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I was wondering if there are some particularly important mechanism by which one can break electronics, when undervolting it. Its pretty obvious that lots of electronics will not work properly if undervolted, but what about permanent damage? The question was motivated by repair work. I was wondering about what sorts of secondary effects one should look for when a damaged power supply was involved.

I imagine motors could be damaged if they stalled due to undervolting.

So what are specific mechanisms for permemenant damage due to undervolting (or better put undersupplying)? Are there even any?

To add to the question, what are components or simple circuits that fail when undersupplied?

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  • \$\begingroup\$ There are some electronic parts that can be damaged by under voltage. For example, if the filament voltage on a Magnetron is too low, it can be damaged. \$\endgroup\$ – Suirnder Apr 16 '13 at 15:11
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Damage by undervoltage is not as common as by overvoltage, but it is not unheard of.

An example: a simple circuit that has a power mosfet driving a motor. The intention is that the mosfet is either completely on or completely off. In both cases the power dissipated by the mosfet is very low:

  • when it is on the power is low because the o-fully-on resistance of the mosfet is very low, hence the voltage across it is also very low, so the power (V*I) is low.
  • when it is off the full voltage of the power is across the mosfet, but the current is almost zero, hence the power is almost zero too.

A mosfet needs a certain voltage at its gate to turn fully on. 8V is a typical value. A simple driver circuit could get this voltage directly from the power that also feeds the motor. When this voltage is too low to turn the mosfet fully on a dangerous situation (from the point of view of the moseft) can arise: when it is half-on, both the current through it and the voltage across it can be substantial, resulting in a dissipation that can kill it. Death by undervoltage.

Note that I started by assuming a simple circuit. In practice a serious circuit like this would have an undervoltage protection.

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    \$\begingroup\$ Magnetic Ballasts (not electronic) used in HID or similar lighting can be also be damaged if under powered for long periods of time. Usually causing it to overheat and burn out. \$\endgroup\$ – Piotr Kula Apr 16 '13 at 15:56
  • \$\begingroup\$ And a Transistor as well, though by undercurrent causing it not to saturate, instead of undervolting it. \$\endgroup\$ – Passerby Apr 16 '13 at 16:16
  • \$\begingroup\$ @Passerby that is indeed possible, but much less likely than for a mosfet: a transistor that is designed to be fully on is usually overdriven by some large factor (by choosing a suitable resistor), so a few volts less won't matter much. A mosfet is voltage-driven, so there is no easy way to keep a safety margin when the supply voltage the circuit was designed for is fixed. Plus, the safety margin for the gate voltage is not nearly as large as for base current: a common maxiumum is 18V. \$\endgroup\$ – Wouter van Ooijen Sep 13 '16 at 12:57
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Wouter has some good information, but there are more scenarios where not providing a high enough voltage can damage a device.

Some higher end display screens require multiple voltage sources, and failing to power one source to a high enough level, or fast enough, before a second source, can cause damage to the screen or controller.

Some devices with internal mosfet can be damaged by underpowering the source. As was explained by a TI employee about a current controlled led driver, if the VLed source is too low to provide the selected current through a channel, the logic in that channel will try to drive the channel's mosfet harder to try to sink more current. Eventually, the mosfet will burn out, if not other parts of the chip. I wish I could find that discussion and link it.

While not directly causing damage to the device being underpowered, failing to provide the right voltage to a heating element could cause what is being heated to not heat up correctly/fast enough. Winter Water Pipe heaters, electric stoves, microwaves (for a loose meaning of "heater"), certain car parts. Worse, medical devices or heating in artic environments. Same for cooling solutions, like fans or ACs or pelters. A underperforming fan due to voltage issues can cause it's target to overheat. Water pumps as well. And all three can be damaged by the side effects of it. Water pumps normally use the moving water to cool themselves. A lower voltage will cause it to move water, but might not be fast enough to cool itself down. Underperforming fans might be cooked by the device it could not cool down. Heaters themselves might freeze if they cannot get hot enough.

And last I can think of, battery chargers. A malfunctioning charger, or simply badly designed one, as part of a larger circuit, could cause a lower voltage in a charging state. A battery could feed back into the circuit when it shouldn't.

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  • \$\begingroup\$ I'm assuming this is probably why a spot near the inverter on an LCD controller for a project of mine caught fire when a ribbon cable came out? Was never sure exactly why that happened since I didn't have any datasheets and it came from a no-brand dvd player. \$\endgroup\$ – Wyatt8740 Jan 18 at 16:43
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It depends on your load.

If it's a resistive load, lowering voltage means it will conduct less current and dissipate less heat. Nothing wrong here.

If you drop the voltage on the gate/base of a transistor and it may not fully saturate and have a larger voltage drop. As power dissipation is P=U*I; the voltage drop on the transistor could double (from 0.5V to 1V) while the current may stay more or less the same (for e.g. 1000mA to 800mA). You effectively doubled the power dissipation and that could lead to damage!

If the device uses a linear regulator, the regulator will have to regulate less voltage. This will lead to lower power dissipation. Of course there is a limit at which the regulator cannot maintain regulation anymore and the output voltage will drop too. This output may shutdown or stop working at a certain point.

Switch mode power supplies are a constant power load. If you assume the output to draw a constant power ; for example 3.3V 1A. This equals to 3.3W which means whatever the input voltage is, it will always draw 3.3W. In practice you have efficiency (which can vary) and limits to the voltage region, but it will try to draw 3.3W.

If the input voltage drops the input current increases. If parts like inductors, diodes or MOSFETs cannot handle the higher current (heat dissipation or exceeding saturation/peak currents) it can cause damage.

However, in that case you're probably exceeding a certain operation window. For example, a product may have a input voltage requirement of 9-15V. Although the switching regulator would work fine on (for example) 7V, it may exceed the current on some part and become unreliable.

Sometimes you see "Undervoltage lock-out" on these devices. This is a voltage at which the switch mode supply will shutdown because it cannot guarantee reliable operation.

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  • \$\begingroup\$ You suggest that a linear load (resistive) is safe - agreed. Some resistive loads are monotonic (like a lightbulb) but still require a supply of over-current while it lights up. If the supply is too feeble to supply this brief overcurrent, it can come to grief. Especially true of quartz-halogen bulbs. \$\endgroup\$ – glen_geek Sep 13 '16 at 14:07
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One example of a specific failure mode of certain electronic systems is Latch-Up.

https://en.wikipedia.org/wiki/Latch-up

Quote from the above link...

This frequently happens in circuits which use multiple supply voltages that do not come up in the required sequence on power-up, leading to voltages on data lines exceeding the input rating of parts that have not yet reached a nominal supply voltage.

Often this can be resolved by simply power-cycling the system, but if that system is controlling some other mechanism it can cause further failure or even physical damage as an indirect side-effect.

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The general term for low-voltage events is "brownout"; there are lots of ways of incorporating brownout prevention into your power supply design.

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    \$\begingroup\$ While this is a valid comment, it does not answer the question in any way. \$\endgroup\$ – Olin Lathrop Apr 16 '13 at 15:32
  • \$\begingroup\$ I thought brownouts were short time intervals of under voltage. not constant under voltage. browouts occur when a normally powered device suddenly draws a higher current and causes a voltage dip- which can be counteracted using filter capacitors. I don't think an under powered device can be called as a brownouted device?! \$\endgroup\$ – Piotr Kula Apr 16 '13 at 15:53
  • \$\begingroup\$ @ppumkin what if the brownouts are constant, like a sine wave? At what point does a brownout become undervolting? \$\endgroup\$ – Passerby Apr 16 '13 at 16:14
  • \$\begingroup\$ Well under volting would result in the device not function at all. Where as some devices can deal with minor brownouts and function normally(have internal filters) other devices just reset because it is up to the designer to implement brownout protection. so brownouts != undervolting - even if it is sinusoidal and the wave dives into a brownout then to an under volt then comes back in as a start up voltage and all over again? \$\endgroup\$ – Piotr Kula Apr 16 '13 at 16:19
  • \$\begingroup\$ @ppumkin brown out results if you dip but not far enough to cause reset, and you can hang down there for a long time without changing system response and when you recover voltage it will be stuck in brown out. I agree this is not a correct answer. \$\endgroup\$ – Kortuk Apr 16 '13 at 17:59

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