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Anonymous
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July 20, 2002 1:34:27 PM

Upping Vcore, CPU Overvolting Explained Mar 06, 2002, 07:30 AM
Introduction
By: Dan Mepham

If you’ve ever overclocked a processor, or been around people who have, odds are you’ve heard people talk about 'upping the core-voltage' before. In fact, nowadays it’s pretty much a no-brainer. Just about anyone who overclocks anything knows that, often, an extra bit of voltage will help a component run at speeds it wouldn’t before, be it a CPU, video card, or even memory. But do you know why?

It’s one of those things we tend to take for granted. We don’t know why it works, all we know is that it does work, and that’s good enough for us. And that’s just fine, because in most cases, it really doesn’t matter why it works, just that it does (that’s engineering in a nutshell).

Unless, of course, you’re curious, in which case, we’ve got an answer for you! A common misconception is that increasing the voltage gives the CPU the added ‘power’ it needs to run at higher speeds. While it's true that power varies with voltage, it’s actually the current demands of the processor that increase with speed, not voltage demands (when you overclock a CPU, it requires more current at the same voltage, not more voltage with the same current). It doesn't need more power, it needs more current, so feeding it more voltage in an effort to increase power won't help. So why, then, does increasing the voltage a bit often help the CPU run faster, and with more stability? Read on, and we’ll explain.
Next >>

1. Introduction
2. How Your CPU Thinks
3. It's All About Tolerance
4. Transient States
5. Putting it All Together
6. Summary

the explanation continue <A HREF="http://www.hardwareanalysis.com/content/article/1482/" target="_new">here</A>. :smile:


i've plugged my home blower to my case ... dunno what happen ... that works?!?

More about : hardware gurus

July 20, 2002 1:58:50 PM

Actually that article mentions CMOS and TTL. Those are not types of transistor but those are ways that the transistors are coupled together.

MOS and Bipolar are types of transistor, hence the name C(MOS).
You can also buy DMOS, ECL and a few other types of logic I can't remember all the names.

Now the types of logic aare rated with speed, starting with the slowest and lowest power:
CMOS
TTL
ECL

Now too long to explain here but ECL being the fastest and highest power drawing is used in PC's however they use a mix of the two types of transistor I believe.
Anonymous
a b à CPUs
July 20, 2002 4:05:34 PM

good article.
Related resources
July 20, 2002 4:23:41 PM

I like that closing door metaphor.

Question: The author indicated the higher voltages are needed to overcome internal resistance and allow switching to reach proper threshholds when overclocking. My question is that once voltages are raised, if one then lowers the clock speed the processor no longer has the switching obstacles, the voltage is now surpassing the thresholds, does this cause damage to the processor? or can it cause errors?

Incidentally, I found the explanation very interesting of how 10% and 20% voltage increases allow higher levels of overclock but 30% shows almost no improvement. I was wondering why I needed 1.95 volts to get my 1.0 ghz Tbird to 1.56 ghz but going has high as 2.25 volts didn't help at all. Now I know why.

<b>I have so many cookies I now have a FAT problem!</b><P ID="edit"><FONT SIZE=-1><EM>Edited by phsstpok on 07/20/02 12:32 PM.</EM></FONT></P>
July 20, 2002 5:35:45 PM

yeah good article, sorry forgot to say :-P
Anonymous
a b à CPUs
August 3, 2002 10:17:41 PM

huh?!? ya:p 


i've plugged my home blower to my case ... dunno what happen ... that works?!?
a b à CPUs
August 4, 2002 1:28:32 AM

Not seeing the forest for the trees. The easiest explanation for the need for added voltage is this: a CPU is made of a SEMICONDUCTOR. Now, you need more current, how do you get it? Look at Ohms law. If you increase the voltage, you overcome resistance to current flow!

They mention the point of diminishing returns, again, the fact is that we're dealing with a semiconductor: The hotter it gets, the more the resistance to current flow. There comes a point when, at a certain temperature, more voltage simply allows more current which results in a proportional increase in resistance.

Keeping it simple.

<font color=blue>By now you're probably wishing you had ask more questions first!</font color=blue>
August 4, 2002 2:39:06 AM

Quote:
Not seeing the forest for the trees. The easiest explanation for the need for added voltage is this: a CPU is made of a SEMICONDUCTOR. Now, you need more current, how do you get it? Look at Ohms law. If you increase the voltage, you overcome resistance to current flow!

They mention the point of diminishing returns, again, the fact is that we're dealing with a semiconductor: The hotter it gets, the more the resistance to current flow. There comes a point when, at a certain temperature, <b>more voltage simply provides a proportional increase in resistance</b>.

That was almost believable until your last sentence. Voltage does not affect resistance in any way, proportionally or otherwise. Resistance remains constant regardless of voltage.

You're right about temperature and resistance, though.


<b>I have so many cookies I now have a FAT problem!</b><P ID="edit"><FONT SIZE=-1><EM>Edited by phsstpok on 08/03/02 10:40 PM.</EM></FONT></P>
a b à CPUs
August 4, 2002 3:11:01 AM

Dood, obviously I lost you somewhere, so let me put it this way-extra voltage helps more current overcome more resistance, creating more heat. Therefore, more voltage ~ more current = more heat = more resistance!

<font color=blue>By now you're probably wishing you had ask more questions first!</font color=blue>
August 4, 2002 4:24:49 AM

...but the voltage alone does not cause the resistance change, nor the temperature change for that matter. There are time delays involved.

As current flow increases you increase the amount of power consumed. (Power=volts X amperes) The increase in power is what increases the heat (and temperature).

None of what you said explains why overclocking fails. So you increase voltage, you get more current, and you increase resistance (a little). However, you have a cooling system and within bounds your temperature remains steady. The voltage doesn't change, the temperature is at steady state and doesn't change (so resistance can't increase any further due to heat), and the current is available.

Ohm's Law alone does not explain why overclocking fails and, in fact, Ohm's Law doesn't have any provision for temperature effects.

<b>I have so many cookies I now have a FAT problem!</b>
August 4, 2002 5:16:34 AM

...An interesting fact - Silcon has a negative temperature coefficient. This means as temperature increases electrical resistivity decreases. (see <A HREF="http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/rstiv..." target="_new">here</A>).

This has no bearing on this discussion as semiconductors do not consist of pure silcon but are formed on a lattice based on silicon and other chemicals.

<b>I have so many cookies I now have a FAT problem!</b>
a b à CPUs
August 4, 2002 6:03:35 AM

OK, to begin with, we're working with a SEMICONDUCTOR, like a toaster element. It's also a closed circuit, like a toaster element. You add power, you get heat, like a toaster element. You add more power, you get more heat. Now, add an poor heat insulator, say, a piece of glass. Sandwich the toaster element between two pieces of glass. The glass gets hot, but not as quickly as the toaster element. Now add a heatsink to the glass. The toaster element will always be hotter than the heatsink. To the extreme, you could have an ice cold heatsink and a scalding hot toaster element.

Now imagine that is your CPU circuit. You get to a point where you get so much resistance that no more current can flow. You can add more voltage, but it only returns more resistance. Add even more voltage and it pops. Add more cooling, and it will go a little further, but how much cooling can you add?

Were this NOT the case, a Peltier wouldn't help. The fact is, a Peltier helps overclocking by LOWERING RESISTANCE in the circuit! You effectively raise the point of deminishing returns with a Peltier, but once the core temp reaches that magic place, bang, you cannot overcome the resistance, any attempt only yields more heat and more resistance.

You can add LIQIUD NITROGEN cooling! But that again only raises the point of diminshing returns.

Ohms law tells you why it takes more voltage to overcome more resistance.

<font color=blue>By now you're probably wishing you had ask more questions first!</font color=blue>
a b à CPUs
August 4, 2002 6:09:28 AM

Well, if the reverse is true, a pure silicon transistor (not possible, but in theory) would no longer work at too high a temperature, not becuase of too much resistance, but too little! In that case, the transistor would be shorted!

So thanks for pointing that out! But it still would show that it's a heat problem!

<font color=blue>By now you're probably wishing you had ask more questions first!</font color=blue>
August 4, 2002 7:03:09 AM

What I was trying to point out (and doing a poor job of it) is that different materials react differently to heat. Many materials require very large temperature changes for very small changes in resistance, inconsequential amounts in some cases.

To make the blanket statement that an increase in voltage causes in increase in current flow which results a temperature increase and then some increase in resistance and therefore leads failed overclocking is not entirely true.

You/We don't even know how much the resistances within a CPU change with temperature. It could be 2% for each 10 degrees C or it could be 0.00002% for that 10 degrees. We don't know the temperature coefficient. It could even be negative (ok so I doubt that, myself). Although, some semiconductors have negative thermal coefficients, thermisters for instance.

Don't get me wrong. I know heat is involved but not from increased voltages and corresponding current flow. The majority heat comes from the increased energy of higher frequencies.

Think about it if the problem was just excessive electrical current then you would have the same problem when you raise the voltage at low frequencies as well as high frequencies. Same processor, same voltage levels, same Ohm's Law, why wouldn't you have the same problem?


<b>I have so many cookies I now have a FAT problem!</b>
a b à CPUs
August 4, 2002 9:49:04 AM

Hmm, I droped a K6-2 from 2.2v to 1.9v at the same speed in a laptop and saw a huge drop in heat production. The automatic fan went from being on all the time to being on 5 minutes every half hour! So I may be wrong about the increased resistance (thanks for pointing out the evidence) but nonetheless heat causes a change in the electrical properties of the semiconductor in a way that causes the chip to stop functioning properly.

<font color=blue>By now you're probably wishing you had ask more questions first!</font color=blue>
August 4, 2002 11:58:02 AM

Are you guys agreeing that it is the heat (from the higher frequency and increased voltage combined) and the corisponding resistance that is causing the problems? Or what was it you guys actually decided. I am interested to know.
August 4, 2002 3:17:47 PM

I have no doubt that heat is an issue, a big one. I just don't believe that increased resistance part of the equation.

<b>I have so many cookies I now have a FAT problem!</b>
Anonymous
a b à CPUs
August 4, 2002 3:47:06 PM

"My question is that once voltages are raised, if one then lowers the clock speed the processor no longer has the switching obstacles, the voltage is now surpassing the thresholds, does this cause damage to the processor? or can it cause errors?"

That's several questions.

The author didn't make it clear, but anytime you raise the voltage you the stress on the part due to a correspondingly higher power consumption, among other things. It all depends on the heat generation, dielectric strength, doping quality etc. as to whether or not it's more damaging. If you cause physical damage to the circuitry or are running it out of tolerance (tolerance being very case specific) you can most likely expect some kind of malfunctioning and errors.
August 4, 2002 3:49:52 PM

Quote:
Hmm, I droped a K6-2 from 2.2v to 1.9v at the same speed in a laptop and saw a huge drop in heat production. The automatic fan went from being on all the time to being on 5 minutes every half hour! So I may be wrong about the increased resistance (thanks for pointing out the evidence) but nonetheless heat causes a change in the electrical properties of the semiconductor in a way that causes the chip to stop functioning properly.

You've definitely got my curiosity up regarding this subject. I honestly don't know the answer.

Let us consider something simplistic. Assume the resistance doesn't change and for argument's sake the current didn't change either. By decreasing the voltage you have greatly decreased the amount of power consumption, P=EI (volts x amperes). By reducing the voltage from 2.2 volts to 1.9 volts you have reduced the power consumption by almost 14%. This, all things equal, should reduce heat production in a proportional fashion. (I'm not sure it's 1:1 but I think it is). Less heat means lower temperatures. The temperature is just dropping below the fan sensor's threshold more often.


<b>I have so many cookies I now have a FAT problem!</b>
August 4, 2002 4:14:28 PM

Quote:
"My question is that once voltages are raised, if one then lowers the clock speed the processor no longer has the switching obstacles, the voltage is now surpassing the thresholds, does this cause damage to the processor? or can it cause errors?"

That's several questions.

The author didn't make it clear, but anytime you raise the voltage you the stress on the part due to a correspondingly higher power consumption, among other things. It all depends on the heat generation, dielectric strength, doping quality etc. as to whether or not it's more damaging. If you cause physical damage to the circuitry or are running it out of tolerance (tolerance being very case specific) you can most likely expect some kind of malfunctioning and errors.

OK, I wasn't considering total heat production when I asked the question(s). The article claimed that higher voltages were needed because signaling thresholds won't be reached using stock voltage settings. I was just wondering if the higher voltage alone could be damaging. I didn't state my question well. I'll try again.

For example, I'm overclocking my system. I find it necessary to raise the voltage to obtain higher levels of overclock with stabilty. (If I understand the article, the higher voltage allows internal processor signaling to reach the necessary threshold levels). Now, for some reason, I decide to reduce my clock speed to stock settings but I forget to reduce the voltage.

My question is this. Is the higher voltage more harmful at stock speeds than at the overclocked speed? At normal speeds the signaling is now well beyond threshold levels and this is my concern.

<b>I have so many cookies I now have a FAT problem!</b><P ID="edit"><FONT SIZE=-1><EM>Edited by phsstpok on 08/04/02 01:20 PM.</EM></FONT></P>
Anonymous
a b à CPUs
August 4, 2002 4:17:10 PM

If I may be so bold. If I am understanding everyone here.

Crashman validly points out that gains from increasing voltage might be overcome by the negative effect of extra heat now being produced because of the corresponding increase in power consumption.

phsstpok makes the also valid observation that the rise resistitvity of conductors due to temperature rise is not a very significant effect. Then he also goes on to point out the interesting fact that silicon has a negative temperature coefficient.

My opinion on the subject: I think that what the author is pointing out is due to the capacitive nature of CMOS inputs. Notice the curviness of the clock waveform in his example waveforms(which ideally would be a square wave). This represents the nonidealness of the real world situation. CMOS inputs are most certainly capacitave. Therefore our nice square wave gets distorted into the curvy thing.

Brief electronics review: a capacitor is like a charge storage tank. In other words to see any appreciable voltage across the leads it must first draw some current in order to charge it.

The authors whole point is that at the higher clock speeds the voltage is not large enough to charge the capacitive CMOS inputs quickly enough in order to create the conditions for valid logic levels at those inputs. Increasing the voltage is in effect allowing the capacitive inputs to get more charge per unit of time, and thus allow a greater opportunity to reach valid logic levels.

Both those guys are right. I don't know why they are arguing. But then again as you have seen cowgoesmoo, I really don't have much room to talk.
Anonymous
a b à CPUs
August 4, 2002 4:21:43 PM

the same amount of harmful due to the stresses on the PN junctions from higher voltages. but in general less harmful due to the correspondingly less heat/power from operating at lower clock speeds.

sorry it's at least a two headed beast.
August 4, 2002 5:21:50 PM

Thanks.

<b>I have so many cookies I now have a FAT problem!</b>
Anonymous
a b à CPUs
August 4, 2002 5:30:49 PM

right on!
August 4, 2002 6:11:22 PM

<P=EI (volts x amperes).>

P is also equal to V^2/R.
So, the voltage is the dominant factor in this equation.

As for the waveform, it is a lot closer to a distorted sinewave, rather then the stupid picture of that bended squarewave.
August 4, 2002 6:43:06 PM

Good point! So power increases (or decreases) proportionally with the square of the voltage per given resistance!

<b>I have so many cookies I now have a FAT problem!</b>
Anonymous
a b à CPUs
August 4, 2002 6:44:49 PM

I wish I understood what you are trying to say.

It did make me laugh though, I don't know why.

edit that is understand the part about the waveforms...<P ID="edit"><FONT SIZE=-1><EM>Edited by knewton on 08/04/02 11:46 AM.</EM></FONT></P>
August 4, 2002 7:03:42 PM

LMAO! Well the simple fact is, around here you ask a question you get an opinion. You start an argument then you get information (sometimes more than you want but not in this case. I'm learning a lot).

Thanks for all your info, by the way!

p.s. Now I'm way out my league but I thought capacitive devices aren't greatly affected by temperature.

I do understand the deformed waveforms. (Might be the only thing I understand).

<b>I have so many cookies I now have a FAT problem!</b>
Anonymous
a b à CPUs
August 4, 2002 10:06:41 PM

Capacitance changes with temp more or less depending on dielectric quality. For the dielectrics we use and over the temperature range we are talking about it's pretty significant. I can't speak for AMD and Intel but typical values are 5-20% change in Capacitance over 0-85 degrees Celsius for dielectrics you can afford. I'm imagining the situation may be even worse in silicon wafer land, I'm used to hearing values for discrete parts.

This is why SOI is a big buzzacronym lately. Think of the I (for insulator) as just another way of saying we are finally upgrading to a better dielectric. It should allow for some nice improvements.

Just to clarify, the capacitance I'm referring to is a by product of the architecture of CMOS electronics. I'm not saying that the circuits are merely capacitors. Obviously things are a bit more complicated, but that is a good way to think about it while trying to understand the waveforms in the article.

The waveform distortion comes in when we consider that in reality any output (in this case the clock signal) is not a perfect voltage source. In other words while trying to produce that nice square clock signal, the capacitance I spoke of causes the output to produce the smushed up and rounded version of it. This due to the fact that an output can only provide a certain amount of current.

An engineer might say it's all due to outputs not being near the ideal zero output impedance and the inputs are not near the ideal infinite input impedance.
a b à CPUs
August 4, 2002 11:15:51 PM

You are correct!

It's important to remember that while the number of cycles, aka work performed by the processor, aka data, increases heat with increased cycles, it's also important to remember that passing a current through a resistor also generates heat regardless of data, aka work performed=heat. Raising the voltage without raising the clock rate has the affect of passing more power through a resistor, more heat. The K6-2 in this case was operating at a higher than needed voltage and was producing heat as work more than data as work.

<font color=blue>By now you're probably wishing you had ask more questions first!</font color=blue>
a b à CPUs
August 4, 2002 11:18:59 PM

Heat is the main issue. Whether it causes increased resistance where the thing can't flow enough current to go higher without burning, or decreased resistance where the transistors become too conductive and quit working, would depend on the exact material. In either case, heat is the main factor.

<font color=blue>By now you're probably wishing you had ask more questions first!</font color=blue>
August 4, 2002 11:27:17 PM

eh? Thanks for the vote of confidence but I think I was wrong on that one. Although, I'm now seeing how really complicated this is. I'm wondering if Era's prefered equation, Power=voltage**2/resistance, may be a much better predictor of heat production. Unfortunately, I don't have a way to test it, no way to measure actual heat. I can only measure temperature and very inaccurately at that.

Anyway, this is still a very interesting discussion.


<b>I have so many cookies I now have a FAT problem!</b>
a b à CPUs
August 4, 2002 11:37:03 PM

Quote:
The temperature is just dropping below the fan sensor's threshold more often.

You're still correct in your conclusion. More precisely, it takes longer to heat up the large heatsink now, but the same idea.

<font color=blue>By now you're probably wishing you had ask more questions first!</font color=blue>
August 5, 2002 11:39:59 AM

Heat being the major overclockers problem? I think that is not entirely true, as the author of the article that started this discussion points out. Btw, I think the resistance-increase with a higher temperature is neglectible, and it would even surprise me if silicon used in chips has a increasing resistance with a higher temperature. Even more: The silicon is not used for conduction: There are metal masks on top of the semiconductor that are used to interconnect the trasistors. But anyway ...
Think of it: transistors need a (damn, these terms are hard to think of in English ...) certain tension (maybe svol can help me out: een 'drempelspanning' in Dutch) to go into conduction, in plain old TTL PNP or NPN transistors that is around .7 V. Then, in a recent CPU, like the P4 northwood, you have a VCC of around 1.7 V, so you would have 1 V you'd need to lose in the connection between one transistor with another one, assuming that recent manufacturing processes still need that .7 V tension, though I'm quite sure that has lowered by a lot tenths of volts. With the extreme small sizes (in length) of the conductors and the extreme small currents flowing, you'd need a very high resistance to obtain such a loss, according to that infamous Ohm's law. Considering the material used to interconnect the transistors (silver, copper, ...) I think that at no temperature whatsoever they would reach such resistances.
The other reason you give, the fact that the transistors go conducting randomly because of the high temperature seems more realistic as the cause of a chip failing at high temperatures. I'm thinking of the Brown-movement or the kinetic energy atoms and electrons get when at a certain temperature to explain that, but I'm far from sure about this one. Maybe knewton, who seems quite educated in these subjects can help me out ...

Anyway, I guess I'm going back to the studying Gauss-Jordan method for solving linear equations ... *Sigh*

Greetz,

Bikeman
PS: Concerning the temperature: I know somebody who had a P3 450 (is that a Katmai, the pre-Coppermine PIII?) running without heatsink at full load at ... 113° Celsius for more than 24 hours without crashing!!! Hmmm ... Those were chips :-)



<i>Then again, that's just my opinion</i>
August 5, 2002 12:53:15 PM

All these small transistors fets ets are devices that need energy to change state. Never mind if the cicuit has to work
out the rising or falling square wave.Each change of state
invoves a small amout of energy.If your CPU has lets say one
million devices,what is the the amount of energy if each device swiches at 66Mz,133Mhz,1GHz,2.4gHz.All answers sent
to avandonk@bigpondnet.au


avandonk
August 5, 2002 1:13:45 PM

Sorry avandonk@bigpond.net.au
OK assuming each flip flop bistable cicuit consumes one microwatt!

avandonk
a b à CPUs
August 5, 2002 6:46:19 PM

Temperature may not be the only factor, but it's such a large factor that it's probably all the typical overclocker needs to be concerned about.

<font color=blue>By now you're probably wishing you had asked more questions first!</font color=blue>
August 5, 2002 6:53:49 PM

You're probably right ...

But you did make me think how the higher temperatures can cuase problems ... I wonder what the exact causes are ...

Greetz,

Bikeman

<i>Then again, that's just my opinion</i>
Anonymous
a b à CPUs
August 5, 2002 7:48:55 PM

ooo, device physics. now were really getting to the nasty nitty gritty.

I think bikeman may be refferring to Thermal (Johnson) noise content. Caused by the thermal agitation of electrons in a conductor. Related to temperature and conductor resistance.

I had to look that up BTW. But deffinitely noises are not good, and are just as capable of causing problems as any other unwanted signal.

In case you were wondering the following list is where I got this info. My expertise are certainly not device physics, by any means.

1. Shot (Schottky, Quantum): Caused by the random fluctuations in the motion of charge carriers in a conductor. Related to current.

2. Thermal (Johnson): Caused by the thermal agitation of electrons in a conductor. Related to temperature and conductor resistance.

3. Flicker (1/f): Not fully understood. Believed to be casused by charge carrier across crystalline structure defects. Related to current flow.

4. Burst (popcorn): Caused by imperfections in semiconductor material and heavy Ion implants. Related to purity of material.

5. Avalanche: With a PN junction operated in strong reverse breakdown, electrons with enough Kinetic Energy that collide with lattice atoms create an additional electron-hole pair (which in turn may cause another collision). Related to materials and application.
August 6, 2002 5:45:10 AM

great article it helped explain alot. I had upped the voltage when i had oced my cpu but had no idea why it helped until now :) .

The only thing i know...

is that i know nothing.
August 6, 2002 11:57:08 AM

Hmm ... Do they call that device physics? Okay, didn't know that. I didn't know that the quite intuitive reasoning I was doing actually had a name ... I'm sorry, I really am not the physics-know-it-all I'd want to be. I'm just a below-mediocre student of civil engineering with a good bunch of inuition ...

But this is an interesting subject, though, something I'd like to have a little more than qualitative explanations about. Do you have a link to a document about the stuff you were talking about, the causes for flaws in semiconductors? Though I could also look it up using a search engine, though ... Well, a professor once told us that the biggest motivation in engineering science is laizyness ...

Anyway, thanks for the expalation ...

Greetz,
Bikeman

<i>Then again, that's just my opinion</i>
August 6, 2002 5:27:35 PM

"The authors whole point is that at the higher clock speeds the voltage is not large enough to charge the capacitive CMOS inputs quickly enough in order to create the conditions for valid logic levels at those inputs. Increasing the voltage is in effect allowing the capacitive inputs to get more charge per unit of time, and thus allow a greater opportunity to reach valid logic levels. "

This would seem to be the key point of the article but this brings up more questions in my mind. If the key factors are time and voltage tolerances then there must be other factors pertaining to successful overclocking.

Consider two processors are made in the same manufacturing process. Due to testing, binning, and what not one processor is designated to operate at say 2.0 ghz the other 1.6 ghz. Let us say that the 2.0 processor only overclocks to 2.2 ghz but requires some voltage help to do it. The 1.6 ghz processor overclocks to 1.8 ghz without changing the voltage but overclocks to 2.0 ghz with voltage increases and will not overclock any further.

What factors are at play resulting in the different outcomes? Why aren't the timing and voltage demands identical with two processors? What else is different? If one processor has defects, what could they be?

<b>I have so many cookies I now have a FAT problem!</b>
August 6, 2002 6:58:44 PM

Also, you have to consider that clock pulses are not perfect, there is a ripple effect after each transition from high to low and low to high, if the ripple is big enough it can sent a false signal change to the circuit and cause an error. Upping the core voltage adds a little more overhead to compensate for ripples.


Edit: Sorry, phsstpok, not necissarily directed at you, I tend to just click on the last post. :) 
<P ID="edit"><FONT SIZE=-1><EM>Edited by bardic on 08/06/02 03:53 PM.</EM></FONT></P>
August 6, 2002 7:24:55 PM

I can't agree with this nor dispute it because I don't know anything about "ripple effects". I just find it curious that performance can vary so much from one processor to the next. The differences seem so much greater than one might expect considering the extremely tight tolerances in manufacturing.

You can't call the differences defects as the processors do still function.

<b>I have so many cookies I now have a FAT problem!</b>
Anonymous
a b à CPUs
August 6, 2002 7:40:45 PM

You know when I read back through that quote I realized I didn't say that quite right. To be more correctI would have said the extra voltage allows the clock outputs to be driven harder which should allow them to supply more current. Maybe no one else noticed, but I did not like reading back through it like that.

Now on to your questions: I think the main reason for variations you mention is due to manufacturing tolerances. For example, earlier we were talking about dielectric quality. Two batches of the same dielectric material might have slightly different dielectric properties (which gets into it's manufacturing). Another example might be trace material purity, a different purity level might correspond to different etching speeds, and could result in varying tracewidths from one wafer to the next. I'm sure the list goes on and on. That doesn't even consider temp fluctuations, humidity changes, air cleanliness etc.

All things are conspiring in attempt to get these little microscopic circuits to behave differently, or not at all. I'm sure there are folks here who would have much more insight to the follies of wafer production than I would, but maybe that'll get you started.

edit: caught one grammatical error<P ID="edit"><FONT SIZE=-1><EM>Edited by knewton on 08/06/02 12:42 PM.</EM></FONT></P>
August 6, 2002 8:10:57 PM

Quote:
You know when I read back through that quote I realized I didn't say that quite right. To be more correctI would have said the extra voltage allows the clock outputs to be driven harder which should allow them to supply more current. Maybe no one else noticed, but I did not like reading back through it like that.

Didn't notice a thing! I'm pretty much only vaguely understanding the technicals, anyway.
Quote:
Now on to your questions: I think the main reason for variations you mention is due to manufacturing tolerances. For example, earlier we were talking about dielectric quality. Two batches of the same dielectric material might have slightly different dielectric properties (which gets into it's manufacturing). Another example might be trace material purity, a different purity level might correspond to different etching speeds, and could result in varying tracewidths from one wafer to the next. I'm sure the list goes on and on. That doesn't even consider temp fluctuations, humidity changes, air cleanliness etc.

All things are conspiring in attempt to get these little microscopic circuits to behave differently, or not at all. I'm sure there are folks here who would have much more insight to the follies of wafer production than I would, but maybe that'll get you started.

I gather you are saying that the large variances in performance from one chip to another can be due to the multiplicative effect of slight differences. With enough small differences you end up with one large one. Yet, each small difference is within tolerance and producing no real errors, as in defects.

<b>I have so many cookies I now have a FAT problem!</b>
August 6, 2002 8:26:37 PM

Now I'm lost. Are these explanations of how heat is produced in a CPU or are they explanations of how heat can cause logic failures or both?

<b>I have so many cookies I now have a FAT problem!</b>
August 6, 2002 9:00:21 PM

I think that the things Knewton explained, are explanations to the differences between the 1.6 and the 2.0 you asked about. And actually I wondered, too. But now I see: An impurity in for example the dielectricum that carries the semiconducting silicon, could cause a increase in the capacity, causing a region of the semiconductor, or maybe just even a couple of transistors, to get worse characteristics concerning the things discussed in the article about the Vcore upping. But I can guess that's just one of the many, many causes for the chips to differentiate in the same manufacturing line.
Also, maybe that way, if there are impurities in the conductors, their temperature coefficient could get a bigger infleunce ... And that could be another explanation for the temperature to ben a determining factor when overclocking ... But the picture is far from complete to me ... Keep the info coming, Knewton! We're learning ...

Greetz,
Bikeman

<i>Then again, that's just my opinion</i>
August 6, 2002 11:15:25 PM

I dropped out of Electrical Engineering twenty-four years ago. I'm starting to get that dizzy feeling again from this stuff.

<b>I have so many cookies I now have a FAT problem!</b>
August 6, 2002 11:35:35 PM

Now I have read everyones comments and I have a couple of points:

People have talked about how the that temprature does not affect the resistance too much. Actually in my recent solid state electronics module on my electronics engineering course we were taught about ohms law and how it is derived.

The law is derived using electron mobility which is a function of temperature. Infact a 5C rise in temprature can change the resistance quite rapidly.

The problem arises when:
1. The temprature gets too high
2. The Voltage driving the electrons is too high

This is because the transistors saturate too early or when the noise on the input is such that causes the transistor to turn on when it should be off.

Basically in athlon processors at 95C the processor ceases to function.

The capacitance answers arecorrect however, because the voltage across the capacitor cannot change instantly therefore a square wave becomes sine wave like.

More heat why, think about a switched mode power supply for a second. To provide more current the rate at which it switches on or off is increased. Therefore more current is supplied. So look at it this way, as you ramp up your clock speed more current is driven through the device, therefore using Power = current squared X resistance, the faster the processor operates the more power would be dissapated, and this power increases the processor temperature, this increases the resistance. But ofcourse then the amount of current decreases, therefore you need more voltage to drive the current through the resistance. etc etc etc, but ofcourse due to the heat sink the heat dissapation is stabilised, therefore this settles down. This is the exact reason why without a heatsink the processors explode and reach 400C.

I cant be sure if what I said is clear as it is late when I am typing. Sorry if it is not clear but I hope it helps.
!