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Can low power damage components?

Frick

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#1
It basicly comes from a statement made by drdeathx:

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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#2
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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#3
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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#4
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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#5
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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#6
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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#7
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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#8
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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#9
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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#10
^ yeah, i too must say that i have noticed drdeathx to have poor sense of logic on many an occasion!
 

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#11
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:
Code:
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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#12
Aquinus, you said the opposite of me, but I thought the culprit is vdroop.
 

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#13
1freedude - Please do not double post. Use the "Edit, Quote and/or Multi-Quote" features...
 

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#14
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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#15
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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#16
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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#17
12v nominal. 14.4v hot off the charger. Not even close to the topic.
 

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#18
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!


Ohms law can be describe by this equation:
Code:
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.
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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#19
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. :p

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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#20
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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#21
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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#22
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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#23
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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#24
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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#25
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. :p
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.