I guess you should just read this or try searching with google.com for some information on CPU power usage. You obviously assume.
... and no/yes Watts = Volts * Amps. In a 3-phase circuit, though, don't forget to multiply line current by the square-root of 3 and multiply by the Power Factor.
Enough said.
A few of us have tried to explain these technical issues in plain English, so those without an advanced degree in engineering or sciences can understand them. However, all we get from you so far is the reciting of a few formulas. You have not explained why you believe a CPU does not obey the basic rule for power in a DC supply condition. In other words, what makes you think power does not equal current (Amps) times electromotive force (Volts)? Unless you can explain it, we may just conclude that you do not understand the topic and were simply trying to impress us.
You mentioned AC circuits, which is one case where power may be less than Amps times Volts. You didn't explain why this is so, however. It happens when the voltage and current variation are not synchronized, so part of the time the device is actually giving back power to the generator. Of course this example has nothing to do with DC power supply circuitry, so why bring it up?
One thing that does affect a CPU a little bit is its many inputs and outputs. If some of these are adding to the CPU power without drawing current from the CPU, the total power will be a little higher than what is calculated from Amps times Volts. But this has nothing to do with the formulas that you have been posting. Also, the link you provided has little to do with this discussion, as far as I can tell.
I know that the current numbers he's mentioned sound huge, but as he's said, at low voltages, the resulting power is not as big as what most are used to at household voltages (120V, America, 100V Japan, 220V in much of Europe I think)
However, the amperages that he has listed would be the average current drawn. Peak currents would be higher, and one could use some much more complicated equations to estimate those peaks. The peaks will happen when the most transistors are switching simultaniously in the chip. There will also be periods of lower current draw.
However, as snoopy said, in the end the average current will obey the P=VI law.
And if you think the aforementioned currents are big, listen to this:
During one part of the CPU fabrication cycle, called Burn-In, the currents drawn by a single CPU chip can reach up to 300 Amps and 400 Watts
Seriously. This is due to the dramatically increased leakage currents (as someone else mentioned earlier), and elevated core voltages.
If anyone wants to know more about Burn-In, let me know.
Sorry about that. I was really responding to a reply to your post, which interpreted you to mean we would not see the 970 until September. In my haste, I thought it would be better to quote your post too, but as soon as it appeared on the message board I realized I was wrong. I apologize. I should have quoted what I was replying to.
Fool me once shame on you. Fool me twice shame on me. Herb Brooks '81
I wouldn't trust this shiester farther than I could throw him if he was glued to the floor. He already has a track record of playing fast and loose when big $$ are on the table and his referenced posts smell like yesterdays catch left in the sun.
He's all about a quick buck before the well goes dry and his false bravado is one of the pre-requisites to entering that line of work--the (almost) con job...
Yea. I mean he may have contacts, he may know a few people since he deals with distribution of Apple compatible merchandise, but I think there's something missing and that's the motivation. So while some of his info may become reality, I won't follow his words with bated (?) breath as I do Think Secret's.
I know that the current numbers he's mentioned sound huge, but as he's said, at low voltages, the resulting power is not as big as what most are used to at household voltages (120V, America, 100V Japan, 220V in much of Europe I think)
However, the amperages that he has listed would be the average current drawn. Peak currents would be higher, and one could use some much more complicated equations to estimate those peaks. The peaks will happen when the most transistors are switching simultaniously in the chip. There will also be periods of lower current draw.
However, as snoopy said, in the end the average current will obey the P=VI law.
And if you think the aforementioned currents are big, listen to this:
During one part of the CPU fabrication cycle, called Burn-In, the currents drawn by a single CPU chip can reach up to 300 Amps and 400 Watts
Seriously. This is due to the dramatically increased leakage currents (as someone else mentioned earlier), and elevated core voltages.
If anyone wants to know more about Burn-In, let me know.
The CPU is an ?LC circuit that has frequency, capacitance and inductance. The power useage is proportional to Voltage^2 and frequency. It's not like one wire having 300 amps going through it, there are many wires and many circuits tied together. An as I stated above Power = Volts* Amps substitute the appropriate terms in the Power equation I gave above and re-arrange as you like.
The CPU is an ?LC circuit that has frequency, capacitance and inductance. The power useage is proportional to Voltage^2 and frequency. It's not like one wire having 300 amps going through it, there are many wires and many circuits tied together. An as I stated above Power = Volts* Amps substitute the appropriate terms in the Power equation I gave above and re-arrange as you like.
(edit: that's an Omega-L-C circuit)
As long as it doesn't burn my lap, I am ok with it...
The power useage is proportional to Voltage^2 and frequency.
A small calculation for the 970:
IBM said its power usage is 19w for 1.2GHz@1.1v and 42w for 1.8GHz@1.3v.
This gives 19*(1.8/1.2)*(1.3/1.1)^2 = 40
or 42*(1.2/1.8)*(1.1/1.3)^2 = 20
Close enough to estimate the power usage at a given frequency.
For example it showes us that a 1.4 GHz 970 @1.3v uses about 33w, much more than the 1.2 GHz part @1.1v - just for those dreaming about a 1.4 GHz PowerBook. It will not happen until IBM can deliver a 970 1.4 GHz @1.1v which would only use 22w.
The CPU is an ?LC circuit that has frequency, capacitance and inductance. The power useage is proportional to Voltage^2 and frequency. It's not like one wire having 300 amps going through it, there are many wires and many circuits tied together. An as I stated above Power = Volts* Amps substitute the appropriate terms in the Power equation I gave above and re-arrange as you like.
(edit: that's an Omega-L-C circuit)
Further to what snoopy has said.
No, it isn't. The inductance (L) in a CPU is tiny and mainly comes from the bond wires. There will be some small efect from the LC circuits which cause current to flow back into the power supply, but this would only be from a few tiny elements, and the measurement of average current takes them into account anyway, they simply slightly reduce the current drawn at certain times. The instantaneous current draw may not be purely resistive (like a simple wire), but the average current draw is.
The current is formed by a combination of leakage (a perfectly straight forward resistive power dissipation) and active current. The active current is also effectively resistive since it is formed from capacitors being charged up to the supply voltage through the channels of transistors (the channel of a transistor is resistive) and then discharging through the channels of other transistors to ground, hence all the current flows through resistive transistor channels from the power supply to ground. Any stray capacitances outside this chain merely swap charge around and do not contribute to the average current drawn.
It is true that the formula Power = afCV^2 is used, but this is because afCV is the average current flowing through the device, which follows from the explanation above.
The calculation using power factor is only valid when you have an alternating voltage, which is not the case here as the CPU has a constant DC voltage power supply, hence omega is not applicable.
The introduction of three phase calculations was wilfully misleading as nothing anywhere within a standard computer uses a three phase power supply.
For example it showes us that a 1.4 GHz 970 @1.3v uses about 33w, much more than the 1.2 GHz part @1.1v - just for those dreaming about a 1.4 GHz PowerBook. It will not happen until IBM can deliver a 970 1.4 GHz @1.1v which would only use 22w. [/B]
In one of the IBM PDFs that I can no longer find on the net there were (estimated) power consumption figures for the 980 chips which were scheduled to begin production next year. I do not recall the precise numbers, but the 980 was to be better in every way than the 970. When it moves from the 130 nm process to the 90 nm process with related internal voltage changes and such the power consumption was projected to drop. This will benefit IBM as much as Apple as IBM's use of the 9xx series of chips in blades will benefit from reduced power consumption and heat issues.
If the 970 production is on schedule or a bit ahead of schedule it will be interesting to see if any information about revised 980 production schedules makes its way out later this year.
Also, there was a news item several months ago about a DOD sponsored project that was working on a passive cooling device, small enough to fit in a laptop, to cool CPUs in difficult environments...if that works out it may get incorporated into off the shelf products, but that would probably be some time away from widespread implementation.
Just a little FYI here: The 980 is not a die-shrunk 970. It's the successor to the 970, derived from the POWER5.
The 90nm version of the 970 is referred to generally (if colloquially) as the 970+.
I'm not surprised that the 980 will beat the 970, because the 970 is almost a proof of concept, not especially well-integrated (but not at all bad). The 980/POWER5 will be IBM's first mature foray into this style of CPU, and it should be righteous.
This is not to pooh-pooh the 970 at all, of course.
IBM unveils new-chip details. Thursday - May 15, 2003
From the article:
Quote:
The FSB implementation is crucial. The processor drives this at up to 900MHz. The Pentium 4 offers 533MHz and the Athlon XP gets 333MHz.The report offers several comparisons with competing processors, and the 970 seemingly surpasses or competes with x86-based competition. Its adoption would help Apple remain competitive in the market.Halfhill concludes: "It's a good bet the 970 will also end up in a Mac - unless Apple's thinking is even more different than advertised."
Just a little FYI here: The 980 is not a die-shrunk 970. It's the successor to the 970, derived from the POWER5.
The 90nm version of the 970 is referred to generally (if colloquially) as the 970+.
I'm not surprised that the 980 will beat the 970, because the 970 is almost a proof of concept, not especially well-integrated (but not at all bad). The 980/POWER5 will be IBM's first mature foray into this style of CPU, and it should be righteous.
This is not to pooh-pooh the 970 at all, of course.
Amorph,
That is my point exactly. The 970 is really just something to open the door for the 980. That is not to say it is bad, all indications are that it will be a very worthwhile improvement over what Apple have been using, but that it is a transitional CPU. IBM *really* wants to get the 980 out the door. It is their future, not the 970.
You are, of course, correct in saying that the 980 is an entirely new processor, not merely a "die shrink" of the 970. However the "die shrink" is an important part of the commercial production process from a cost and production perspective which is probably as important as the technical improvements along the way.
The 970+ is an add-on to the product life cycle as near as I can make out as there were no plans originally to take the 970 to the 90 nm process. There are several ways of interpreting this and I do not know which one is more probable. It could be that IBM are moving more rapidly than anticipated to the 90 nm process and the 980 will be introduced with than process rather than the planned 130 nm process. This would be very good as it would accelerate the rest of the development process and save some money in the long run. Another possibility is that the 980/Power 5 development is taking longer than anticipated and going to the 970+ is buying some time to work on it. Of course there is a possibility that the two are actually one, that is to say IBM may have had success with the 90 nm process and decided to skip the 980 130 nm step (this is purely speculation), but the production dies for the 90 nm process may take a bit longer to have ready than was projected for the 130 nm process 980 and so they are preparing the 970+ to sort out any issues with the 980 90 nm process as they are working on it.
I tend to favor the good news scenarios as IBM seems to be on a roll, but there is always the possibility that they are simply having problems.
The 970 is really just something to open the door for the 980.
Umm. using that logic, the same could be said for any chip from any manuf.
the 970 is the first new chip in a family new chips. the is nothing "hold over" it has a limited life span. like all chips. it will be eventually replaced. like all chips.
By your logic the 980 is just a holder over for the 990
That is my point exactly. The 970 is really just something to open the door for the 980. That is not to say it is bad, all indications are that it will be a very worthwhile improvement over what Apple have been using, but that it is a transitional CPU. IBM *really* wants to get the 980 out the door. It is their future, not the 970.
I don't think you can say that any single processor is the future per se. When the 980 ships, the 990 will be the future. When then 990 ships the whatever will be the future. In any way the 970 is an enormous improvement over any current offering that Apple can choose from. It will be mean a huge performance gain, parity or almost parity with the wintel platform (being conservative here) and bragging rights about 64-bitness and so on. So in many ways the 970 is indeed the future for Apple. Then 980 will be more of an evolutionary step no matter how much it pushes performance (that will be quite a lot i recon), while the 970 will bring Apple and the mac back in the performance game. And another advantage is that Apple can tailor the next version of os X spesificially for this processor, while MS has to support a lot of different processors, being both 32 bit and 64 bit, which will take a lot of resources I think.
Quote:
You are, of course, correct in saying that the 980 is an entirely new processor, not merely a "die shrink" of the 970. However the "die shrink" is an important part of the commercial production process from a cost and production perspective which is probably as important as the technical improvements along the way.
Both die shrink and architectural improvements are equally important in the making of a new chip. New chips has often a lot more transistors making them bigger and hotter. A die shirnk can balance that. However a die shrink of a current processor will only lead to improved speed per increased clock cycle, while an improvement of the arcitechture can add more power per clock cycle. So they are both equally important. Increasing clock speed and reducing heat will be increasingly more difficult as processes get smaller and smaller.
Quote:
The 970+ is an add-on to the product life cycle as near as I can make out as there were no plans originally to take the 970 to the 90 nm process.
Why not. The 970 is an impressive processor now, and still will be in a year, especially at 2.5 ghz. The 90nm process will make perfect processors for the powerbooks and imacs.
Quote:
There are several ways of interpreting this and I do not know which one is more probable. It could be that IBM are moving more rapidly than anticipated to the 90 nm process and the 980 will be introduced with than process rather than the planned 130 nm process.
Your probably right about this. But we don't have any exact information about his yet.
Quote:
This would be very good as it would accelerate the rest of the development process and save some money in the long run. Another possibility is that the 980/Power 5 development is taking longer than anticipated and going to the 970+ is buying some time to work on it. Of course there is a possibility that the two are actually one, that is to say IBM may have had success with the 90 nm process and decided to skip the 980 130 nm step (this is purely speculation), but the production dies for the 90 nm process may take a bit longer to have ready than was projected for the 130 nm process 980 and so they are preparing the 970+ to sort out any issues with the 980 90 nm process as they are working on it.
Not very likely I think. I think they had meant to make the 970+ for a long time. The 970 was a "we have to do it in so-and-so many days", so we take it to the 130nm process because that make it easier and faster. Plus the 90nm facility hasn't been finalized yet, what would have meant a long wait for 970 powermacs for Apple. I think Apple had a hand in a lot of these decisions.
Quote:
I tend to favor the good news scenarios as IBM seems to be on a roll, but there is always the possibility that they are simply having problems.
Of course. Making modern processors aren't easy, and there could always be problems. But I think IBM is one of the best out there, and I'm fairly confident in them working problems out in a far more "polished" manner than motorola. And they already have made the Power5 which make them almost experts on this processor already. That has to mean a lot when they are making the 980. Well, just had to say a few words here.
The 970+ is an add-on to the product life cycle as near as I can make out as there were no plans originally to take the 970 to the 90 nm process.
All of the IBM literature I've read has talked about a debut at 130nm, with a move to 90nm "shortly". 90nm and smaller lithography is what Fishkill is really set up for, and IBM has consistently voiced an intention to move to 90nm as quickly as they can. Fishkill cost a mint, and it's gotta pay for itself.
The die-shrunk 970 should arrive some time before we see the 980, and it'll be a great PowerBook/iMac CPU.
Quote:
I tend to favor the good news scenarios as IBM seems to be on a roll, but there is always the possibility that they are simply having problems.
Or, their schedule was always 970 @ 130nm, 970+ @ 90nm, 980 at 90nm, etc. That's the impression I've had since last fall from the available documentation.
Or, their schedule was always 970 @ 130nm, 970+ @ 90nm, 980 at 90nm, etc. That's the impression I've had since last fall from the available documentation.
Hmm... for some reason this "schedule" is imprinted into my brain
Or, their schedule was always 970 @ 130nm, 970+ @ 90nm, 980 at 90nm, etc. That's the impression I've had since last fall from the available documentation. [/B][/QUOTE]
I know that I saw a PDF (these things take up drive space, but I really have to start saving them) that had the 980 being introduced with a 130 nm process, but it may be that that plan was scrapped and I did not see the revised plan to introduce the 980 at the 90 nm process. Still, I think I would have heard something as this would be a pretty important change in their development roadmap.
Uncertainties in the construction and outfitting of the production facility could have been the holdup that was being covered with the (contingency plan?) roadmap to intro the 980 at 130 nm. (Few people complain when a project comes in early.)
Comments
Originally posted by Bigc
I guess you should just read this or try searching with google.com for some information on CPU power usage. You obviously assume.
... and no/yes Watts = Volts * Amps. In a 3-phase circuit, though, don't forget to multiply line current by the square-root of 3 and multiply by the Power Factor.
Enough said.
A few of us have tried to explain these technical issues in plain English, so those without an advanced degree in engineering or sciences can understand them. However, all we get from you so far is the reciting of a few formulas. You have not explained why you believe a CPU does not obey the basic rule for power in a DC supply condition. In other words, what makes you think power does not equal current (Amps) times electromotive force (Volts)? Unless you can explain it, we may just conclude that you do not understand the topic and were simply trying to impress us.
You mentioned AC circuits, which is one case where power may be less than Amps times Volts. You didn't explain why this is so, however. It happens when the voltage and current variation are not synchronized, so part of the time the device is actually giving back power to the generator. Of course this example has nothing to do with DC power supply circuitry, so why bring it up?
One thing that does affect a CPU a little bit is its many inputs and outputs. If some of these are adding to the CPU power without drawing current from the CPU, the total power will be a little higher than what is calculated from Amps times Volts. But this has nothing to do with the formulas that you have been posting. Also, the link you provided has little to do with this discussion, as far as I can tell.
I know that the current numbers he's mentioned sound huge, but as he's said, at low voltages, the resulting power is not as big as what most are used to at household voltages (120V, America, 100V Japan, 220V in much of Europe I think)
However, the amperages that he has listed would be the average current drawn. Peak currents would be higher, and one could use some much more complicated equations to estimate those peaks. The peaks will happen when the most transistors are switching simultaniously in the chip. There will also be periods of lower current draw.
However, as snoopy said, in the end the average current will obey the P=VI law.
And if you think the aforementioned currents are big, listen to this:
During one part of the CPU fabrication cycle, called Burn-In, the currents drawn by a single CPU chip can reach up to 300 Amps and 400 Watts
Seriously. This is due to the dramatically increased leakage currents (as someone else mentioned earlier), and elevated core voltages.
If anyone wants to know more about Burn-In, let me know.
Originally posted by snoopy
Sorry about that. I was really responding to a reply to your post, which interpreted you to mean we would not see the 970 until September. In my haste, I thought it would be better to quote your post too, but as soon as it appeared on the message board I realized I was wrong. I apologize. I should have quoted what I was replying to.
No biggie
Originally posted by AirSluf
Fool me once shame on you. Fool me twice shame on me. Herb Brooks '81
I wouldn't trust this shiester farther than I could throw him if he was glued to the floor. He already has a track record of playing fast and loose when big $$ are on the table and his referenced posts smell like yesterdays catch left in the sun.
He's all about a quick buck before the well goes dry and his false bravado is one of the pre-requisites to entering that line of work--the (almost) con job...
Yea. I mean he may have contacts, he may know a few people since he deals with distribution of Apple compatible merchandise, but I think there's something missing and that's the motivation. So while some of his info may become reality, I won't follow his words with bated (?) breath as I do Think Secret's.
Originally posted by Transcendental Octothorpe
Look guys, snoopy is right.
I know that the current numbers he's mentioned sound huge, but as he's said, at low voltages, the resulting power is not as big as what most are used to at household voltages (120V, America, 100V Japan, 220V in much of Europe I think)
However, the amperages that he has listed would be the average current drawn. Peak currents would be higher, and one could use some much more complicated equations to estimate those peaks. The peaks will happen when the most transistors are switching simultaniously in the chip. There will also be periods of lower current draw.
However, as snoopy said, in the end the average current will obey the P=VI law.
And if you think the aforementioned currents are big, listen to this:
During one part of the CPU fabrication cycle, called Burn-In, the currents drawn by a single CPU chip can reach up to 300 Amps and 400 Watts
Seriously. This is due to the dramatically increased leakage currents (as someone else mentioned earlier), and elevated core voltages.
If anyone wants to know more about Burn-In, let me know.
The CPU is an ?LC circuit that has frequency, capacitance and inductance. The power useage is proportional to Voltage^2 and frequency. It's not like one wire having 300 amps going through it, there are many wires and many circuits tied together. An as I stated above Power = Volts* Amps substitute the appropriate terms in the Power equation I gave above and re-arrange as you like.
(edit: that's an Omega-L-C circuit)
Originally posted by Bigc
The CPU is an ?LC circuit that has frequency, capacitance and inductance. The power useage is proportional to Voltage^2 and frequency. It's not like one wire having 300 amps going through it, there are many wires and many circuits tied together. An as I stated above Power = Volts* Amps substitute the appropriate terms in the Power equation I gave above and re-arrange as you like.
(edit: that's an Omega-L-C circuit)
As long as it doesn't burn my lap, I am ok with it...
Originally posted by Bigc
The power useage is proportional to Voltage^2 and frequency.
A small calculation for the 970:
IBM said its power usage is 19w for 1.2GHz@1.1v and 42w for 1.8GHz@1.3v.
This gives 19*(1.8/1.2)*(1.3/1.1)^2 = 40
or 42*(1.2/1.8)*(1.1/1.3)^2 = 20
Close enough to estimate the power usage at a given frequency.
For example it showes us that a 1.4 GHz 970 @1.3v uses about 33w, much more than the 1.2 GHz part @1.1v - just for those dreaming about a 1.4 GHz PowerBook. It will not happen until IBM can deliver a 970 1.4 GHz @1.1v which would only use 22w.
Originally posted by Bigc
The CPU is an ?LC circuit that has frequency, capacitance and inductance. The power useage is proportional to Voltage^2 and frequency. It's not like one wire having 300 amps going through it, there are many wires and many circuits tied together. An as I stated above Power = Volts* Amps substitute the appropriate terms in the Power equation I gave above and re-arrange as you like.
(edit: that's an Omega-L-C circuit)
Further to what snoopy has said.
No, it isn't. The inductance (L) in a CPU is tiny and mainly comes from the bond wires. There will be some small efect from the LC circuits which cause current to flow back into the power supply, but this would only be from a few tiny elements, and the measurement of average current takes them into account anyway, they simply slightly reduce the current drawn at certain times. The instantaneous current draw may not be purely resistive (like a simple wire), but the average current draw is.
The current is formed by a combination of leakage (a perfectly straight forward resistive power dissipation) and active current. The active current is also effectively resistive since it is formed from capacitors being charged up to the supply voltage through the channels of transistors (the channel of a transistor is resistive) and then discharging through the channels of other transistors to ground, hence all the current flows through resistive transistor channels from the power supply to ground. Any stray capacitances outside this chain merely swap charge around and do not contribute to the average current drawn.
It is true that the formula Power = afCV^2 is used, but this is because afCV is the average current flowing through the device, which follows from the explanation above.
The calculation using power factor is only valid when you have an alternating voltage, which is not the case here as the CPU has a constant DC voltage power supply, hence omega is not applicable.
The introduction of three phase calculations was wilfully misleading as nothing anywhere within a standard computer uses a three phase power supply.
michael
Originally posted by smalM
[B
For example it showes us that a 1.4 GHz 970 @1.3v uses about 33w, much more than the 1.2 GHz part @1.1v - just for those dreaming about a 1.4 GHz PowerBook. It will not happen until IBM can deliver a 970 1.4 GHz @1.1v which would only use 22w. [/B]
In one of the IBM PDFs that I can no longer find on the net there were (estimated) power consumption figures for the 980 chips which were scheduled to begin production next year. I do not recall the precise numbers, but the 980 was to be better in every way than the 970. When it moves from the 130 nm process to the 90 nm process with related internal voltage changes and such the power consumption was projected to drop. This will benefit IBM as much as Apple as IBM's use of the 9xx series of chips in blades will benefit from reduced power consumption and heat issues.
If the 970 production is on schedule or a bit ahead of schedule it will be interesting to see if any information about revised 980 production schedules makes its way out later this year.
Also, there was a news item several months ago about a DOD sponsored project that was working on a passive cooling device, small enough to fit in a laptop, to cool CPUs in difficult environments...if that works out it may get incorporated into off the shelf products, but that would probably be some time away from widespread implementation.
The 90nm version of the 970 is referred to generally (if colloquially) as the 970+.
I'm not surprised that the 980 will beat the 970, because the 970 is almost a proof of concept, not especially well-integrated (but not at all bad). The 980/POWER5 will be IBM's first mature foray into this style of CPU, and it should be righteous.
This is not to pooh-pooh the 970 at all, of course.
From the article:
The FSB implementation is crucial. The processor drives this at up to 900MHz. The Pentium 4 offers 533MHz and the Athlon XP gets 333MHz.The report offers several comparisons with competing processors, and the 970 seemingly surpasses or competes with x86-based competition. Its adoption would help Apple remain competitive in the market.Halfhill concludes: "It's a good bet the 970 will also end up in a Mac - unless Apple's thinking is even more different than advertised."
http://www.macworld.co.uk/news/top_n...fm?NewsID=6332
--
Ed
Originally posted by Amorph
Just a little FYI here: The 980 is not a die-shrunk 970. It's the successor to the 970, derived from the POWER5.
The 90nm version of the 970 is referred to generally (if colloquially) as the 970+.
I'm not surprised that the 980 will beat the 970, because the 970 is almost a proof of concept, not especially well-integrated (but not at all bad). The 980/POWER5 will be IBM's first mature foray into this style of CPU, and it should be righteous.
This is not to pooh-pooh the 970 at all, of course.
Amorph,
That is my point exactly. The 970 is really just something to open the door for the 980. That is not to say it is bad, all indications are that it will be a very worthwhile improvement over what Apple have been using, but that it is a transitional CPU. IBM *really* wants to get the 980 out the door. It is their future, not the 970.
You are, of course, correct in saying that the 980 is an entirely new processor, not merely a "die shrink" of the 970. However the "die shrink" is an important part of the commercial production process from a cost and production perspective which is probably as important as the technical improvements along the way.
The 970+ is an add-on to the product life cycle as near as I can make out as there were no plans originally to take the 970 to the 90 nm process. There are several ways of interpreting this and I do not know which one is more probable. It could be that IBM are moving more rapidly than anticipated to the 90 nm process and the 980 will be introduced with than process rather than the planned 130 nm process. This would be very good as it would accelerate the rest of the development process and save some money in the long run. Another possibility is that the 980/Power 5 development is taking longer than anticipated and going to the 970+ is buying some time to work on it. Of course there is a possibility that the two are actually one, that is to say IBM may have had success with the 90 nm process and decided to skip the 980 130 nm step (this is purely speculation), but the production dies for the 90 nm process may take a bit longer to have ready than was projected for the 130 nm process 980 and so they are preparing the 970+ to sort out any issues with the 980 90 nm process as they are working on it.
I tend to favor the good news scenarios as IBM seems to be on a roll, but there is always the possibility that they are simply having problems.
The 970 is really just something to open the door for the 980.
Umm. using that logic, the same could be said for any chip from any manuf.
the 970 is the first new chip in a family new chips. the is nothing "hold over" it has a limited life span. like all chips. it will be eventually replaced. like all chips.
By your logic the 980 is just a holder over for the 990
Originally posted by RBR
Amorph,
That is my point exactly. The 970 is really just something to open the door for the 980. That is not to say it is bad, all indications are that it will be a very worthwhile improvement over what Apple have been using, but that it is a transitional CPU. IBM *really* wants to get the 980 out the door. It is their future, not the 970.
I don't think you can say that any single processor is the future per se. When the 980 ships, the 990 will be the future. When then 990 ships the whatever will be the future. In any way the 970 is an enormous improvement over any current offering that Apple can choose from. It will be mean a huge performance gain, parity or almost parity with the wintel platform (being conservative here) and bragging rights about 64-bitness and so on. So in many ways the 970 is indeed the future for Apple. Then 980 will be more of an evolutionary step no matter how much it pushes performance (that will be quite a lot i recon), while the 970 will bring Apple and the mac back in the performance game. And another advantage is that Apple can tailor the next version of os X spesificially for this processor, while MS has to support a lot of different processors, being both 32 bit and 64 bit, which will take a lot of resources I think.
You are, of course, correct in saying that the 980 is an entirely new processor, not merely a "die shrink" of the 970. However the "die shrink" is an important part of the commercial production process from a cost and production perspective which is probably as important as the technical improvements along the way.
Both die shrink and architectural improvements are equally important in the making of a new chip. New chips has often a lot more transistors making them bigger and hotter. A die shirnk can balance that. However a die shrink of a current processor will only lead to improved speed per increased clock cycle, while an improvement of the arcitechture can add more power per clock cycle. So they are both equally important. Increasing clock speed and reducing heat will be increasingly more difficult as processes get smaller and smaller.
The 970+ is an add-on to the product life cycle as near as I can make out as there were no plans originally to take the 970 to the 90 nm process.
Why not. The 970 is an impressive processor now, and still will be in a year, especially at 2.5 ghz. The 90nm process will make perfect processors for the powerbooks and imacs.
There are several ways of interpreting this and I do not know which one is more probable. It could be that IBM are moving more rapidly than anticipated to the 90 nm process and the 980 will be introduced with than process rather than the planned 130 nm process.
Your probably right about this. But we don't have any exact information about his yet.
This would be very good as it would accelerate the rest of the development process and save some money in the long run. Another possibility is that the 980/Power 5 development is taking longer than anticipated and going to the 970+ is buying some time to work on it. Of course there is a possibility that the two are actually one, that is to say IBM may have had success with the 90 nm process and decided to skip the 980 130 nm step (this is purely speculation), but the production dies for the 90 nm process may take a bit longer to have ready than was projected for the 130 nm process 980 and so they are preparing the 970+ to sort out any issues with the 980 90 nm process as they are working on it.
Not very likely I think. I think they had meant to make the 970+ for a long time. The 970 was a "we have to do it in so-and-so many days", so we take it to the 130nm process because that make it easier and faster. Plus the 90nm facility hasn't been finalized yet, what would have meant a long wait for 970 powermacs for Apple. I think Apple had a hand in a lot of these decisions.
I tend to favor the good news scenarios as IBM seems to be on a roll, but there is always the possibility that they are simply having problems.
Of course. Making modern processors aren't easy, and there could always be problems. But I think IBM is one of the best out there, and I'm fairly confident in them working problems out in a far more "polished" manner than motorola. And they already have made the Power5 which make them almost experts on this processor already. That has to mean a lot when they are making the 980. Well, just had to say a few words here.
Originally posted by RBR
The 970+ is an add-on to the product life cycle as near as I can make out as there were no plans originally to take the 970 to the 90 nm process.
All of the IBM literature I've read has talked about a debut at 130nm, with a move to 90nm "shortly". 90nm and smaller lithography is what Fishkill is really set up for, and IBM has consistently voiced an intention to move to 90nm as quickly as they can. Fishkill cost a mint, and it's gotta pay for itself.
The die-shrunk 970 should arrive some time before we see the 980, and it'll be a great PowerBook/iMac CPU.
I tend to favor the good news scenarios as IBM seems to be on a roll, but there is always the possibility that they are simply having problems.
Or, their schedule was always 970 @ 130nm, 970+ @ 90nm, 980 at 90nm, etc. That's the impression I've had since last fall from the available documentation.
Originally posted by Amorph
Or, their schedule was always 970 @ 130nm, 970+ @ 90nm, 980 at 90nm, etc. That's the impression I've had since last fall from the available documentation.
Hmm... for some reason this "schedule" is imprinted into my brain
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I know that I saw a PDF (these things take up drive space, but I really have to start saving them) that had the 980 being introduced with a 130 nm process, but it may be that that plan was scrapped and I did not see the revised plan to introduce the 980 at the 90 nm process. Still, I think I would have heard something as this would be a pretty important change in their development roadmap.
Uncertainties in the construction and outfitting of the production facility could have been the holdup that was being covered with the (contingency plan?) roadmap to intro the 980 at 130 nm. (Few people complain when a project comes in early.)
<http://www.macworld.co.uk/news/top_n...fm?NewsID=6332>