+12v possible problem?

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ok for as long as i can remember i've been having problems with my computer rebooting, i've swapped basically every piece in my machine but to no avail. i was looking at my asus probe and i noticed that my +12v was reading at 13.1, and it sets my alarm sometimes when i have the probe running. now all my other voltage settings, vcore, +3.3v, +5v all run lower then what that is , could this be causing my machine to randoming reboot? also, how can i manually set how much the +12v pulls? thanks!
 
I'm not an expert on this, but since no one else has said anything, I'll give you my 2 cents. I would guess that your power supply should put out at least 16A if not more. You've got a lot of stuff running on that rail. You say "as long as I can remember", does that mean it did not always give you the alarm?
 
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as long as i remember, meaning i can't remember when my computer didn't reboot, cause if i could i might be able to single out the problem by remember what i had just added that caused it to reboot. i haven't always run the probe, and just recently i noticed that it was running at 13.1, sometimes 13.2, i was told this is not healthy for my mobo. could this be causing it ? is there anyway to adjust it ?
 
Firstly, it is unlikely that +13.2 volts is going to damage anything (and you haven't reported any damage yet) and it at least shows that your PSU is able to supply enough current on the +12 volt line - if it wasn't able to supply enough current, then the voltage would drop, not increase. Also, your motherboard will have regulators on board to help keep the +12 volts at a suitable level, so if the voltage did get a lot higher, it would be your IDE devices that would be the first to fail, as these are connected directly to the +12 volts from the PSU.

Secondly, the voltage sensors on your motherboard are unlikely to be very accurate, so the +13.2 volts indicated may be incorrect. This voltage is out of spec, but most components are normally fine with some overvoltage. If your computer is spontaneously rebooting, and you suspect the PSU, then the cause is most likely due to unstable, or too little, voltage output rather than the voltage being a bit high. If it got much higher, then you should start to worry, as components may be damaged.

However, the fact that you say "... all my other voltage settings, vcore, +3.3v, +5v all run lower then what that is ..." could potentially be a problem, as the voltages may be significantly lower than the required values, thus leading to instability. Note that the required tolerance for these voltages (and also the +12 volt line) is around +/- 5%, so you can do the calcs yourself. To be certain you would need to check the voltages with a multi-meter, or some other accurate measurement device. Try substituting the psu if you can.
 
13.2 is is within 10% tolerence, it is borderline, but still within tolerence. Voltage and current do not necessarily rise and fall together, therefore Nic, your assumption is wrong.

You need to check the current of the 12v line, it should be 16A or higher. You then need to check what your devices are using combined, this is where the problem would lie if it is in the PSU. Your voltages look fine, you need to check the current, this is usually where the problem comes in.

People always assuming that voltage and current are the same and behave the same, is the major reason the PSU is blamed even when it is not at fault, it is also the reason it is dismissed when it is the cause.
 
As far as I know all mainstream electrical devices still fall within the standard tolerance range of +/- 10%, not 5%, and are designed as such. Any device that has a lower tolerance range is going to be a lot more expensive. Even the most simplistic of devices can double, triple, and even quadruple in price with tolerance ranges of 5% or lower.

"People always assuming that voltage and current are the same and behave the same"

Yes, I notice that too.


A cheap investment that can help out quite a bit is a DMM. Buy a DMM that can check voltage, current, and resistance, and with some basic training you can learn how to accurately test devices like that. I love mine.
 
If you guys are talking about me "people always assuming------" I'm not offended, I should have stayed out of this thread anyway. Thanks for setting me straight. Thanks.
 
Olefarte, if you read the posts more thoroughly (which is one of my flaws), you would have noticed that Soul and Storm were not referring to you. It is obvious that they were setting Nic straight about the 5% tolerance which is indeed 10%. As with many of my posts, they may sound offensive, but it is not my intention to offend anyone. I would just like to make that clear before someone bites off my head about it.
 
Olefart, I wasn't refering to you, I didn't see anything wrong with what you said either BTW. I was refering to statements made by Nic, but was also talking about general misconceptions and assumptions made by many people dealing with PSUs.
 
OK folks, I guess I am a little guilty, so apologies for any misunderstandings. :)

With regard to voltage tolerances, if you look over here ...

ATX 2.01 Spec [pdf]

you'll see that voltage specs are quoted at +/- 5% (3.3 volts @ +/- 4%, but close enough).

However, if you now look over here ...

ATX 2.03 Spec [pdf]

you can see that some of the tolerances have been changed to +/- 10%.

Needless to say I was looking at the older spec when I made my post, but its not really relevent, as most devices will operate outside these ranges (especially on the '+' side) without much complaint - overclockers do this all the time.

With regard to voltage and current not necessarily rising together, well thats true, because it all depends on the resistance. Remember that simple litte formula they taught you at high school (voltage = current * resistance), well that shows that for any given resistance (notice I'm making an assumption that the resistance, stays the same, then current does in fact increase with voltage.

Now, given that the problem computer in question is as stated running normally, though unstable, and the the PSU is working (we are in windows running asus probe) and power is being supplied, the expected current draw would be as normal. Note that due to the way switching PSUs work to maintain output voltages, and relating to the issues raised in this thread (voltage @ 13.2 volts does not necessarily mean that the PSU is supplying enough current to the computer) ...

1) If the PSU was NOT able to supply enough current, i.e. the load was demanding more than the PSU could provide, then the output voltage would decrease and NOT be maintained at 13.2 volts. This is typically what happens when you use a PSU that is of too low wattage rating for a PC - the computer won't be able to power up at all.

2) The only other way that 13.2 volts could be present, with low current output, would be if the load was very high (low resistance). This is why lower rated PSUs can be used in smaller PCs that have lower power requirements.

Finally, as we already know that the PC in question has booted into windows and all the peripherals and components are running (i.e. the load is normal), then the measured voltage of 13.2 volts suggests that the PSU is meeting all the demands for current and that is why the voltage is able to remain high.

Try putting in some high current drain devices (low resistance - e.g. short circuit) and you are certain to see the voltage drop followed shortly after by the PC shutting down (assuming that it had already booted before we add the extra load).

Also note that the current demand is greatest at start up, therefore given that the PC has booted into windows it would seem reasonable to believe that the PSU is in fact easily capable of supplying the power required (to maintain its current state).

Now I am NOT perfect, and I DO get things wrong from time to time, but mostly I try to be as accurate as I can. Its true that the atx specs I refered to were older versions (his PSU might very well be atx 2.01), but it doesn't change the reasoning behind my comments, which are valid and still stand.

If anyone believes otherwise, then please feel free to post your reasoning here so others may comment. Hope this helps clarify things a little, so don't be afraid to post if you disagree and wish to say so. :)

One last thing, there is one case where I can see that the PSU in question, is unable to supply enough power, despite having 13.2 volts output, and that is if there is some device present that can suddenly increase the load on the PSU by a significant amount, such that the output voltage would momentarily drop below the required threshold. But then that should be noticeable, and would not be random as it could be replicated quite easily. Also, its hard to imagine any device that could achieve this, though its not impossible.
 
Here's some additional info on PSUs (for anyone interested). The link below is to an article in Ton's Hardware Guide, where he stress tests 21 PSUs to see if they live up to their specifications. I've linked to a page that shows PSU output voltage *dropping* when the PSU is unable to meet the current required by the load.

Here it is ...

Maximum Output Overload Test [Comparison of 21 Power Supplies]
 
Nic, some people do use +/-5%, but this tolerence is generally reserved for highly precision instruments such as test equipment. The industry standard for most consumer components is +/-10%. A computer PSU is not considered a precision device, therefore is is governed by 10% tolerence.

The other thing you are completely missing is the relationship among power, voltage and current. Voltage multiplied by current = Power. Voltage drops to create current flow. A component could be drawing more current, you were almost on to something there but were mistaken in part of what you said. Motors tend to draw more current on startup than they use while running, therefore this could be posing a problem if startup load is above the current rating of the 12v rail.
 
Storm, just to clear up a few things ...

1) I didn't write the atx specs, I only quoted them.

2) Yes, P=V*I, just as you say.

3) Voltage does NOT drop to create current flow, but drops because the device supplying the current has reached its limit and the output voltage thus sags, due to low impedence of the load (low resistance).

The voltage has to drop because most power supplies have a limit on the power they can supply, so as current goes up, voltage goes down, and power output remains roughly the same (max for PSU). P=V*I, to maintain 'P' at max (e.g. 450w), if 'I' increases then 'V' decreases. Nuff said.

When you put a short across a power supply - the voltage will drop to something approaching ZERO volts and the current will be VERY high - remember the resistance of a short is close to ZERO and I=V/R. Note that I=VERY_HIGH, V=LOW, but R=VERY_LOW_INDEED.

This is just as you said: High Current + Low Voltage, both at the same time due to extremely Low Resistance. But remember that as was mentioned in the problem scenario, voltage output was remaining high, therefore as I had said, and as your own argument points out, the PSU is in fact meeting the power demands of the system, so the high voltage was a sign that the PSU is NOT underpowered.

So as you can see, all my previous statements were in fact correct. Say no more. Apology accepted. ;) :p
 
Originally posted by Nic
Storm, just to clear up a few things ...

3) Voltage does NOT drop to create current flow, but drops because the device supplying the current has reached its limit and the output voltage thus sags, due to low impedence of the load (low resistance).

The voltage has to drop because most power supplies have a limit on the power they can supply, so as current goes up, voltage goes down, and power output remains roughly the same (max for PSU). P=V*I, to maintain 'P' at max (e.g. 450w), if 'I' increases then 'V' decreases. Nuff said.

When you put a short across a power supply - the voltage will drop to something approaching ZERO volts and the current will be VERY high - remember the resistance of a short is close to ZERO and I=V/R. Note that I=VERY_HIGH, V=LOW, but R=VERY_LOW_INDEED.


I don't know if you are just mistaken or being coy. What are you talking about? Voltage must drop for there to be current. No ifs, ands, or buts about it. Even a short will have a voltage drop which can be read in at different points. Voltage does not drop because of excess load. What i see you doing is applying series circuit theory to a parallel device, they aren't the same and do not behave as such. As you increase load in a parallel circuit, moreso with low resistance devices, current goes up and source voltage remains the same, period.
 
Nic, it looks to me that you are confused. Series circuits and parallel circuits do not behave and react the same.

I don't know how much experience you have in this stuff, but I've been working in electronics for over 10 years. I think I know the theory of operation of electronic devices. What you said was partially correct but partially incorrect as well. Voltage must drop before current can flow, not the other way around. The load decides the drop as well as the current. The source voltage will remain the same.
 
Looking at the statements that have been made since my last post, it seems that we are all of us confused and misinterpreting what each of us are trying to say. :confused:

As for voltage having to drop for current to go up, this is what *can* happen in real life, and is a symptom of the fact that all power sources have a limit on the power that they can supply (as I stated), and is due to the internal impedence of the power source. So voltage does not *have* to drop, it drops because the power supply is unable to maintain it (bigger PSU - less voltage drop).

Remember, that in the case of a voltage *regulated* PSU, the voltage will only drop when the PSU has reached the limit of how much power it is capable of supplying (voltage does NOT drop as current increases - unless the power output limit has been reached).

In the case of an *unregulated* device, such as a 1.5 volt dc battery, then yes, the voltage will drop as more current is supplied.

As to measuring the voltage across a short, then if it is a true *short*, the voltage will always measure ZERO. We are talking theoretically here, since we all know that in the real world there is no such thing as a true short, only very very low impedence.

It's sometimes easy to forget that electronic and electrical devices can change the way we expect things to operate, and thats why we find these things so useful. In the case of the voltage *regulated* psu mentioned earlier, it will only start behaving as a linear power source, when it has reached its limit for power output. :)

Hopefully things are a little bit clearer now, though I'm not counting on it. :blush:

PS: I am not under any misconceptions whatsoever. :)
 
SOTLordRahl

My quick answer is 'most likely not'. The +12V Line to the motherboard is mainly used for motherboard fan connectors, serial ports and possibly to voltage regulators, all which should have a wide latitude of input voltage with no ill effects. The low +5V line could be a problem if it's down around 4.80V and 3.3V line if it's down below 3.25V . You might check this thread for our comments on the same subject....

https://www.techspot.com/vb/showthread.php?s=&threadid=5174

The most likely causes, after looking at your profile to see your hardware, would be drivers (update or roll back), messed up registry (reinstall recommended), applications, or ram (get memtest86 and test).

OK, back to the thread hijackers!:p
 
Nic, you are still making a mess with your explanations. It is circuit voltagew that behaves the way you are describing, source voltage stays the same unless the source changes. You start out talking about a device using more power than it is rated to use, then you go off on a tangent about the PSU being flawed. If the problem isd in the device, that would be circuit voltage, but if the source is changing because of a flaw in the PSU, that is a different story. I'll leave it at that. This argument is not even relavent to this problem. You seem to have taken it personally that some of us here who have extensive background in electronics have found your amateur explanation flawed. You should just move on.
 
Storm, I didn't mention any names in my last post as it is not my intention to make others feel inferior. However, the comment you made about my 'amateur flawed explanation' is a touch offensive.

Firstly, I have an honours degree in electronics engineering (I was top of my class), and secondly I have around 10 years experience in electronics, working in the defense industry.

I feel I should like to point out that it is in fact your own comments that are flawed and amateurish, though I would never have said that before (I always give others the benefit of the doubt), buts its true. If you are unable to understand the arguments I set forth, then it is simply because you are merely pretending to understand the subject, and have lost the plot when it comes to really knowing what you are talking about.

Sorry, if I seem a little annoyed, but I am just a wee bit.

Please define what you mean by 'source' as you put it.

Here's a link that might refresh your memory a little ...

Kirchoff's Laws

PS: I 'll reply further if you wish, but for now just as you said, we as off topic, and its time to move on. ;)
 
One further comment ...

I seems that when you are refering to voltage drop you are in fact talking about potential difference across a component (actually I am not sure about this as you are not being clear), whereas I am talking about the voltage output by the PSU. It seems that you were off topic, as the original subject was about the voltage output by the PSU, so it looks like you got a little confused.

Enjoy. :)

PS: Apologies SOTLordRahl, this type of 'debating' was not expected. Hope you resolve your issue.
 
I'm done trying to explain myself to someone who cannot understand the things I'm trying to say. You have twisted what I have said around every time you have replied to it. I don't know where you get your facts but I think you are missing the difference in voltage and voltage drop. The drop is the difference of what goes into the load and what comes out. This is where the entire argument seems to stem from. The voltage drop has to happen to have current flow.

This is all I have to say on the subject.
 
Storm, lets just agree that we are just misunderstanding each other. I'm sure we can still get along with one another despite the heated debate.

I didn't intend to start 'twisting' any of your arguments, so if you feel I did, then I apologise. It was remarks that my original post was misguided that started it all.

Don't let yourself get frustrated, I know exactly how you feel.

PS: Your last statement about voltage drops is absolutely correct, so it seems that we were just misunderstanding one another after all. :)
 
I'll agree with that and I think that we are both probably trying to say generally the same thing. Even as I look back at your other posts, you have many of the facts correct, I'm thinking we are each misunderstanding some of what the other is saying and we are probably both having some trouble relaying our intention to the other concerning some parts of this. Maybe it is the difference in the ways the theory was explained to us, or maybe it is just that these things are much easier to understand than they are to explain. This is pretty in depth stuff we are talking about here and goes much deeper than this guy's problem anyway.
 
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