Please Explain Micron

[lazerin]

What do people mean when they say something uses this particular micron techonology? For example, the NV30 uses 0.13 micron process.

I have no idea what that means :confused:.

Please explain? :)

[HardwareJedi]

I'm sure some engineer will correct me if I'm wrong... But The micron size you refered too is the width of the circuit traces used inside the cpu. Hence they are able to shrink the die size(CPU core) as the process reduces in size... from .25 to .15 to .13 to .09 micron which I believe is going to be the next step in the chain. You can pack more traces and transistors into a given space as the size of the Process goes down.

[Edward2]

A Micron is a measurement of distance. 1 Micron = 1 Micrometer = 0.000039 Inches. So when manufacturers talk about .25, .18, or .13 Micron technology that is telling you how small the circuitry is in that particular computer chip.

[macklin01]

Let me elaborate on a few points from HardwareJedi (good points!)

1) A smaller micron size lets the core fit in a smaller space, just as he said.

2) Fitting the core in a smaller space cuts down on the loss of the circuit and allows one to run the same core at lower voltages. (For example, the .18 micron PIII (Coppermine) ran at voltages near 1.65V, but the .13 micron PIII (Tualatin) uses a voltage of 1.45V - 1.475V)

3) Because the core can run at a lower voltage, its power consumption is reduced, making it more economical and also better suited for laptop / portable solutions. Also, the lower power consumption reduces the heat production. Again, the PIII 1.2 GHz tualatin produces at most 29.9W of heat, wheras the PIII 1000 coppermine produces 29W at 200MHz slower speeds.

4) The smaller core size allows Intel and AMD to fit more cores per silicon wafer. This helps to increase yields and make chips less expensive.

5) Having a smaller core size also makes room for additional features on the die. For example, the PIII-S, being a .13micron die shrink, has an additional 256KB of L2 cache. According to my testing, that extra cache gives the PIII-S at least a 5% speed increase over PIII's of identical clock speed.

I'm sure there are other aspects as well, but hopefully this should be a start. Good question! -- Paul

[JimmyG]

HardwareJedi was almost right. The "micron" number refers to the size of a circuit feature, such as a tranistor. That is, a .13 micron chip contains features ( transistors, resistors, etc.) that are .13 microns across each way.

[macklin01]

Ah, I thought it was something more to that tune ... thanks for clarifying! -- Paul

[lazerin]

Cheers guys :) Appreciated it!

[David]

For comparison, a human hair is approx 100microns wide.

David

[HardwareJedi]

So .10 micron would be abour 1/1000th the width of a Human hair. That is small...

Thanks to JimmyG and macklin01 for fleshing out my explanation and filling in the gaps.

[David]

Originally posted by HardwareJedi
So .10 micron would be abour 1/1000th the width of a Human hair. That is small...

Thanks to JimmyG and macklin01 for fleshing out my explanation and filling in the gaps.

yup its small. And 0.09um will be even smaller... :eek:

David

[macklin01]

Originally posted by HardwareJedi
So .10 micron would be abour 1/1000th the width of a Human hair. That is small...

Thanks to JimmyG and macklin01 for fleshing out my explanation and filling in the gaps.
My pleasure; it's a fascinating topic! -- Paul

[NookieN]

As several others have said, the .NN micron figure usually refers to the transistor size. Most often that means the width of the transistor. Their length (top to bottom, source to drain) is usually several times larger. However the gate, the "switch" on the transistor, is about half the transistor width.

As another point of interest, in order for chip makers to achieve circuits this small they must use extreme ultra violet and X-ray radiation. Normal light can't give fine enough details.

[macklin01]

Yet some more good points. Thanks! -- Paul

[OC-Master]

I'd just like to add that with the advancement to 0.09 Microns, Intel will be able to have Pentium CPUs run at 4GHz+ with only 1.25V. That is very effecient in terms of heat production.


OC-Master

[macklin01]

Now that will be something! Thanks for the additional info! :) -- Paul

[larva]

Although it was touched upon in an earlier post, it merits some addtional verbage to note that even if we where not decrease the operating voltage heat production on smaller processes is less simply due to the now shorter paths in the circuits of our cpu. When we pass a current through a conductor, heat is generated by the electrical resistance of the conductor. The other factor is the length of the conductor, and ours get shorter on the newer and smaller fabrication processes.

It is also true that the resulting "shrunken" chip will operate on less voltage if the new process is mature. This is good because the physically smaller elements of our processor have a lesser tolerance for voltage. PC processors prior to the 486DX4/100 ran on the same 5V as the rest of the semiconductors in the system. We all know what would happen if we put 5V on any modern chip... smoke. The DX4 was the advent of the 3.3V processor, and voltages have been on a fairly steady decline ever since.

[NookieN]

Originally posted by larva
Although it was touched upon in an earlier post, it merits some addtional verbage to note that even if we where not decrease the operating voltage heat production on smaller processes is less simply due to the now shorter paths in the circuits of our cpu. When we pass a current through a conductor, heat is generated by the electrical resistance of the conductor. The other factor is the length of the conductor, and ours get shorter on the newer and smaller fabrication processes.


That's a good point. I think voltage would still have a larger impact on heat dissipation though.

CMOS transistors really only have current flowing through them when they switch from a '0' to a '1' or vice-versa. During this switch, small capacitors on the transistors either charge or dischange and current flows through the semiconductor. Shrinking the circuits means that less current needs to flow during these transistions and, combined with the lower voltage, the power dissipation is less.

Yes, shortening the trace lengths helps some, but since the traces become narrower their resistance doesn't change much.

[macklin01]

Hmm,


R = L / (sigma A), where
L = length of wire
sigma = conductivity
A = cross-sectional area

As A =pi r^2, where r is the trace radius, if supposing for the sake of argument that a p die shrink (where the new die size is p times the old die size, e.g., .09 = .6923 * .13) reduces all length measurements linearly by p, then the new resistance is
R_new = ( pL) / ( sigma pi R^2 p^2) = 1/p R_old.

As 0 < p < 1, the resistence for each individual trace would actually increase in the die shrink, under these circumstances. The length of each trace would have to be decreased by a factor of p^2 just to maintain the same resistence.

Now, power loss is P = V^2 / R.

So, on a per-trace basis,

P_new = V_new^2 / R_new = (p / R_old) V_new^2.

Let v be the ratio v = V_new / V_old. Then

P_new = (p / R_old) * V_old^2 *v^2
= (V_old^2 / R_old) *( p v^2 ) = P_old * (p v^2).

As 0 < p < 1 and 0 < v < 1, this shows that the new power emitted by each trace is significantly lower than that of the old trace, despite its smaller diameter and higher resistence.

So, that should give a rough estimate of the power loss (which would become heat) in each trace of a processor after a die shrink.

So, in our case, V_old = 1.525V, V_new = 1.25V , v = .820, and p = .69, so in each trace, the power loss is

p v^2 = (.69)*(.82)^2 = .46, so there is 46% of the original power loss in each trace.

As for the transistors, I have no idea. :) -- Paul

Source: Serway, Physics for Scientists & Engineers w/ Modern Physics, 4th Ed. pp 777ff

EDIT: There is one flaw here: That power loss formula requires a voltage drop of V across the trace. This is hardly the case!!! (Otherwise, each transistor would never receive any + voltage) However, since we just worked out a percentage of prior power loss, if we assume that the new voltage drop across the trace is proportional to the old voltage drop across the trace, which is entirely reasonable, then the calculation above still holds. -- Paul

EDIT: One more caveat: this probably only holds for equal clock speeds, multipliers, etc.

[HardwareJedi]

Good thread... Larva, one point, 5V was used up to and including the 60 and 66Mhz Pentium. They switched to 3.3V with the P-75 and on up to the Pentium MMX where they started using the Split Voltage thing... and 2.8V core.

[JimmyG]

On a historical note, back when logic circuits were made with discrete components (transistors, resistors, diodes, etc) signals were sent down wire buses that were hundreds of feet long. These required bus terminating resistors to reduce signal reflections backwards on the bus. Now, even inside the CPU chips, they are still using bus terminating resistors to do the same job because some of the traces inside the chip are long enough to generate these reflections. Long....nowadays measures in microns......LOL

[larva]

Originally posted by HardwareJedi
Good thread... Larva, one point, 5V was used up to and including the 60 and 66Mhz Pentium. They switched to 3.3V with the P-75 and on up to the Pentium MMX where they started using the Split Voltage thing... and 2.8V core.

The 486DX4/100 debuted long before the first 3.3V pentiums. I had one. It was the first 3.3V processor. Due to the fact that AMD had copied the i486 exactly Intel was in a rush to get the pentium to market to try to make AMD look second best. As such they sold the 5V 60 and 66MHz pentium chips at a stage of development that was almost embarrasing. I had one...

As well, the debut 3.3V pentium chips were the 90 and 100MHz speeds. I had one. As noted above this occured (long) after the introduction of the 3.3V DX4. Many months later the 3.3V P-75 was introduced to fill the low end gap left by the early cease in production of the DX4. Again, AMD by this time had its own copy of the DX4 in production so Intel killed its own and replaced it with the more expensive and no faster P-75.

[lazerin]

Great work guys!

I liked your post macklin01, equations are a universal language! :)

[macklin01]

Glad to oblige. ;) Turned out to be an interesting, fun thread! :D Thanks for asking a good question! -- Paul

[larva]

Originally posted by macklin01

EDIT: There is one flaw here: That power loss formula requires a voltage drop of V across the trace. This is hardly the case!!! (Otherwise, each transistor would never receive any + voltage) However, since we just worked out a percentage of prior power loss, if we assume that the new voltage drop across the trace is proportional to the old voltage drop across the trace, which is entirely reasonable, then the calculation above still holds. -- Paul



I'm not sure the assumption rings true. If indeed the new trace's resistance is greater then the voltage drop across it must be greater as well. As I understand it your power loss calculation assumes it to be less.

[larva]

Originally posted by macklin01
[2) Fitting the core in a smaller space cuts down on the loss of the circuit and allows one to run the same core at lower voltages.

Also, the conclusion that the resistance of the trace has actually increased seems to contradict your statement above. The only reason to use more voltage is to overcome the resistance present. If the new trace resistance is greater, then the core could not operate on the even the same voltage, much less less voltage.

I guess the answer must lie in the new properties of our shrunken transistors. Perhaps their power requirements are so lessened that the lower operating voltage is possible in spite of changes in the trace resistance.

[macklin01]

Hey, larva, thanks for writing!

The statement I made earlier was prior to these "back-of-the-envelope" calculations. At any rate, those calculation are very rough, so I wouldn't put much stock in them ...

Now, as for the resistence being higher: there's little to doubt about this. Is the wire thinner? Yes. So does the cross-sectional area decrease as r^2? Yes. So, the resistence increases as 1/r^2 (r is decreasing) due to these effects. But there's a counter-acting shortening of trace length, which tends to decrease resistence as L. (L is decreasing) The net effect is increasing resistence as 1/p, where p is the proportion mentioned above.

This resistence figure IS correct if the die shrink is a proprtional shrink of all components. No if's, and's, or but's about it. Its formula is based entirely upon length scales and conductivity, which are all independent of the voltages. So, that part of the calculation is sound.

The result does contradict my earlier statement. But this one is correct. It's interesting that the effects of shrinking the radius of the traces dominate those of shrinking the lenght of the traces. But that's the way it is. Sometimes, doing the math will give you an interesting surprise. :)

It's the power loss formula that I'm more worried about.

Without a better understanding of what the transistors do as they scale, this part is pretty dicey. Suppose, though, that in some particular part of the circuit (a closed circuit), transistors acount for 98% of the voltage drop and traces lengths for 2%. Then if the new voltage applied to the entire circuit is 80% of the old voltage, the 98% stretch of the new circuit receives 80% voltage, as well as the 2% stretch. So, from that standpoint, the 20% voltage drop along the traces would still hold. (Notice, we're not talking about absolute values of voltage drop across each trace length--we're talking about percentages of previous values. It's the fact that we divide them that makes it okay to talk about them. This is a very important point: we're talking about relative change.)

These assumptions are valid enough.

When you combine all the net effects, you do indeed get less power loss on the traces after a die shrink, just as I mentioned. The mechanisms are more complicated that I had originally anticipated. That's when it's time to pull out the ol' physics books ... :)

But at the end of the day, I would surmise that the majority of the power consumption comes not from loss over the traces lengths, but rather from the transistors themselves. So, while an estimate on the reduced power loss due to the traces is great and all, it probably is not a good estimate on the overall effects on power loss, just because it's harder to predict what those transistors will do. (After all, they're nonlinear devices.)

I guess the answer must lie in the new properties of our shrunken transistors. Perhaps their power requirements are so lessened that the lower operating voltage is possible in spite of changes in the trace resistance.

Absolutely! Without that part of the picture, we don't get a good grasp of what's going on.

I guess I should have prefixed my post with, "Well, let's see what kind of estimate we can get with our scanty details" and concluded with "Well, about all we can do is estimate the power loss along the traces, but this is pretty rough, and we're missing the major effects of the transistors. Too bad, but a nice try and a fun exercise."

Thanks for making these good points. -- Paul

[larva]

Originally posted by macklin01

It's the power loss formula that I'm more worried about.

Without a better understanding of what the transistors do as they scale, this part is pretty dicey. Suppose, though, that in some particular part of the circuit (a closed circuit), transistors acount for 98% of the voltage drop and traces lengths for 2%. Then if the new voltage applied to the entire circuit is 80% of the old voltage, the 98% stretch of the new circuit receives 80% voltage, as well as the 2% stretch. So, from that standpoint, the 20% voltage drop along the traces would still hold. (Notice, we're not talking about absolute values of voltage drop across each trace length--we're talking about percentages of previous values. It's the fact that we divide them that makes it okay to talk about them. This is a very important point: we're talking about relative change.)

These assumptions are valid enough.



Yes I agree that the trace resistance has actually grown rather than lessened, contrary to both our preconceptions of this factor. That part of the mathematical analysis rings true.

The power loss figure is a bit off in my estimation though. Your statement about the distribuition of voltage drop between the traces and the transistors is of course correct, but these figures are not constant. As proven the trace resistance has increased, and as such the voltage drop increased over the traces. As well the voltage drop of the transistors has decreased, as evidenced by our new chip's ability to run on lower voltages. If the distribution may have been 20%/80% on the old chip, it is bound to be say 25/75 for the new one.

I would think the power loss over the traces would still be lessened (if we do indeed achieve as drastic a drop in the operating voltage as in the example), but I feel the magnitude of the difference is less than stated as a result in the different distribution of voltage drop noted above.

It would appear our original concern of the effect of the difference in trace resistance was somewhat academic anyway. It seems to me that difference in the properties of our transistors is the dominant factor here, and by no small amount. I hadn't ever calculated the trace resistance, but your analysis of this point is good and makes it clear the improved transistor properties are the real improvement in function of the new chip.

Of course the manufacturers are not pursuing this goal so much as to improve production yield. New technology often brings performance improvements, but generally it's adoption is spurred more by the economic advantage it offers the manufacturer. I see the performance improvements as great and we all like this aspect of the new technologies employed, but I can't help but think this is as much a side effect of pursuing more economic mass production of these devices as it is the prime motivation for their adoption.

Your input on this subject is appreciated. My last meaningful discussion of this topic was spurred by the introduction of the DX4 processor, and it's attendant voltage drop to 3.3V. I was still in school at the time pursing an EE degree, and my father (an EE and senior design engineer for Philips) and I discussed it over breakfast at Micky D's. Our back-of-the-napkin figures did not approach the depth of your trace resistance calculation, as neither of us is in the habit of packing the necessary reference materials for a trip to McDonalds. Most see me as a pretty strange individual, but even I am not that strange :)

[NookieN]

In theory, at least, the input of one transistor should present an infinite resistance to the output of another. Having current flow into the gate of a transistor is bad, but there is usually a small about of leakage (pico-amps) so there's some power loss there.

Most of your power loss will be when the transistor switches and current flows across the diffusion (i.e. from Vdd to output or from output to Gnd). Since even doped silicon isn't a great conductor, the parasitic resistance should be much greater than the trace resistance. Unfortunately, I'm not having much luck in my google searchs for some numbers on transistor resistance.

Obviously, keeping the traces as short as possible helps with both speed and power loss though.

[macklin01]

Originally posted by larva
Yes I agree that the trace resistance has actually grown rather than lessened, contrary to both our preconceptions of this factor.

Thanks!

If the distribution may have been 20%/80% on the old chip, it is bound to be say 25/75 for the new one.
I think you're right on. This is the very sort of issue that I'm concerned about -- without knowing more about the transistors, etc., I don't feel very confident about any estimates of % of voltage drop being along the traces and % on the transistors.

...but I feel the magnitude of the difference is less than stated as a result in the different distribution of voltage drop noted above.
If it's as you stated, you're absolutely right.

It would appear our original concern of the effect of the difference in trace resistance was somewhat academic anyway. It seems to me that difference in the properties of our transistors is the dominant factor here, and by no small amount.

I agree 100%. That's just what I was trying to say -- I think the dominant effects lie with the transistors.

I hadn't ever calculated the trace resistance, but your analysis of this point is good and makes it clear the improved transistor properties are the real improvement in function of the new chip.

Thanks! Yes, if anything has really come of those scribblings, it's that the effects are more complicated than can be understood at first glance, and the transistor properties are the most important.

Of course the manufacturers are not pursuing this goal so much as to improve production yield. New technology often brings performance improvements, but generally it's adoption is spurred more by the economic advantage it offers the manufacturer. I see the performance improvements as great and we all like this aspect of the new technologies employed, but I can't help but think this is as much a side effect of pursuing more economic mass production of these devices as it is the prime motivation for their adoption.

By and large agreed. Intel made great gains by moving to a smaller process in terms of productivity.

On the other hand, if they don't improve their thermal performance, they won't be able to advance their technology (introduce faster chips) and will stagnate. So, economic concerns will also drive them to address thermal performance issues, as it is one of the primary roadblocks to technological advancement / product development. Also, as the handheld market continues to be more important, decreasing power consumption and increasing battery life will be important to their bottom lines. (If they don't do it, somebody else will.)

Your input on this subject is appreciated. My last meaningful discussion of this topic was spurred by the introduction of the DX4 processor, and it's attendant voltage drop to 3.3V.

Thanks -- I really appreciate it! And I'd have to say that this is my first meaningful discussion of it.

I was still in school at the time pursing an EE degree, and my father (an EE and senior design engineer for Philips) and I discussed it over breakfast at Micky D's.

Sounds like a great time!

Our back-of-the-napkin figures did not approach the depth of your trace resistance calculation, as neither of us is in the habit of packing the necessary reference materials for a trip to McDonalds.

:D Harder with McD's scarcer resources. I'm impressed! :)

Most see me as a pretty strange individual, but even I am not that strange

Saddest part, I used to pack up all my references to study at BK .... there I was, sprawled out with Complex Analysis, real analysis, PDE books, notepads, and the like at BK with my 70 cent coffee and $.99 croissanwich. I can assure you, you'll never be the strangest one out there ... :D

Thanks for the fun discussion!! I really value your input, too, because it has been invaluable! :)

Hope to see more of you in the forums! -- Paul

[macklin01]

Originally posted by NookieN
, the parasitic resistance should be much greater than the trace resistance. Unfortunately, I'm not having much luck in my google searchs for some numbers on transistor resistance.
Hi, and thanks for the additional insight. I'm afraid I don't have the same knowledge of the internals of transistors, etc., as you.

I think we've all come to a similar conclusion: the transistors are much more important to the equation (and also harder to estimate!)

If you do find anything else, please let us know!

Thanks again! :) -- Paul

[larva]

Originally posted by NookieN

Most of your power loss will be when the transistor switches and current flows across the diffusion (i.e. from Vdd to output or from output to Gnd). Since even doped silicon isn't a great conductor, the parasitic resistance should be much greater than the trace resistance. Unfortunately, I'm not having much luck in my google searchs for some numbers on transistor resistance.

Obviously, keeping the traces as short as possible helps with both speed and power loss though.

Good info there. I'm not much on the internal workings of transistors. But it is obvious to me that their power consumption dwarfs that of the traces due to the net effect that our chip can run on lower voltage despite the increasing trace resistance now established.

Another benefit of short traces is to control RFI and other properties I am probably unfamiliar with. As the trace length increases so does its ability to act as an antenna, an effect exaserbated by increasing clock speeds. For any number of reasons, short traces are better.

I would also like to mention how beneficial this thread is. At no point do the holes in your own understanding become clearer than when you try to explain something to others. I absolutlely love helping others, especially where understanding is involved. But just as great a reward lies in the advancement of your own understanding. Although it is not what motivates me to try to help in the first place, it is a significant benefit those that strive to help their fellow man enjoy. All too rare is an acceptant audience for instruction present, much less one that thirsts for it as do most members of this board. Some aspects of "Internet Life" may pale in comparison to "real" life, but others are a vastly enhanced by the free exchange of ideas with as large and diverse a body of people as we have access to in this day and age. Kudos to all that seek to better themselves and others, and to the providers of this forum that so ably accomodates our desires.

[lazerin]

This thread should be stickied :)

[macklin01]

Second that. ;) -- Paul

[s[H]sIkuA]

lazerin your 3dmark is awesome!

[lazerin]

lol. cheers :)

[larva]

Perhaps this thread could be made a sticky in the "Technical Discussions" section. The information and reasoning present here certainly qualifies and all would do well to read it.

[NookieN]

Yeah, this thread alone would increase the post count by 250% in that forum!

[macklin01]

LOL, I didn't even know we had a Technical Discussions section .. :D -- Paul