.

So it is possible to damage 1000V 200A device with momentary combination of much lesser current and voltage, because energy will dissipate in parts/places of structure different to normally expected.

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  • \$\begingroup\$ How does dV/dt damage the device though? Say that you have a few microamps flowing through it, would it still be damaged, even though the current is very low? \$\endgroup\$ – Thomas O Nov 15 '10 at 20:56
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    \$\begingroup\$ Damage is always thermal. Its several watt or kilowatt of peak power, which irreversibly ruins the part. But the cause of applying such power (say in bridge with hight side shorted to low by accident) is unexpected opening of side, which supposed to be closed. The opening can happen with help of microamp current if it exceeded the dv/dt. It sure, makes no damage for first nanoseconds because its microamperes, but after parasitic thyristor (or parasitic inverse BJT) opens wide, the avalanche current is counted in amperes \$\endgroup\$ – user924 Nov 16 '10 at 1:30
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Reverse biased PN junctions can only take so much voltage before they start to conduct. Sometimes they are designed for this, like zener diodes, but more often they're not. When multiple transistors are fab'd into an integrated circuit, reverse biased junctions can be used to isolate them. If you get one of these normally reverse biased junctions conducting, for example by exceeding the peak reverse voltage it can take, all sorts of unintended conduction paths can be opened, a cook the IC.

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The most common form of electrical damage to things is overheating caused by total power dissipation. In many cases, one can safely get by either limiting voltage to a very low level and not worrying about current, or limiting current to a very low level and not worrying about voltage. There are some exceptions, though:

  1. It is possible for excessive voltage to cause a sudden current flow, or for excessive current to cause a sudden voltage drop, and these currents or voltage drops may be sufficiently localized that damage can occur with very small total power dissipation. In some cases, particularly with overvoltage, it may possible for localized capacitance to hold enough energy to damage the device even if current is externally limited.
  2. As others have noted, excessive voltage or current applied to a pin of a device which is powered may cause the device to enter a mode (such as latch-up) that converts a lot of supply power into heat. Even if the power into the over-voltage or over-current pin is limited, the supply may feed enough power to totally destroy the device.
  3. Overvoltage and overcurrent conditions may accelerate physical or chemical changes in a device sufficiently to cause it to fail prematurely or go out of spec; an electrolytic capacitor which is charged beyond its rated voltage, for example, may have its dielectric gradually get thicker as a result, reducing its capacitance. Note that such effects can cause damage even if power dissipation is slight and cooling is sufficient to prevent overheating.
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Answer is, it depends on the device and how the voltage/current is applied.

If you put too much voltage on a CMOS transistors gate, then it'll punch through damaging it. Maybe enough that the circuit doesn't work, or maybe not. It's classic problem with analog IC's, they get zapped, then they get noisy. Same thing can happen to bi-polar transistors.

Classic failure in CMOS IC's is latch up, where a current spike flips on parasitic SCR's associated with CMOS transistors. Current then flows from VCC to ground, potentially overheating the device. And also high currents frying the protection diodes on inputs, causing them to leak.

And as the man said, dV/dt tends to kill power devices. Often because it causes them to partially turn on in localized areas, which then overheat and punch through. Which is why rolling your own motor controller usually results in big smoke.

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I know this is gross simplification, but the way I always looked at it was; over voltage breaks down the insulation layer between conductors and damages devices, over current damages the conductors themselves primarily through over heating.

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