Low voltage damages RAM?

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Warpspasm

Warpspasm

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May 28, 2003
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Here's a lot of background before I get to the actual question. I finished building my PC about 3 weeks ago and it was running great and totally standard without any overclocking. Here are the specs:

Motherboard: Gigabyte P35 D3SL
CPU: E4500
CPU COOLER: ROSEWILL|RCX-Z775-EX
Memory: 1Gx2|CRUCIAL BL2KIT12864AA804 Ballistix
Power Supply: OCZ|OCZ700GXSSLI 700W GamerX
Graphics Card: EVGA 512-P3-N802-A1 8800GT 512M
Case: Cooler Master 830 Evo

I decided to try overclocking a bit, so I increased the CPU FSB to bump it from 2.2 to 2.75 Ghz. I did nothing to the RAM for overclocking. Temps seemed fine and I decided to let Prime95 run all night. The next morning the PC was frozen with the XP load screen in mid load. I managed to get to the desktop and looked at the temp log generated. The core temps never got higher than 70c. The system however was really flakey and I began getting reboots etc. I put everything back to stock settings and it seemed a bit better, but still really unstable. I ran MemTest86+ from a CD for 8 hours and got something like 1400 errors. Rebooted the PC, went to BIOS and changed the setting from the default of Turbo to Standard. Ran MemTest for 8 hours with no errors. Rebooted, was able to surf the net but when I tried to burn a Lightscribe disk the PC crashed hard. Ran MemTest with bunch of errors again.

I called Crucial and the guy in tech support was very friendly. He asked what voltage I was running on my RAM and I told him the motherboard had it set at 1.9v. He said that Ballistix was designed for overclocking and should be run at 2.2v. He also said that running at the lower voltage may have actually damaged the RAM creating the issues I was experiencing. He was willing to RMA the sticks, but suggested that I first try seeing if the system became more stable if I bumped up the voltage. I did that and it really wasn't any better. I also changed the timings from motherboard assigned 5-5-5-18 to the chip rated 4-4-4-12. Now things are getting worse even when I set things back to standard safe settings. Crucial is sending me new sticks without any hassle, but I want to make sure I don't have the same problem. Can that lower voltage actually cause RAM corruption the way the tech said?

Yeah, I know it was wordy, but too many times it seems like posters don't give enough background.

Thanks guys.


 
My crucial ballistix ran fine at 1.8, I have overclocked mine now to 2.0 so I can run them at a solid stable 4-4-4-12. They are even made to be run at 2.2 or higher but since mine run stable at 2.0 I don't need the extra heat on them. as soon as you get the new ones run them at 2.2 then back the voltage down a bit till you see what you can run stable at.
I personally like the higher voltage, low timings and would sacrifice some mhz if you have to for the lower timings.
 
If you have chips that arent capable at running lower voltages sure but realistically it will just be unstable until it gets into its targeted range. Memory fails and thats all there is say about that, if the chips on the memory began to leak Ie overvoltage or defective manufacturing they will fail. Trust me that tech support guy is talking out his ass when he said that.
 
nope, low voltage can cause memory read/write errors (thus possibly corrupting your windows installation and causing all sorts of funny problems even after the rams themselves are fixed) but it shouldn't damage the rams themselves.
 
I'm wondering if perhaps I overheated the RAM when I overclocked the CPU. I think I let the motherboard link the RAM to the CPU when it was overclocked. I can't swear to it because I've made so many adjustments I can't remember what I did, when! I remember at one point seeing the RAM at a higher Mhz in CPUID. I don't know if it adjusted the voltage as well.
 
If the RAM was at a 1:1 ratio you would not have gotten over the rated speed even at 2.75 GHZ. According to my caclulations your RAM would have been running at 500MHz, well below the rated speed.
 
low volatage "can" force it to draw more current. BUT Ive never EVER seen this happen to ram. Infact many baords default to 1.8v min eis 1.95 so they had to have that in mind at design time if the engineers worth their pay that is. I mean it should just be unstable till voltage applied.

Anyways corsair told us at their forums that even though its tested stable, most failures happen in 30 days of being new and no way to tell till in use. A weak chip prob showed its flaws after being used etc. Im sure ALL ram co's have this problem just like cpu and chipset co's do.
 
I wouldnt doubt it.

I did a simple mobo swap (work) theold mobo died. Well new conroe board started fine- then all assudden update stopped. Hm odd well reinstall and look into it. Well I had to disable floppy since it had no floppy port figured thats odd. XP installed. THEN the sound stopped working -at this point I knew board was bad. Owner wanted to try with pci sound so I did. ANd 100% load came back. Turned off usb 2.0 in bios and back to normal. Hmmmm next thing you know it stops picking up the USB devices at all!

In the end the Southbridge chip was bad and the more we used it the more it failed cause more and more to die on the board (anything it controlled really). Newegg replaced and its back working.

Long story I know but it shows that just last week I expereinced similar failure. Point is that if its weak it can fail in those first few hours or days even though it passed testing.
 
I think running at low voltage wont damage RAM but rather Higher voltage could damage it since it will create more heat.

Overclocking a RAM and running it at stock voltage may make it unstable like unpredictable restarts or hangups other than that I think it wonk cause a permanent damage.
 
well, in most electronic a too low voltage situation WILL cause the device to draw more current as WATTS is the only constant.

By this I mean say the ram is 7watts. You give it 2.2volts. IT will then draw 3.18amps (volts x amps=watts)
So same 7watt mem at say 1.8v would draw 3.8amps. While voltage is the measure of presure and amps is the rating of x amount of electrons moving from one given point to another in x amount of time. A higher amperage still can raise temp, just not as quickly as volts. Since presure = higher friction remember electrons are FORCED though a conductor and have skin effect -that is electrons like to travel on the outside of conductors very little travels inside it.

However pushing mroe electrons through said condcutor at a lower current rating, means it is forcing it to move more. This is how fueses work they overheat from too much current (Amps). So amps CAN heat it up.

Its just I never seen such a small diff of 4 tenths of a volt cause such a rise in temp to fail! I dont think it would I think it was just bad mem being written off as user error by the Tech you talked to.
 
The first thing I'm going to do when I get this new RAM is set it for 2.2 volts with 6-6-6-12 timing. I'm going to leave everything else on the motherboard at the default. I'll then boot to MemTest 86+ and let it run for 8 hours or so. Hopefully, that'll help identify any problems in the RAM early on. Does MemTest86 stress the RAM enough to unearth problems?
 
ya I think memtest86 puts the most stress on it. A few tests it does are cpu only but very few. I use Orthos to stress ram though. Its prime95 but runs on multi cores using them all. It has a ram stress test -cpu stress and blend.
 
Only if it has a active power stage that regulates the voltage and current that is fed to the rest of the system. To my knowledge rams dont have such a system onboard. So lower voltage means lower current and lower power consumption, less heat.

When OC'ing you are effectively shorting the timespan when current can carry a charge to the memorycells. If the charge isn't big enough you'll get errors. So upping the voltage increases the current and thus increases the charge thats being carried to the memorycells.....

And nope WATTS arent constant...
 
Only if it has a active power stage that regulates the voltage and current that is fed to the rest of the system. To my knowledge rams dont have such a system onboard. So lower voltage means lower current and lower power consumption, less heat.
Click to expand...

aha you make a good point, one that I didnt think about. True the regulators wouldnt allow excessive draw. Hm. I think we pretty much can sum this up to bad ram then.

By constant I ment power rating. As volts and current can change but be same watts overall. Tech this is diff depending on what your talking about I agree. Watts was first used to measure steam- so the rules arent the same for everything of course. Heck human energy is even measured in watts.
 



well, lets break this down a bit more.. Actually WATTS (or power disappated) is not constant. it depends soley on the amount of voltage and the amount of current consumed by the device. Now, lets make the appropriate analogy here...

a resistor (10 ohm) will allow whatever amount of current flow thru it that the voltage (higher=more pressure) will allow by ohms law. So, if you have 10v across the 10 ohm resistor, you have 1 amp of current flow (10v/10ohm=1A). you also have the power (watts) the resistor has to disappate) which in this example is 100W (10v*10A=100w).

Now, take that voltage and cut it in half.. 5v in the same circuit. 5v/10ohm=0.5A, and 5v*10ohm=50W.

you can plainly see that lower voltage does NOT increase amp or power disappated.

yes, this is an overly simple example, but EVERY electronic circuit can eventually be simplified enough to equate to that example.

The ONLY way that lowering the voltage will increase the current draw is in a voltage regulator circuit. such as power supplies, the Vregs on the mobo, vid card, ect. if you lower the incomming voltage, the circuit MUST increase the amount of current used in order to maintain the proper output voltage (assuming load voltage and current are constant).

To the best of my knowledge, todays common ram do NOT have onboard voltage regulators!

Also, Kari's explanation is also correct, as it can be broken down to the same simple ohms law equation
 
watts also rate heat though too. Watts disapated as heat is waisted energy and that differs from one component to another of course by how efficient it is. The common resistor is SUPER loosy and makes tons of heat. Which is why its being moved away from in power circuits in favour of fast switching regulator IC's. Ive made quite a few fan controler designed around such IC's.

I get your point about the watts though, just thought Id mention the diff you have to specific about watts.... watts what? Watts energy in general? Watts electrical? Watts heat? Watts steam? Like amd and intel TDP (thermal Design Power) refers to heat.

Its easy to just jumble them all together and confuse (perticular me LOL)

"Actually WATTS (or power disappated)"

eh? watts rates heat and electrical power. The heat dissapated IS rated in watts. But so can the POWER consumed by the device. A device that consumes 8 watts WONT dissapate 8watts of heat or it would be 100% wasited as heat energy and not used in the device at all! Energy can not be created nor destroyed only change form (till proven otherwise). Ergo some of the POWER watts is used to make the device work the rest is waisted as WATTS energy in heat.
 
By constant I ment power rating. As volts and current can change but be same watts overall.
Click to expand...
But this applies only to things with active power stage, (choppers and whatnots). Those choppers can draw more current if the input voltage is low, and use the same amount of power from the source. But the circuit the chopper is feeding doesnt know the choppers input voltage, it only sees the choppers output voltage that can be higher or lower than the choppers input voltage depending of the choppers type (and the circuits needs of course). So lets say the circuit needs 5V input and at that voltage it draws 1A, so 5W. Choppers input is 2,5V and because it needs to give the circuit total of 5 watts it draws 2 amps from the source, if the input voltage drops to 2 volts it would draw 2,5A (If the input voltage is still high enough for the chopper to actually work). Sorry if that sounds too lecturing, you obviously know something about electronics allready.

But with RAMs the constant is the circuits resistance/impedance, so higher voltage causes higher current. But the funny thing is that the current isnt flowing constantly, it changes even within a clockcycle. In the beginning of a cycle its higher and decreases (exponentially?) as time passes. It might even drop to zero when the cell it's feeding gets charged up to full potential within the same clock cycle. When OC'ing the number of cycles in a second increases and the current flows more frequently. This and the needed increase in voltage that resulted in higher current both contribute to the incraesed power consumption during OC.
 
A device that consumes 8 watts WONT dissapate 8watts of heat or it would be 100% wasited as heat energy and not used in the device at all
Click to expand...
Actually, it will (eventually). The point is what it can do with the energy during its transformance into heat inside the device or elsewhere. I said elsewhere, because for example the fans push some air out of the comp but that air eventually comes to standstill because of friction and all of its energy will be transformed in to heat... But your normal electrical components will turn the energy in to heat at surprisingly high effiency 😛
 

:heink: I think you got your power calculations a bit wrong.
10V over 10 ohms results in 1A, yep
but the power is P=UI=10V*1A=10W
and P=UI=U^2/R=5^2/10=2,5W

If the resistance is kept constant but the voltage changes, it changes the current aswell, thus the power changes 'more' than just the change in the current... man, I suck at explaining things in english... :cry:

edit: this thread turned me into an enthusiast, yay 😀
 


Thank you for correcting my boo boo.

You know, I went over them many times saying to myself "self, that aint right".. its been about 20 years since I actually HAD to use ohms law, so I admit, I goofed.. hell, I even looked it up again and got it wrong! LOL
 


This is not true; V=IR (voltage = current x resistance). now, if your resistance stays the same (which it does) and you lower your voltage, your current will be less.

Current is directly proportional to the voltage.

Current goes up as voltage goes up as well with semiconductors (diodes, transistors, etc) as well, but the relationship isn't as easily shown.

electronic devices have a power rating. power = V x I. so, if you lower the voltage, your current becomes less, therefore your power is proportional to the square of the voltage. if you exceed your maximum power rating for the device, you'll burn it up, but if you're lower than the power needed, it just won't work as expected

well, in most electronic a too low voltage situation WILL cause the device to draw more current as WATTS is the only constant.

By this I mean say the ram is 7watts. You give it 2.2volts. IT will then draw 3.18amps (volts x amps=watts)
So same 7watt mem at say 1.8v would draw 3.8amps.
Click to expand...



power is not the constant, it is the manufactures rating for the device. power is a product of current and voltage AND current is directly proportional to the voltage.

Power = V * I

I = V/R

therefore P = V*V/R

looking at this, you easily see that as you lower your voltage, your power through the device lowers.
 
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