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Please Explain Micron

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lazerin

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Joined
Jul 13, 2002
Location
Australia
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? :)
 
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

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Apr 22, 2002
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[email protected] in Ball Ground, GA
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

Computational Oncologist / Biomathematician / Mode
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Location
Bloomington, IN
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

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Apr 21, 2001
Location
Michigan
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.
 

David

Forums Super Moderator
Joined
Feb 20, 2001
For comparison, a human hair is approx 100microns wide.

David
 

David

Forums Super Moderator
Joined
Feb 20, 2001
HardwareJedi said:
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

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Location
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HardwareJedi said:
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

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Jul 14, 2002
Location
The West
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.
 

OC-Master

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Jul 14, 2001
Location
Edmonton, Alberta
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
 

larva

Inactive Moderator
Joined
Jul 12, 2002
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

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larva said:
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

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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.
 
Last edited:

JimmyG

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Apr 21, 2001
Location
Michigan
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