I found your post to be very instructive, especially about the L3 cache and PowerTune. However, this one sentence makes no sense to me, and I am not sure what you are trying to say?
Are you confused about electrical network analysis? Each transistor has a resistance associated with it, but when you add more transistors to a processor its resistance does not go up. It goes down! The transistors are added in "parallel" with the supply voltage, not in series. What produces higher power dissipation when you add transistors is the increased current due to those added transistors.
saying that the total current running through resistors increased is as pointless as saying that the total number of resistors having current being run through increased. The reason that there is more switching current is only because there is more capacitances to charge and discharge. it is an effect, not a cause of the higher power dissipation.
. . . I dont think you really understand this stuff. go read a book.
I'm puzzled why you would be having such a problem understanding my posts if you truly do understand the topic? Now you bring up my reply to the fellow who wrote, quote:
"Transistor count is very important as the more transistors the greater the resistance which means that you have to run the chip at a higher voltage."
He appears to believe that adding more transistor gates will increase the resistance. (You just mentioned the "equivalent" resistance for each transistor yourself. The concept is not that outlandish.) Clearly he doesn't understand that these devices are in parallel across the supply, and adding more would make the "equivalent" resistance of the chip less, not greater. So then, I responded as much as possible on his own level and using his imagery and terms. Now, if you really know your stuff, you should have no problem with that. I was getting at the basics of networks, not addressing insignificant (for the issue at hand) details.
Yet you were able to brilliantly tear my discussion apart and point out innumerable errors in my logic and deficiencies in my knowledge. I must bow to your superior intellect. I'm a slow learner, but it's obvious to me now that no one in these forums can possibly match your depth of knowledge and understanding of any topic you care to address. Au Revoir, and don't forget to take your own advice.
So now that the PPC chips in Macs have come up to or in some cases surpassed PC performance, you are still complaining???
Now that we have power AND design, and rock solid stability, and some of the best intergration of any computer, you are still complaining???
Apple makes computers that work. that run forever, that look great, that intergrate well with other hardware, that are solid and stable.
Now that they ONLY match PC's speed (if you don't count crashing and restarting or application start up) you are still complaining???
If you want a 3 ghz processor go get a PC.
If you want a Mac, you have a great selection to choose from. Do you need cheap? emac. do you need iconinc and good? iMac. do you need the best portable ever designed? powerbook. Do you need the most power for number crunching? PowerMac g5.
There will be a 3ghz g5. So you can either shut and wait for a mac with 3ghz, get a dual 2ghz, or get a PC with a 3ghz in it today.
I'm puzzled why you would be having such a problem understanding my posts if you truly do understand the topic? Now you bring up my reply to the fellow who wrote, quote:
"Transistor count is very important as the more transistors the greater the resistance which means that you have to run the chip at a higher voltage."
He appears to believe that adding more transistor gates will increase the resistance. (You just mentioned the "equivalent" resistance for each transistor yourself. The concept is not that outlandish.) Clearly he doesn't understand that these devices are in parallel across the supply, and adding more would make the "equivalent" resistance of the chip less, not greater. So then, I responded as much as possible on his own level and using his imagery and terms. Now, if you really know your stuff, you should have no problem with that. I was getting at the basics of networks, not addressing insignificant (for the issue at hand) details.
Yet you were able to brilliantly tear my discussion apart and point out innumerable errors in my logic and deficiencies in my knowledge. I must bow to your superior intellect. I'm a slow learner, but it's obvious to me now that no one in these forums can possibly match your depth of knowledge and understanding of any topic you care to address. Au Revoir, and don't forget to take your own advice.
it is pointless to talk about the equivalent resistance of a chip through the parallel devices anyways because no substantial current is flowing directly from Vdd to ground. If you kno your stuff you wont talk about the equivalent resistance that way. you should treat each gate separately. within a gate, if you have large fan in for example in a nand gate, your pull down resistance will increase a lot.
the reason im talking like this is because you were nitpicking at random errors from stingerman; errors that werre not relevant to the discussion at all. whats worse is that you reply him by saying somethign also totally strange.
If IBM has made 2.6Ghz chips, then couldn't they put them into powerbooks, by down clocking them to say 1.6-1.8Ghz, wouldn't that probably reduce the power needed and heat generated?
I know it isn't as simple as this but it's a strategy that is used, isn't it?
I don't get it. Are you trying to say that the number of transistors in a chip have nothing to do with how much power is used? You mean it doesn't take current to charge a capacitor? There is always the resistance of the path that charges the capacitor, the more the capacitors, the more current paths and therefore the more current needed. More current implies more voltage given the same path resistances...or reduce the path resistances (90nm chip) and you reduce the voltage needed (or current) to power the chip (except for the fact that wires closer together have their own capacitance and there is additonal leakage at smaller chip process sizes)...
I don't get it. Are you trying to say that the number of transistors in a chip have nothing to do with how much power is used? You mean it doesn't take current to charge a capacitor? There is always the resistance of the path that charges the capacitor, the more the capacitors, the more current paths and therefore the more current needed. More current implies more voltage given the same path resistances...or reduce the path resistances (90nm chip) and you reduce the voltage needed (or current) to power the chip (except for the fact that wires closer together have their own capacitance and there is additonal leakage at smaller chip process sizes)...
of course # of transistors will increase power used. it'll just increase the capacitances. the point is that you dont quantify the power used by the chip by analyzing the total current running through the total path resistance. That is very hard to analyze. The current doesnt even flow through a path. It either flows from Vdd to the separate capacitances or from the separate capacitances to ground. There are many different paths, with lots of different currents and different directions of flow. It is much more convenient to just look at the capacitances and voltages. without doubt, the total current required by the chip will increase, but that is a result of the power consumption. you dont analyze the power consumption by saying that the current has increased. you analye it by saying that since total switching capacitance increase, you require more current. for power consumption, you dont give about resistances either. they are only important iin speed.
90nm chip does not reduce device eq resistance since the W/L ratio is still the same. what it does is reduce capacitances.(and also wire resistance, as wires in smaller chips tend to be taller). interwire crosstalk capcitance is increased, but interlayer wire capacitances are reduced.
I would imagine if apple took a 90nm 2.6Ghz G5 and scaled the speed down to 2Ghz they would not have any trouble putting it in a powerbook. I don't see any reason why G5 PowerBooks should not be released for the next upgrade. Heck if they needed they could scale the chip down lower. the next powerbooks could be 1.4, 1.6 and 1.8...
Anyway, I am very much interested in buying a PowerBook G5 when they come out. I have a 1Ghz Tibook right now so no big rush, but I am still looking forward to getting one.
In a nut shell what I think Apple did was to make alow noise computer instead of a high performance computer. Every indication is there that they could have introduced even faster machines but didn't. Thus my indications that the G5 is the result of twisted marketing practices. Apple delievered just enough computer to allow them to make some claims that would cause continuous conversation in the trades.
There is all sorts of information floating around that could be considered supporting evidence. The least of which is the public announcement that they would be at 3GHz in a year. For a CEO of ay publicly traded company to say such things, he must have had very solid evidence that this was possible whitout the risk of new technology.
The issue then becomes in my mind is this: is Apple in competition with the rest of the PC world or are they just manageing the publicity that comes from their claims? It is begining to look more and more like the G5 is a marketing effort more than a high performance machine it could be. Once at 90nm what is really keeping this processor from running at greater than 3GHz? If we see anything less than 3GHz in January then it will be the result of Apple trying to control the growth of the line instead of offering us all they can offer performance wise.
Obviously this is my perspective, but I think the market has become so use to no growth in the G4 series that they have little in the way of expectations for the 970 machines. Low expectations along with people ignoring the realities of information floating around makes it easy for Apple to keep the G5 slower than they could be.
Thanks
Dave
Quote:
Originally posted by Eric_Z
Well it's not just the info from the German pdf that tells you this, it's allso the fact that IBM representetives went over and met Chinese goverment officials and claimed that they could produce 2.5 Ghz 970 processors, on the 130nm process.
These processors, however, have got lower yields then the 2Ghz and they produce close to 100 Watts of heat. Apple, me thinks, did not choose the 2.5 because they thought that the 2.0 was/is good enough. Good enough to substantiate the PR claims about the Power Mac being the fastest personal computer in the world, and this without having to deal with the problems that the 2.5 poses in terms of heat and low yields.
While I can agree that Apple needs high volumn there is ample evidence that the only reason we are at 2GHz is for other design issues. IBM's ability to provide the 2 GHz processors even when demand became an issue kinda points out that there is little in the way of yield issues at this speed. We are talking 5 to 8 hundred MHz up side potential here with the current 970, that is an awful lot of margin.
I know ths sounds cynical but I don't see the 2GHz G5 as being a high performance effort on Apples part. More of a low noise machine than anything else.
Thanks
dave
Quote:
Originally posted by Programmer
Apple sells units in the hundreds of thousands. They simply could not meet the demand if they went with the lower yield speeds. They will always skip the leading edge speeds available because they need serious volume (unless they come out with a high end workstation and price it accordingly).
of course # of transistors will increase power used. it'll just increase the capacitances. the point is that you dont quantify the power used by the chip by analyzing the total current running through the total path resistance. That is very hard to analyze. The current doesnt even flow through a path. It either flows from Vdd to the separate capacitances or from the separate capacitances to ground. There are many different paths, with lots of different currents and different directions of flow. It is much more convenient to just look at the capacitances and voltages. without doubt, the total current required by the chip will increase, but that is a result of the power consumption. you dont analyze the power consumption by saying that the current has increased. you analye it by saying that since total switching capacitance increase, you require more current. for power consumption, you dont give about resistances either. they are only important iin speed.
90nm chip does not reduce device eq resistance since the W/L ratio is still the same. what it does is reduce capacitances.(and also wire resistance, as wires in smaller chips tend to be taller). interwire crosstalk capcitance is increased, but interlayer wire capacitances are reduced.
Nr9, I'm afraid you're making a lot of typical undergraduate mistakes in understanding a very complex process. The oversimplification is especially invalid at these very short gate lengths. You also seem to be deliberately misunderstanding other posters points.
Nonetheless it is axiomatic that all power dissipation is resistive, how you analyse it depends on how much you know about the circuit and how it operates, but that is a different matter.
Nr9, I'm afraid you're making a lot of typical undergraduate mistakes in understanding a very complex process. The oversimplification is especially invalid at these very short gate lengths. You also seem to be deliberately misunderstanding other posters points.
Nonetheless it is axiomatic that all power dissipation is resistive, how you analyse it depends on how much you know about the circuit and how it operates, but that is a different matter.
michael
eh.. short channel effects do not change the fact that you determine quantitatively the power dissipation in a CMOS device by the capacitance. to the first order, its always analyzed by the capacitances.
the first order effects are always the most important.
In a nut shell what I think Apple did was to make alow noise computer instead of a high performance computer. Every indication is there that they could have introduced even faster machines but didn't. Thus my indications that the G5 is the result of twisted marketing practices. Apple delievered just enough computer to allow them to make some claims that would cause continuous conversation in the trades.
Or it could be that the 2.5GHz chips weren't available in sufficient quantity, or that they required 100W each and an even less space-efficient design, or they were too expensive, or any combination of these things.
Remember, we were only promised 1.8GHz from IBM. Apple overdelivered. (Now, of course, IBM is promising 3GHz. Draw your own conclusions. )
Quote:
The issue then becomes in my mind is this: is Apple in competition with the rest of the PC world or are they just manageing the publicity that comes from their claims? It is begining to look more and more like the G5 is a marketing effort more than a high performance machine it could be. Once at 90nm what is really keeping this processor from running at greater than 3GHz? If we see anything less than 3GHz in January then it will be the result of Apple trying to control the growth of the line instead of offering us all they can offer performance wise.
I don't believe that any claim that the G5 is a "marketing effort" stands up to an even cursory analysis of the machine itself. Apple did not pull off the near-miraculous Elastic Bus for marketing reasons. They don't have dual channel DDR400 for marketing reasons, nor do they have world-class motherboard bandwidth for marketing reasons. The G5 is designed around leading-edge technology to go incredibly fast.
That notwithstanding, you have this idea that all PowerMac customers want as much power as possible regardless, and that's simply not true. Many things are required of a professional machine. Apple had just had its ass handed to it by its professional customers for releasing a noisy machine - audio professionals couldn't use them for what I hope are obvious reasons, and graphics pros weren't enthusiastic about working in the vicinity of a huge roaring distraction.
So: Apple shipped a machine with faster processors than IBM had promised, built for speed, and built for quiet. If you want to conclude that Apple is holding back, you're welcome to, but I see absolutely nothing to support that conclusion. They do have to take other things into account than absolute CPU speed (such as whether those chips are available by the hundreds of thousands, whether they're so hot as to force a redesign), but given that they've exceeded any published expectations.
We'll see 3GHz when IBM produces 3GHz CPUs in quantity, not sooner. I think it's pretty clear from the G5 introduction keynote that Steve loves blowing the competition away, and I don't see how he'll turn down an opportunity to remove as much ambiguity around that claim as possible. If the hardcore PC performance writers extolled the top-end PowerMac as the machine to beat, period, Apple would gain credibility they haven't had in about a decade.
I don't believe that any claim that the G5 is a "marketing effort" stands up to an even cursory analysis of the machine itself. Apple did not pull off the near-miraculous Elastic Bus for marketing reasons. They don't have dual channel DDR400 for marketing reasons, nor do they have world-class motherboard bandwidth for marketing reasons. The G5 is designed around leading-edge technology to go incredibly fast.
. . . Transistor count is very important as the more transistors the greater the resistance which means that you have to run the chip at a higher voltage. . .
stingerman,
I owe you a sincere public apology for the way you have been dragged into a futile discussion. I have always appreciate your postings and have learned something on many occasions. You seem to keep well informed about computers, which is my interest in reading these forums. So, after you made the above statement, I realize you know a lot, but electrical network analysis is not your forte. It appears you had a basic misunderstanding when you wrote that post, and you may like to know about it. I certainly would if it were me.
Unfortunately, your post became part of a debate here, so I think it is only fair for me to give you a much more complete answer than my original reply. You appear to simplify the processor as a "black box" to facilitate discussion. I find that to be a typical way engineers get cumbersome details out of the way so they can focus on simple issues. Otherwise, we are caught up in technicalities that don't get to the heart of the matter.
In this case, the black box is a collection of transistors, represented by connected resistors. It obeys a simple law, E = I x R, voltage is current times resistance. You appear to believe that adding transistors would increase resistance (R) and therefore the processor voltage (E) would need to be higher to supply the required current (I). This would be true if the resistors were in series, or connected in a daisy chain fashion. In fact, they are connected in parallel, each one going across the power supply terminals. In this case, total resistance is not the sum of the individual resistors.
A resistor has another characteristic that is called conductance, represented by G. It is equal to the reciprocal of resistance, G = 1 / R. In the case of parallel connect components, the total conductance (G) is equal to the sum of the individual elements. When you go through the math, you discover that with parallel connected components the resistance goes down when you add more parts. E = I x R still holds. So when R is less, and the supply voltage for the chip (E) stays the same, the current (I) increases, which increases power. Power = E x I.
I hope this is helpful. It is my interpretation of what you posted, although you may mean something entirely different that I don't see.
I owe you a sincere public apology for the way you have been dragged into a futile discussion. I have always appreciate your postings and have learned something on many occasions. You seem to keep well informed about computers, which is my interest in reading these forums. So, after you made the above statement, I realize you know a lot, but electrical network analysis is not your forte. It appears you had a basic misunderstanding when you wrote that post, and you may like to know about it. I certainly would if it were me.
After reading this board for a long time, I must take a moment to thank you for your decorum and respectful post. I don't know if it is just the holiday season or what... but I, for one, appreciate your kind and tactful approach to expressing your views.
"Kind words can be short and easy to speak, but their echoes are truly
Comments
Originally posted by snoopy
I found your post to be very instructive, especially about the L3 cache and PowerTune. However, this one sentence makes no sense to me, and I am not sure what you are trying to say?
Are you confused about electrical network analysis? Each transistor has a resistance associated with it, but when you add more transistors to a processor its resistance does not go up. It goes down! The transistors are added in "parallel" with the supply voltage, not in series. What produces higher power dissipation when you add transistors is the increased current due to those added transistors.
saying that the total current running through resistors increased is as pointless as saying that the total number of resistors having current being run through increased. The reason that there is more switching current is only because there is more capacitances to charge and discharge. it is an effect, not a cause of the higher power dissipation.
Originally posted by Nr9
. . . I dont think you really understand this stuff. go read a book.
I'm puzzled why you would be having such a problem understanding my posts if you truly do understand the topic? Now you bring up my reply to the fellow who wrote, quote:
"Transistor count is very important as the more transistors the greater the resistance which means that you have to run the chip at a higher voltage."
He appears to believe that adding more transistor gates will increase the resistance. (You just mentioned the "equivalent" resistance for each transistor yourself. The concept is not that outlandish.) Clearly he doesn't understand that these devices are in parallel across the supply, and adding more would make the "equivalent" resistance of the chip less, not greater. So then, I responded as much as possible on his own level and using his imagery and terms. Now, if you really know your stuff, you should have no problem with that. I was getting at the basics of networks, not addressing insignificant (for the issue at hand) details.
Yet you were able to brilliantly tear my discussion apart and point out innumerable errors in my logic and deficiencies in my knowledge. I must bow to your superior intellect. I'm a slow learner, but it's obvious to me now that no one in these forums can possibly match your depth of knowledge and understanding of any topic you care to address. Au Revoir, and don't forget to take your own advice.
start of round three...
why can't we all just be friends?
Nr9, you're too aggressive - chill out ...
now FIGHT!
Now that we have power AND design, and rock solid stability, and some of the best intergration of any computer, you are still complaining???
Apple makes computers that work. that run forever, that look great, that intergrate well with other hardware, that are solid and stable.
Now that they ONLY match PC's speed (if you don't count crashing and restarting or application start up) you are still complaining???
If you want a 3 ghz processor go get a PC.
If you want a Mac, you have a great selection to choose from. Do you need cheap? emac. do you need iconinc and good? iMac. do you need the best portable ever designed? powerbook. Do you need the most power for number crunching? PowerMac g5.
There will be a 3ghz g5. So you can either shut and wait for a mac with 3ghz, get a dual 2ghz, or get a PC with a 3ghz in it today.
Originally posted by snoopy
I'm puzzled why you would be having such a problem understanding my posts if you truly do understand the topic? Now you bring up my reply to the fellow who wrote, quote:
"Transistor count is very important as the more transistors the greater the resistance which means that you have to run the chip at a higher voltage."
He appears to believe that adding more transistor gates will increase the resistance. (You just mentioned the "equivalent" resistance for each transistor yourself. The concept is not that outlandish.) Clearly he doesn't understand that these devices are in parallel across the supply, and adding more would make the "equivalent" resistance of the chip less, not greater. So then, I responded as much as possible on his own level and using his imagery and terms. Now, if you really know your stuff, you should have no problem with that. I was getting at the basics of networks, not addressing insignificant (for the issue at hand) details.
Yet you were able to brilliantly tear my discussion apart and point out innumerable errors in my logic and deficiencies in my knowledge. I must bow to your superior intellect. I'm a slow learner, but it's obvious to me now that no one in these forums can possibly match your depth of knowledge and understanding of any topic you care to address. Au Revoir, and don't forget to take your own advice.
it is pointless to talk about the equivalent resistance of a chip through the parallel devices anyways because no substantial current is flowing directly from Vdd to ground. If you kno your stuff you wont talk about the equivalent resistance that way. you should treat each gate separately. within a gate, if you have large fan in for example in a nand gate, your pull down resistance will increase a lot.
the reason im talking like this is because you were nitpicking at random errors from stingerman; errors that werre not relevant to the discussion at all. whats worse is that you reply him by saying somethign also totally strange.
I know it isn't as simple as this but it's a strategy that is used, isn't it?
Originally posted by Bigc
I don't get it. Are you trying to say that the number of transistors in a chip have nothing to do with how much power is used? You mean it doesn't take current to charge a capacitor? There is always the resistance of the path that charges the capacitor, the more the capacitors, the more current paths and therefore the more current needed. More current implies more voltage given the same path resistances...or reduce the path resistances (90nm chip) and you reduce the voltage needed (or current) to power the chip (except for the fact that wires closer together have their own capacitance and there is additonal leakage at smaller chip process sizes)...
of course # of transistors will increase power used. it'll just increase the capacitances. the point is that you dont quantify the power used by the chip by analyzing the total current running through the total path resistance. That is very hard to analyze. The current doesnt even flow through a path. It either flows from Vdd to the separate capacitances or from the separate capacitances to ground. There are many different paths, with lots of different currents and different directions of flow. It is much more convenient to just look at the capacitances and voltages. without doubt, the total current required by the chip will increase, but that is a result of the power consumption. you dont analyze the power consumption by saying that the current has increased. you analye it by saying that since total switching capacitance increase, you require more current. for power consumption, you dont give about resistances either. they are only important iin speed.
90nm chip does not reduce device eq resistance since the W/L ratio is still the same. what it does is reduce capacitances.(and also wire resistance, as wires in smaller chips tend to be taller). interwire crosstalk capcitance is increased, but interlayer wire capacitances are reduced.
Anyway, I am very much interested in buying a PowerBook G5 when they come out. I have a 1Ghz Tibook right now so no big rush, but I am still looking forward to getting one.
There is all sorts of information floating around that could be considered supporting evidence. The least of which is the public announcement that they would be at 3GHz in a year. For a CEO of ay publicly traded company to say such things, he must have had very solid evidence that this was possible whitout the risk of new technology.
The issue then becomes in my mind is this: is Apple in competition with the rest of the PC world or are they just manageing the publicity that comes from their claims? It is begining to look more and more like the G5 is a marketing effort more than a high performance machine it could be. Once at 90nm what is really keeping this processor from running at greater than 3GHz? If we see anything less than 3GHz in January then it will be the result of Apple trying to control the growth of the line instead of offering us all they can offer performance wise.
Obviously this is my perspective, but I think the market has become so use to no growth in the G4 series that they have little in the way of expectations for the 970 machines. Low expectations along with people ignoring the realities of information floating around makes it easy for Apple to keep the G5 slower than they could be.
Thanks
Dave
Originally posted by Eric_Z
Well it's not just the info from the German pdf that tells you this, it's allso the fact that IBM representetives went over and met Chinese goverment officials and claimed that they could produce 2.5 Ghz 970 processors, on the 130nm process.
These processors, however, have got lower yields then the 2Ghz and they produce close to 100 Watts of heat. Apple, me thinks, did not choose the 2.5 because they thought that the 2.0 was/is good enough. Good enough to substantiate the PR claims about the Power Mac being the fastest personal computer in the world, and this without having to deal with the problems that the 2.5 poses in terms of heat and low yields.
I know ths sounds cynical but I don't see the 2GHz G5 as being a high performance effort on Apples part. More of a low noise machine than anything else.
Thanks
dave
Originally posted by Programmer
Apple sells units in the hundreds of thousands. They simply could not meet the demand if they went with the lower yield speeds. They will always skip the leading edge speeds available because they need serious volume (unless they come out with a high end workstation and price it accordingly).
Originally posted by Nr9
of course # of transistors will increase power used. it'll just increase the capacitances. the point is that you dont quantify the power used by the chip by analyzing the total current running through the total path resistance. That is very hard to analyze. The current doesnt even flow through a path. It either flows from Vdd to the separate capacitances or from the separate capacitances to ground. There are many different paths, with lots of different currents and different directions of flow. It is much more convenient to just look at the capacitances and voltages. without doubt, the total current required by the chip will increase, but that is a result of the power consumption. you dont analyze the power consumption by saying that the current has increased. you analye it by saying that since total switching capacitance increase, you require more current. for power consumption, you dont give about resistances either. they are only important iin speed.
90nm chip does not reduce device eq resistance since the W/L ratio is still the same. what it does is reduce capacitances.(and also wire resistance, as wires in smaller chips tend to be taller). interwire crosstalk capcitance is increased, but interlayer wire capacitances are reduced.
Nr9, I'm afraid you're making a lot of typical undergraduate mistakes in understanding a very complex process. The oversimplification is especially invalid at these very short gate lengths. You also seem to be deliberately misunderstanding other posters points.
Nonetheless it is axiomatic that all power dissipation is resistive, how you analyse it depends on how much you know about the circuit and how it operates, but that is a different matter.
michael
Originally posted by mmicist
Nr9, I'm afraid you're making a lot of typical undergraduate mistakes in understanding a very complex process. The oversimplification is especially invalid at these very short gate lengths. You also seem to be deliberately misunderstanding other posters points.
Nonetheless it is axiomatic that all power dissipation is resistive, how you analyse it depends on how much you know about the circuit and how it operates, but that is a different matter.
michael
eh.. short channel effects do not change the fact that you determine quantitatively the power dissipation in a CMOS device by the capacitance. to the first order, its always analyzed by the capacitances.
the first order effects are always the most important.
Originally posted by wizard69
In a nut shell what I think Apple did was to make alow noise computer instead of a high performance computer. Every indication is there that they could have introduced even faster machines but didn't. Thus my indications that the G5 is the result of twisted marketing practices. Apple delievered just enough computer to allow them to make some claims that would cause continuous conversation in the trades.
Or it could be that the 2.5GHz chips weren't available in sufficient quantity, or that they required 100W each and an even less space-efficient design, or they were too expensive, or any combination of these things.
Remember, we were only promised 1.8GHz from IBM. Apple overdelivered. (Now, of course, IBM is promising 3GHz. Draw your own conclusions.
The issue then becomes in my mind is this: is Apple in competition with the rest of the PC world or are they just manageing the publicity that comes from their claims? It is begining to look more and more like the G5 is a marketing effort more than a high performance machine it could be. Once at 90nm what is really keeping this processor from running at greater than 3GHz? If we see anything less than 3GHz in January then it will be the result of Apple trying to control the growth of the line instead of offering us all they can offer performance wise.
I don't believe that any claim that the G5 is a "marketing effort" stands up to an even cursory analysis of the machine itself. Apple did not pull off the near-miraculous Elastic Bus for marketing reasons. They don't have dual channel DDR400 for marketing reasons, nor do they have world-class motherboard bandwidth for marketing reasons. The G5 is designed around leading-edge technology to go incredibly fast.
That notwithstanding, you have this idea that all PowerMac customers want as much power as possible regardless, and that's simply not true. Many things are required of a professional machine. Apple had just had its ass handed to it by its professional customers for releasing a noisy machine - audio professionals couldn't use them for what I hope are obvious reasons, and graphics pros weren't enthusiastic about working in the vicinity of a huge roaring distraction.
So: Apple shipped a machine with faster processors than IBM had promised, built for speed, and built for quiet. If you want to conclude that Apple is holding back, you're welcome to, but I see absolutely nothing to support that conclusion. They do have to take other things into account than absolute CPU speed (such as whether those chips are available by the hundreds of thousands, whether they're so hot as to force a redesign), but given that they've exceeded any published expectations.
We'll see 3GHz when IBM produces 3GHz CPUs in quantity, not sooner. I think it's pretty clear from the G5 introduction keynote that Steve loves blowing the competition away, and I don't see how he'll turn down an opportunity to remove as much ambiguity around that claim as possible. If the hardcore PC performance writers extolled the top-end PowerMac as the machine to beat, period, Apple would gain credibility they haven't had in about a decade.
Originally posted by Amorph
I don't believe that any claim that the G5 is a "marketing effort" stands up to an even cursory analysis of the machine itself. Apple did not pull off the near-miraculous Elastic Bus for marketing reasons. They don't have dual channel DDR400 for marketing reasons, nor do they have world-class motherboard bandwidth for marketing reasons. The G5 is designed around leading-edge technology to go incredibly fast.
And we all know how bad that is for marketing.
Originally posted by stingerman
. . . Transistor count is very important as the more transistors the greater the resistance which means that you have to run the chip at a higher voltage. . .
stingerman,
I owe you a sincere public apology for the way you have been dragged into a futile discussion. I have always appreciate your postings and have learned something on many occasions. You seem to keep well informed about computers, which is my interest in reading these forums. So, after you made the above statement, I realize you know a lot, but electrical network analysis is not your forte. It appears you had a basic misunderstanding when you wrote that post, and you may like to know about it. I certainly would if it were me.
Unfortunately, your post became part of a debate here, so I think it is only fair for me to give you a much more complete answer than my original reply. You appear to simplify the processor as a "black box" to facilitate discussion. I find that to be a typical way engineers get cumbersome details out of the way so they can focus on simple issues. Otherwise, we are caught up in technicalities that don't get to the heart of the matter.
In this case, the black box is a collection of transistors, represented by connected resistors. It obeys a simple law, E = I x R, voltage is current times resistance. You appear to believe that adding transistors would increase resistance (R) and therefore the processor voltage (E) would need to be higher to supply the required current (I). This would be true if the resistors were in series, or connected in a daisy chain fashion. In fact, they are connected in parallel, each one going across the power supply terminals. In this case, total resistance is not the sum of the individual resistors.
A resistor has another characteristic that is called conductance, represented by G. It is equal to the reciprocal of resistance, G = 1 / R. In the case of parallel connect components, the total conductance (G) is equal to the sum of the individual elements. When you go through the math, you discover that with parallel connected components the resistance goes down when you add more parts. E = I x R still holds. So when R is less, and the supply voltage for the chip (E) stays the same, the current (I) increases, which increases power. Power = E x I.
I hope this is helpful. It is my interpretation of what you posted, although you may mean something entirely different that I don't see.
dragging a thread to shit. and stingerman doesnt even care.
Originally posted by snoopy
stingerman,
I owe you a sincere public apology for the way you have been dragged into a futile discussion. I have always appreciate your postings and have learned something on many occasions. You seem to keep well informed about computers, which is my interest in reading these forums. So, after you made the above statement, I realize you know a lot, but electrical network analysis is not your forte. It appears you had a basic misunderstanding when you wrote that post, and you may like to know about it. I certainly would if it were me.
After reading this board for a long time, I must take a moment to thank you for your decorum and respectful post. I don't know if it is just the holiday season or what... but I, for one, appreciate your kind and tactful approach to expressing your views.
"Kind words can be short and easy to speak, but their echoes are truly
endless. - Mother Theresa"
Kudos to you. 8)