# Can low power damage components?



## Frick (Feb 18, 2013)

It basicly comes from a statement made by drdeathx:



drdeathx said:


> Umm yes, that's why manufacturers have minimum specs on amps. My buddy had a GPU go bad with it being underpowered. Just sayin. Don't want to start a debate



As you can see he says a GPU can die if it does not get enough power. I want to know if that is true, and if it is, why? Because it kinda goes against what I know about electronics. I can see why something would be unstable, but flat out damage?

So lets get this rolling! Be very technical if you can!


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## 1freedude (Feb 18, 2013)

Not underpowered, but undervolted.  As the volts go down, the amps will rise to maintain the same, static, power.  This rise in amps at lower volts will do the damage.


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## micropage7 (Feb 18, 2013)

electronic stuff has range where it could operate nicely, if you give them too low i guess it would affect it
its like you pushed to work hard but you dont have enough food to give you power


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## W1zzard (Feb 18, 2013)

1freedude said:


> As the volts go down, the amps will rise to maintain the same, static, power. This rise in amps at lower volts will do the damage.



i thought about that, too, but i'm not sure about it.

what mechanism do you propose causes damage if power is equal but current is higher?


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## MyTechAddiction (Feb 18, 2013)

Usually low power by undervolting is considered to expand the life of electronics devices.However undervolting CPU GPu and or memory (below specs) might cause data coruption.


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## Frick (Feb 18, 2013)

MyTechAddiction said:


> Usually low power by undervolting is considered to expand the life of electronics devices.However undervolting CPU GPu and or memory (below specs) might cause data coruption.



That I can understand, but permanently damaging it?


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## de.das.dude (Feb 18, 2013)

i dont think it is possible. current is dependent on the voltage. if the volts go down, the amps will go down too. (amps = volts/resistance)

unless there is some way that the resistance decreases with voltage, it cannot happen otherwise. Also semiconductors have a negative temperature coefficient to resistance. more temp, = less resistance, and to get to high temperatures, it would mean high current. so.. this says its not possible too.

now if some kind of load line caliberation is employed that raises the amps on loading, then it would be somewhat possible. but i think most LLCs are voltage based as voltage is much easier to control than current.


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## 1freedude (Feb 18, 2013)

W1zzard said:


> i thought about that, too, but i'm not sure about it.
> 
> what mechanism do you propose causes damage if power is equal but current is higher?



Heat.


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## Aquinus (Feb 18, 2013)

This sound, not like an under-current problem but an over-voltage problem. If it is a power supply problem and the PSU can't provide the proper current at the right voltage, it may attempt (or forced) to compensate by over-volting the rail to attain the same power output. (Example: 12v @ 2A has the same power output as 13.3v @ 1.8A.)  This in itself will damage components, not because the current is too low, but because the voltage is too high. I suspect drdeathx has once again (I could be wrong,) evaluated the problem and jumped to the wrong conclusion since multiple things could be going on. The drop in current might be a symptom of the issue, not the issue itself.


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## de.das.dude (Feb 18, 2013)

^ yeah, i too must say that i have noticed drdeathx to have poor sense of logic on many an occasion!


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## Aquinus (Feb 18, 2013)

W1zzard said:


> what mechanism do you propose causes damage if power is equal but current is higher?



That's impossible, a circuit (as a standard temperature, but in normal operating conditions impedance of the circuit doesn't change much,) has a fixed impedance. Lower voltages could *never* result in higher currents.

Let's have a physics lesson, shall we?

Ohm's law states!


			
				Wikipedia said:
			
		

> the current through a conductor between two points is directly proportional to the potential difference across the two points.



Ohms law can be describe by this equation:

```
I = V / R
```

Since we know that R is a fixed value since the resistance (impedance in our case, we have alternating voltages,) that means if I (current) were to increase, so would V. If V increases, so does I. Also as a side node, as I (current) increases in a circuit, the heat generated by this said circuit increases exponentially with respect to current. Voltage has no direct impact on heat. It does influence current which does generate heat though.


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## 1freedude (Feb 18, 2013)

Aquinus, you said the opposite of me, but I thought the culprit is vdroop.


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## HammerON (Feb 18, 2013)

1freedude - Please do not double post. Use the "Edit, Quote and/or Multi-Quote" features...


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## Aquinus (Feb 18, 2013)

1freedude said:


> Aquinus, you said the opposite of me, but I thought the culprit is vdroop.



Vdroop protects your CPU by keeping the voltage spike from transience as current lower from a high current to low current state. What happens is as the CPU uses less power, the inductors in the VRM array will resist the change in current and will produce transience. The goal of VDroop is to lower the loaded voltage so the spike from the transience produced isn't as high. This keeps your CPU from being subjected to high voltage spikes.


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## jaggerwild (Feb 18, 2013)

Aquinus said:


> This sound, not like an under-current problem but an over-voltage problem. If it is a power supply problem and the PSU can't provide the proper current at the right voltage, it may attempt (or forced) to compensate by over-volting the rail to attain the same power output. (Example: 12v @ 2A has the same power output as 13.3v @ 1.8A.)  This in itself will damage components, not because the current is too low, but because the voltage is too high. I suspect drdeathx has once again (I could be wrong,) evaluated the problem and jumped to the wrong conclusion since multiple things could be going on. The drop in current might be a symptom of the issue, not the issue itself.



 So in theory why does your car provide 14 volts when its a 12V system? Voltage fluctuates while a PSU  is running, you can verify this in the Bios check the 3V + 5V feeds.


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## Aquinus (Feb 18, 2013)

jaggerwild said:


> So in theory why does your car provide 14 volts when its a 12V system? Voltage fluctuates while a PSU  is running, you can verify this in the Bios check the 3V + 5V feeds.



No, a car produces 14v because the extra voltage is required to charge the lead-acid battery. You can't charge a lead-acid battery up to 12v using 12v, you need a bit more. Completely different reason for having the higher voltage in a car than in a computer.


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## 1freedude (Feb 18, 2013)

12v nominal.  14.4v hot off the charger.  Not even close to the topic.


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## W1zzard (Feb 18, 2013)

Aquinus said:


> That's impossible, a circuit (as a standard temperature, but in normal operating conditions impedance of the circuit doesn't change much,) has a fixed impedance. Lower voltages could *never* result in higher currents.
> 
> Let's have a physics lesson, shall we?
> 
> ...



I think your underlying assumption that resistance is constant [in the case we are discussing] is incorrect.

A 440 W hairdryer will use 2 A at 220 V and 4 A at 110 V


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## Aquinus (Feb 18, 2013)

W1zzard said:


> I think your underlying assumption that resistance is constant [in the case we are discussing] is incorrect.



I'm simplying it, obviously resistance changes as different parts of the CPU are used, but assuming constant load (or idle,) the resistance will generally speaking remain the same. When the resistance changes is when load on the CPU changes, in that case I would say that transience would be more likely to kill a CPU, not low current. High voltages and high currents is what damages devices, not having too little.

The simple point is that low voltage + not enough current will not damage a device. It doesn't harm a transistor if the base doesn't reach its saturation point or anything. The problem is if the voltage is too high and it's over the transistor's breakdown voltage. The real thing to take away from this is that drdeathx is wrong. 

Do you see cell phones failing (physically, not the battery.  ) because you let the battery fully discharge? Common, this argument is really kind of ridiculous.


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## BiggieShady (Feb 18, 2013)

It is certainly possible to design a circuit that overloads one of it's component when input voltage goes below certain point (using positive feedback, delays, negative resistance, transistors used as amplifiers). 
But why would anyone do that in this case? 
Circuits have bunch of extra components to protect it from overload which can happen only and only if input voltage rises. To design a circuit vulnerable to lower input voltage would be a true achievement in unusefulness :shadedshu


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## qubit (Feb 18, 2013)

1freedude said:


> Not underpowered, but undervolted.  As the volts go down, the amps will rise to maintain the same, static, power.  This rise in amps at lower volts will do the damage.



No it doesn't. The current will also drop and there will be less power to use. It's exactly like when batteries run down. Why? Because you cannot increase the current through a fixed resistance without increasing the voltage across it. This is basic physics. EDIT: the equation that governs this is V=IR.

And no, undervolting doesn't cause damage, but it wouldn't surprise me if there's some odd fluky situation where it might, in the case where a connected component is operating at full power, causing an unintended current to flow between them, perhaps.


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## jaggerwild (Feb 18, 2013)

Aquinus said:


> No, a car produces 14v because the extra voltage is required to charge the lead-acid battery. You can't charge a lead-acid battery up to 12v using 12v, you need a bit more. Completely different reason for having the higher voltage in a car than in a computer.



 I was showing that  voltage changes, like erocker posted in the other thread "usually you will get a 3V or 5V rail" out of wack witch in time will cause things to start actting funny.
Heat+ Effecency=over time cause failure.
Carry on!!


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## W1zzard (Feb 18, 2013)

Aquinus said:


> Do you see cell phones failing (physically, not the battery.  ) because you let the battery fully discharge?



a li-ion battery will die if it's fully discharged. additional cellphone logic does prevent that. but that's not related to the discussion here.
a battery is different to a psu, the battery will simply run out of charge


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## newtekie1 (Feb 18, 2013)

If low power didn't kill electronics we wouldn't have to worry about brown outs, but we know that brownouts do kill electronics.

In terms of PC components, I don't think it is the actual GPU or CPU that gets damaged, but rather the VRMs.  They are designed to take 12v +/- 10% in and output whatever voltage the GPU/CPU needs.  But if the voltage going into the VRM drops the amps go up, and the VRM has to work harder to supply the component with the power it requires.  This could damage the VRM.

But the more likely reason as to why GPU manufacturers put minimum power requirements on their cards is so the power supply isn't overloaded.  If the power supply is overloaded, then it dies, and when a PSU dies it could take components in the system with it.  Of course an overloaded PSU will also likely have pretty big voltage drops, so if you start feeding the VRM 11v when it is expecting 12v, it could damage the VRM.


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## McSteel (Feb 18, 2013)

Let's not forget that a PSU operates as a voltage source, not a current source. As it's voltage drops due to an inability to provide sufficient power at a given load, the CPU or GPU VRM has to compensate for the drop on it's input. This requires additional work, thus producing excess heat. In theory, using a PSU that will continuously provide voltage significantly below ATX spec but just enough for the PC to function (say 10,8V @ 12V rail), and having inadequate VRM cooling could lead to permanent capacitor damage, as ripple levels increase and so does heat. This is a very specific scenario, one I believe has lower probability of arising than getting struck by lightning...

TL;DR: *drdeathx* is wrong.

## EDIT ##

newtekie1 beat me to the punch... He posted just as I started replying... Hate it when that happens. 
Anywhoo, feeding a PSU low voltage at it's input (say 90V instead of 120V) won't necessarily kill it or cause it to shut down, but it will lower it's efficiency and loosen it's voltage regulation noticeably.


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## xvi (Feb 18, 2013)

jaggerwild said:


> So in theory why does your car provide 14 volts when its a 12V system? Voltage fluctuates while a PSU  is running, you can verify this in the Bios check the 3V + 5V feeds.



Car voltage is higher to charge the battery. Many car accessories have a decently good tolerance to voltage since it is a bit necessary. 13.5v to 16v is intentional. The way I understand it, the alternator will pulse, for example, twice the voltage at 50% duty cycle to equal the correct voltage

On PSUs, voltage is commonly set intentionally high so that the voltage droop under load stays within spec. Take two identical power supplies, one set for 12v and one set for 12.2v, the former will drop out of "normal" 12v range as defined by the ATX specification slightly before the latter.



Aquinus said:


> I'm simplying it, obviously resistance changes as different parts of the CPU are used, but assuming constant load (or idle,) the resistance will generally speaking remain the same. When the resistance changes is when load on the CPU changes, in that case I would say that transience would be more likely to kill a CPU, not low current. High voltages and high currents is what damages devices, not having too little.
> 
> The simple point is that low voltage + not enough current will not damage a device. It doesn't harm a transistor if the base doesn't reach its saturation point or anything. The problem is if the voltage is too high and it's over the transistor's breakdown voltage. The real thing to take away from this is that drdeathx is wrong.
> 
> Do you see cell phones failing (physically, not the battery.  ) because you let the battery fully discharge? Common, this argument is really kind of ridiculous.



Voltage regulators keep the voltage constant. Like W1zzard said, logic on the battery/phone will kill the power before causing physical damage.


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## Aquinus (Feb 18, 2013)

newtekie1 said:


> In terms of PC components, I don't think it is the actual GPU or CPU that gets damaged, but rather the VRMs. They are designed to take 12v  /- 10% in and output whatever voltage the GPU/CPU needs. But if the voltage going into the VRM drops the amps go up, and the VRM has to work harder to supply the component with the power it requires. This could damage the VRM.



Not that you're wrong, but I'm pretty sure the PSU will shut off or fail before the VRMs get exposed to that kind of voltage. Considering the PSU has to draw more current to do the same job, the amount of heat that it generates go up very quicky if the PSU stays active. All in all, I'm pretty sure a brownout would damage your PSU more than any other component. Surges on the otherhand are a different story. Also, I'm pretty sure that modern powersupply need under and over voltage protection by spec.

Yup! Here it is.



			
				ATX Spec said:
			
		

> 1.3.1: The power supply shall contain protection circuitry such that the application of an input voltage below the minimum specified in Section 3.1, Table 1, shall not cause damage to the power supply.


Source

...and that "low voltage" limit is 90v AC on 115v mains iirc.

Here is a snippit from a superuser.com thread:


			
				superuser.com thread said:
			
		

> Typically power supplies have an input section composed of a bunch of interesting circuitry that, at the end of the day, provides about 308 VAC to a transformer, which then powers the regulation and conditioning circuitry. This circuitry actually forms the major basis of the regulation circuitry, and if you are using less than the full wattage of the power supply may be able to manage with significant undervoltage conditions without falling out of regulation on the output side.
> 
> When a brownout occurs, the powersupply will attempt to deliver the rated current for as long as it can (based on the incoming voltage and current) and if it cannot maintain regulation it'll deassert the Power Good signal going to the motherboard. The motherboard is responsible for deasserting the power on signal going to the supply, and if it does so in time, then the supply will drop all its output and turn off.
> 
> ...


Source


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## newtekie1 (Feb 18, 2013)

Aquinus said:


> Not that you're wrong, but I'm pretty sure the PSU will shut off or fail before the VRMs get exposed to that kind of voltage. Considering the PSU has to draw more current to do the same job, the amount of heat that it generates go up very quicky if the PSU stays active. All in all, I'm pretty sure a brownout would damage your PSU more than any other component. Surges on the otherhand are a different story. Also, I'm pretty sure that modern powersupply need under and over voltage protection by spec.



First, not every PSU adhears to ATXSpec.  Those cheap PSUs that weight less than a feather, I guarantee you they aren't following ATX Spec and probably don't have any protection in place for a brownout.

Second, I'm not saying a brownout kills components, they usually kill the PSU, I've seen it, but rarely anything else.  I was just using the brownout as proof that low voltage can in fact kill electronics.

The PWM on the motherboard or graphics card also doesn't have the protection features built in that a PSU does.  So they are even more vulnerable to low voltage situations.  So when a PSU is overloaded to the point that the 12v voltage is below 11v, it could definitely damage something.


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## LAN_deRf_HA (Feb 18, 2013)

This is the wrong way to go about this conversation. You need to find if there's confirmed cases of this or not first. If there aren't you can theorize why, and if there are you do the same until the explanation fits the result. Everything so far is "blind" theorizing which isn't all that productive.


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## TheoneandonlyMrK (Feb 18, 2013)

Imho if a psu is strained to the point of drooping a lot and under volting its main rails its technically possible for damage to occur .
Its unlikely but certainly possible.
Not in my op directly the low voltage though , more the circuit instabilty due to the complex power regulation components overworking or being forced outside there designed operating specs, in the extreme causing power spikes beyond the specs of parts later in the circuit and causing drdeath of the mobo and possibly more.
Oh and im baseing this off having seen it a few times Rip Q6600 oh again,  Before you say it its not prove able easy as dead bits stay dead mpre oft then not and I was using a corsair psu not a cheapo


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