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MacBidouille posts PPC 970 benchmarks - Page 9

post #321 of 666
Quote:
Originally posted by snoopy
power = current X voltage

I haven't done any of these electrical calculations since high school, but it always helped me to think of the units used for the different variables. Using units, the above equation would go like this:

Watt = Ampère x Volt

So if we have a processor drawing 19 Watt at 1.1 Volt, it does indeed draw 17.3 Amps. As pointed out above, a household vacuum cleaner would draw 100 times the voltage of a 1.2 Ghz PPC 970 (200 times in Europe). That's why it would draw about a 100 times more Watts than said PPC 970. Does this make more sense?

Escher
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post #322 of 666
Quote:
Originally posted by KidRed

. . . So it looks like 970 around August, so maybe announced at MWNY and shipping immediately or within weeks.


Why? Here is what he said regarding a question about August. "90% likelihood before then. 100% likelihood within 30 days of then." Looks like he says 'before' August is the most likely.
post #323 of 666
Quote:
Originally posted by snoopy
Why? Here is what he said regarding a question about August. "90% likelihood before then. 100% likelihood within 30 days of then." Looks like he says 'before' August is the most likely.

Question is: ¿Is that when you can order them for 6-8 week delivery or get them off-the-shelf?
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post #324 of 666
Quote:
Originally posted by Escher
I haven't done any of these electrical calculations since high school, but it always helped me to think of the units used for the different variables. Using units, the above equation would go like this:

Watt = Ampère x Volt

So if we have a processor drawing 19 Watt at 1.1 Volt, it does indeed draw 17.3 Amps. As pointed out above, a household vacuum cleaner would draw 100 times the voltage of a 1.2 Ghz PPC 970 (200 times in Europe). That's why it would draw about a 100 times more Watts than said PPC 970. Does this make more sense?

Escher

Try Power ~ ½ CV2Af
C= Capacitance
V= Core Voltage
A= Activity Factor
f= Clock Frequency

It's a dynamic switch and not a heater coil.

edit - that's Voltage squared
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post #325 of 666
Quote:
Originally posted by Leonis
Rage 128 with 8MB VRAM

Yeah...in the top end
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post #326 of 666
Quote:
Originally posted by Bigc
Well sounds like were back to the original schedule August-September release and maybe pre-order in July or at WWDC

"back to"? Some of us never left.
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post #327 of 666
Quote:
Originally posted by Bigc
Try Power ~ ½ CV2Af
C= Capacitance
V= Core Voltage
A= Activity Factor
f= Clock Frequency

It's a dynamic switch and not a heater coil.

edit - that's Voltage squared

Current times voltage works for any black box. It does not matter what is inside. This calculation will give you the power dissipation for any equipment powered by DC voltage sources, as in computers. AC voltage sources are a different matter, and we need a power factor term to take into account phase differences.

The formula you give is a special case to estimate power. The one half C V-squared term is the energy stored in a capacitor. Since it is multiplied by a frequency it has a one over time term, so the units look okay. Power is energy divided by time, or rate of energy delivery.

So your formula estimates energy by the size of capacitor being discharged during a switching cycle, and the duty cycle is what you call the activity factor. I guess if one were to add up all the switches in a processor with some educated guess about how often they switch, you might get in the ball park for power. In the end, they get the real numbers from measurements of the processor as a black box. Power = current X voltage.
post #328 of 666
Quote:
Originally posted by snoopy
Current times voltage works for any black box. It does not matter what is inside. This calculation will give you the power dissipation for any equipment powered by DC voltage sources, as in computers. AC voltage sources are a different matter, and we need a power factor term to take into account phase differences.

The formula you give is a special case to estimate power. The one half C V-squared term is the energy stored in a capacitor. Since it is multiplied by a frequency it has a one over time term, so the units look okay. Power is energy divided by time, or rate of energy delivery.

So your formula estimates energy by the size of capacitor being discharged during a switching cycle, and the duty cycle is what you call the activity factor. I guess if one were to add up all the switches in a processor with some educated guess about how often they switch, you might get in the ball park for power. In the end, they get the real numbers from measurements of the processor as a black box. Power = current X voltage.

not for a capacitive-discharge system which has a frequency of oscillation, which is what a microprocessor is. If there is no oscillation then no power. Under dead short conditions then maybe Power = volts*amps since there is no oscillation at that time.

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post #329 of 666
er, you might want to google for "leakage current". A gate will leak some current even when not undergoing changes of state.
post #330 of 666
Quote:
Originally posted by boots
er, you might want to google for "leakage current". A gate will leak some current even when not undergoing changes of state.

well hopefully it's not much or there will be problems.
I heard that geeks are a dime a dozen, I just want to find out who's been passin' out the dimes
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post #331 of 666
Quote:
Originally posted by KidRed
Very interesting thread. I for one believe the Powerjack (MacWhispers guy) to be legit.

My favourite one from the forum:
Quote:
if you are under NDA, why would you tell us you are under NDA, that makes it more suspicious

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post #332 of 666
Quote:
Originally posted by Bigc
not for a capacitive-discharge system which has a frequency of oscillation, which is what a microprocessor is. If there is no oscillation then no power. Under dead short conditions then maybe Power = volts*amps since there is no oscillation at that time. . .


The laws of physics don't change depending on how you design a system. The formula you give, provided your estimates are all correct, and the formula for power based on measurement of voltage and current to the CPU will both give the same answer. The problem with your formula is that you must know the capacitance and duty cycle of each switch. With millions of switches this is a big task, and it is inaccurate because you do not know the values exactly enough.

If you think about it, you will realize that the very things that increase and decrease power in your formula will also increase and decrease current from the power supply. If the CPU is inactive, the switches have low or zero activity factor in your formula, and current from the supply is also low, resulting in low power either way you calculate it. The 17 Amperes at 1.1 Volts is base on IBM's reference to 19 Watts. This is likely running full out, using some standard conditions for measuring maximum power consumption.

How do you think IBM got that 19 Watts? (Or the 42 Watts?) They measured it. They didn't do some estimate based on capacitance and duty cycle. There is absolutely no way to "measure" power dissipation using your formula.
post #333 of 666
Quote:
Originally posted by snoopy
The laws of physics don't change depending on how you design a system. The formula you give, provided your estimates are all correct, and the formula for power based on measurement of voltage and current to the CPU will both give the same answer. The problem with your formula is that you must know the capacitance and duty cycle of each switch. With millions of switches this is a big task, and it is inaccurate because you do not know the values exactly enough.

If you think about it, you will realize that the very things that increase and decrease power in your formula will also increase and decrease current from the power supply. If the CPU is inactive, the switches have low or zero activity factor in your formula, and current from the supply is also low, resulting in low power either way you calculate it. The 17 Amperes at 1.1 Volts is base on IBM's reference to 19 Watts. This is likely running full out, using some standard conditions for measuring maximum power consumption.

How do you think IBM got that 19 Watts? (Or the 42 Watts?) They measured it. They didn't do some estimate based on capacitance and duty cycle. There is absolutely no way to "measure" power dissipation using your formula.

The confusion here seems to be that people are getting the processor must be drawing 17 amperes directly from the plug in the wall. That is not the case. To power the processor alone, the AC plug in the wall must draw 158 mA, assuming 120 V line voltage. Obviously, this is a small fraction of the current required to power the whole system. The line voltage is transformed and rectified to 1.1 V DC to power the processor. As stated above, the 17 A current through the processor is dominated by the switching rate of capacitive elements. It is not the current through the heating element of a heater or light bulb.
post #334 of 666
Anybody that believes Macwishpers is a fool. That guy has lied about products that he sells. He is at best a con artist and at worst just a theif.

The only rumor sites out their that has seems to have any real information are "Think Secret" and "Mac rumors" and they have said nothing about the 970. Nothing. They were right about the iMacs, they were right about the eMacs and they were even right about the 17 inch powerbooks.
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post #335 of 666
Quote:
Originally posted by anand
The only rumor sites out their that has seems to have any real information are "Think Secret" and "Mac rumors" and they have said nothing about the 970. Nothing. They were right about the iMacs, they were right about the eMacs and they were even right about the 17 inch powerbooks.

Oh trust me Nick (Think Secret) is well informed about the 970, that you can BET ON. The big difference that I see with Nick when compared to just about all the other rumors sites is that Nick works hard on the stories he posts and it reflects in his site. He doesn't jump the gun on anything and as such we see one maybe two items each week (if we're lucky). When you see something posted on TS, 99.9% of the time 'the story is a lock' and that's what makes TS special.

In short Nick knows what the words '2nd source' means and he also knows that lots of stuff that he hears that WERE true can change/morph or even die as time goes on.

Case in point:

Anyone remember those iBrary screen shots? Well here's a little nugget... I've been told that they were 'the real deal' (a skunkworks project at Apple) but since then for reasons unknown the project was put back in the basement... I'm sure tons more projects just like that are going on at Apple each and every day and only the best of em will ever see the light of day.

Had Nick jumped the gun done a story on iBrary (and it never came out) it would have made TS look pretty bad even though the project was REAL and his source was on the up and up...

I take my hat off to very few rumor reporters Nick, Matt Rothenberg and the legend NMR. These folks have GREAT contacts and enough insight to hold back till they have something to say that they KNOW will come true. The rest of the rumor sites are a good read and fun but you can't bank on em.

Dave
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post #336 of 666
Quote:
Originally posted by snoopy
Why? Here is what he said regarding a question about August. "90% likelihood before then. 100% likelihood within 30 days of then." Looks like he says 'before' August is the most likely.

And? 'Around August' is 'before August', 'during August' and even 'after August'. Besides, he's only 90% sure it's before August, the within 30 days of August leads us to the original 'around August'. Not sure what your beef is.
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post #337 of 666
Quote:
Originally posted by snoopy
The laws of physics don't change depending on how you design a system. The formula you give, provided your estimates are all correct, and the formula for power based on measurement of voltage and current to the CPU will both give the same answer. The problem with your formula is that you must know the capacitance and duty cycle of each switch. With millions of switches this is a big task, and it is inaccurate because you do not know the values exactly enough.

If you think about it, you will realize that the very things that increase and decrease power in your formula will also increase and decrease current from the power supply. If the CPU is inactive, the switches have low or zero activity factor in your formula, and current from the supply is also low, resulting in low power either way you calculate it. The 17 Amperes at 1.1 Volts is base on IBM's reference to 19 Watts. This is likely running full out, using some standard conditions for measuring maximum power consumption.

How do you think IBM got that 19 Watts? (Or the 42 Watts?) They measured it. They didn't do some estimate based on capacitance and duty cycle. There is absolutely no way to "measure" power dissipation using your formula.


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.
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post #338 of 666
Quote:
Originally posted by anand
Anybody that believes Macwishpers is a fool. That guy has lied about products that he sells. He is at best a con artist and at worst just a theif.

The only rumor sites out their that has seems to have any real information are "Think Secret" and "Mac rumors" and they have said nothing about the 970. Nothing. They were right about the iMacs, they were right about the eMacs and they were even right about the 17 inch powerbooks.

After reading a macintouch page with emails from quite a few people I changed my stance and opinion. He doesn't sound very knowledgeable at least not in the rumor sence. If anything, it may seem he's in the rumor game to get traffic to sell others and potentially his own products.
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post #339 of 666
Regarding that whole Powerjack post on spymac rumor forums, the only info I have come up with has been this one,
URL here:

ECS reportedly lands Apple 12-inch PowerBook orders Compiled from outside sources; Chinmei Sung, DigiTimes.com [Tuesday 21 January 2003] Elitegroup Computer Systems (ECS) has landed orders from Apple Computer to contract manufacture the 12-inch PowerBook unveiled early this month at the Macworld Conference and Expo, reported the Chinese-language Economic Daily News. Shipment volume is estimated at 300,000 units a year. The company also produces iBooks for the US vendor. Apple Computer:

Taiwan ODM manufacturers

Quanta Computer
- 15.2 PowerBook, 17 flat-panel iMac

Compal Electronics
- 15.4 PowerBook, 17 PowerBook

ECS
- 12 PowerBook, iBook

Foxconn Electronics*
- iMac/eMac

Source: compiled by DigiTimes, January 2003. *The registered trade name of Hon Hai Precision Industry
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post #340 of 666
Quote:
Originally posted by KidRed
And? 'Around August' is 'before August', 'during August' and even 'after August'. Besides, he's only 90% sure it's before August, the within 30 days of August leads us to the original 'around August'. Not sure what your beef is.

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.
post #341 of 666
Quote:
Originally posted by KidRed
After reading a macintouch page with emails from quite a few people I changed my stance and opinion. He doesn't sound very knowledgeable at least not in the rumor sence. If anything, it may seem he's in the rumor game to get traffic to sell others and potentially his own products.

Precisely my point.

8)
post #342 of 666
post #343 of 666
Quote:
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.
post #344 of 666
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.
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post #345 of 666
Quote:
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
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post #346 of 666
Quote:
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.
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post #347 of 666
Quote:
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)
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post #348 of 666
Quote:
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...
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post #349 of 666
Quote:
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.
post #350 of 666
Quote:
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
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post #351 of 666
Quote:
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.
post #352 of 666
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.
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post #353 of 666
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."

http://www.macworld.co.uk/news/top_n...fm?NewsID=6332

--
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post #354 of 666
I was to slow, Amorph did already a better answer to RBR than me
post #355 of 666
Quote:
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.
post #356 of 666
Quote:
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
post #357 of 666
Quote:
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.
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.
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post #358 of 666
Quote:
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.

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.
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post #359 of 666
Quote:
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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post #360 of 666
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]
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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.)
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