It just means a dual channel memory system, similar to what the Tsunami series machines had (and IIRC some of the 680x0 machines in the distant past). Typically DIMMs have to be installed in pairs, although more advanced memory controllers can deal with non-pair DIMMs you just don't get full performance.
A dual channel memory system typically divides memory across two banks of DIMMs in some way such as all the even bytes in one and all the odd bytes in another. Since most (all?) reads & writes are of far more than 1 byte at a time this nets you double the bandwidth. A DDR400 memory system with two channels should approach 6.4 GB/sec of bandwidth.
"At 1.8GHz, the PowerPC 970 will consume 1.3-volts and dissipate 42-Watts. At 1.2 GHz, the PowerPC 970 will consume 1.1-volts and dissipate only 19-Watts. For comparison, a 1GHz G4 consumes 1.6-volts and dissipates 21.3-Watts."
This statement from MacCentral is poorly worded. A processor does not "consume" the voltage it is operating at. Voltage is like an electrical force, and the power supply keeps it constant. Power dissipation follows the law P = VI, where "I" is the electrical current drawn by the device. At higher clock rates the processor draws more current to do the work, and hence power goes up. To reduce power, therefore, the CPU can be operated at a lower clock rate so it draws less current. At lower clock rates the operating voltage can also be set lower, so power goes down even more. Both V and I are less. At 1.1 Volts and 1.2 GHz, the 970 will draw about 17.27 Amperes and dissipate 19 Watts.
"At 1.8GHz, the PowerPC 970 will consume 1.3-volts and dissipate 42-Watts. At 1.2 GHz, the PowerPC 970 will consume 1.1-volts and dissipate only 19-Watts. For comparison, a 1GHz G4 consumes 1.6-volts and dissipates 21.3-Watts."
This statement from MacCentral is poorly worded. A processor does not "consume" the voltage it is operating at. Voltage is like an electrical force, and the power supply keeps it constant. Power dissipation follows the law P = VI, where "I" is the electrical current drawn by the device. At higher clock rates the processor draws more current to do the work, and hence power goes up. To reduce power, therefore, the CPU can be operated at a lower clock rate so it draws less current. At lower clock rates the operating voltage can also be set lower, so power goes down even more. Both V and I are less. At 1.1 Volts and 1.2 GHz, the 970 will draw about 17.27 Amperes and dissipate 19 Watts.
Could you run that by me again? It was my understanding that there would be no math.
17.27 amps seems high to me, just asking. I'm mean that will drive a pretty hefty sized electric motor in our plant. Color me confused.
That's just what the equations show. It's worse at 1.8 GHz and 1.3 Volts, where it draws 32.3 Amperes and dissipates 42 Watts. It is surprising to me too, as I never did the math til now. They will need large gauge wire from the power supply for that, like number 10 or bigger. I haven't verified the 1.3 Volts from another source, but if it is incorrect the current will be a different value. I know the 42 Watts is correct.
The big difference between motors and CPUs is the operating voltage. At 120 Volts and 17 Amperes for example, the power would be about 2000 Watts. On AC sources there is something called power factor, which makes the math more messy. Power is not exactly current times voltage.
Um, the typical full-sized household vacuum cleaner draws 11 amps. A 32-amp laptop would probably have about 30 seconds of battery life before it melted through whatever it was sitting on.
'Course, having a combination laptop and arc welder would be something completely different. Let's see Dell compete with that!
(Edited to get *my* units correct. Jeez, now everyone's doing it!)
Um, the typical full-sized household vacuum cleaner draws 11 amps. A 32-amp laptop would probably have about 30 seconds of battery life before it melted through whatever it was sitting on. . .
The laptop would be 17 Amperes, and 19 Watts. Your 11 Ampere vacuum cleaner is about 1300 Watts. It is the Watts, the power dissipation, that determines the amount of heat generated, not the current. High current requires good size conductors however. If a 1.1 Volt DC source supplies 1 Ampere to the load, that is only 1.1 Watts. The 970 is said to dissipate 19 Watts at 1.1 Volts.
That's just what the equations show. It's worse at 1.8 GHz and 1.3 Volts, where it draws 32.3 Amperes and dissipates 42 Watts. It is surprising to me too, as I never did the math til now. They will need large gauge wire from the power supply for that, like number 10 or bigger. I haven't verified the 1.3 Volts from another source, but if it is incorrect the current will be a different value. I know the 42 Watts is correct.
Huh?
The current G4's electrical specs call for 6.4A max current. You're trying to tell us that the PPC970 will more than triple the maximum current (at the high end)? I don't think so. I don't know what formula you're using, but I doubt your figures.
The current G4's electrical specs call for 6.4A max current. You're trying to tell us that the PPC970 will more than triple the maximum current (at the high end)? I don't think so. I don't know what formula you're using, but I doubt your figures.
The G4 CPU at 1 GHz actually draws 13.3 Amperes, since it is said to operate at 1.6 Volts and dissipates 21.3 Watts. The 6.4 Amperes Maximum on the back of your Mac refers to the most line current it will draw from a 120 VAC power source. That would seem like the Mac dissipates over 700 Watts but it really does not. The max amps on a product is for UL listing. First, it draws the highest current only at low line voltage conditions. Second, there is a power factor, which means the actual power is less than the number you get by multiplying RMS current by the line voltage. Third, the product makers almost always state the number a little high. The number of amps is meant to guide people on what load there will be on the building's electrical circuits.
can anybody provide a link ( for newbies ) for this electric things
don´t get me wrong, I am a newbie !!!!!
faxe
I don't have a link, but what we are talking about is how to calculate power, using the formula:
power = current X voltage
If you look at that formula you will notice something. If the voltage is low, it takes more current for the same power you would get at a higher voltage. If you compare the currents of household devices, which operate at 120 V, to currents of automotive devices, which operate at 12 V, you will see a big difference. Now, when we get down to something like 1.3 V or 1.1 V in the 970 CPU, the difference is even bigger.
Very interesting thread. I for one believe the Powerjack (MacWhispers guy) to be legit. He just sounds too serious to be BS-ing. Besides, he should know if he screws around with us with lies no one will buy his prodcuts to be released later this year.
So it looks like 970 around August, so maybe announced at MWNY and shipping immediately or within weeks.
Very interesting thread. I for one believe the Powerjack (MacWhispers guy) to be legit. He just sounds too serious to be BS-ing. Besides, he should know if he screws around with us with lies no one will buy his prodcuts to be released later this year.
So it looks like 970 around August, so maybe announced at MWNY and shipping immediately or within weeks.
Well sounds like were back to the original schedule August-September release and maybe pre-order in July or at WWDC
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?
Comments
Originally posted by Leonis
Can anyone explain what TwinBank is?
It just means a dual channel memory system, similar to what the Tsunami series machines had (and IIRC some of the 680x0 machines in the distant past). Typically DIMMs have to be installed in pairs, although more advanced memory controllers can deal with non-pair DIMMs you just don't get full performance.
A dual channel memory system typically divides memory across two banks of DIMMs in some way such as all the even bytes in one and all the odd bytes in another. Since most (all?) reads & writes are of far more than 1 byte at a time this nets you double the bandwidth. A DDR400 memory system with two channels should approach 6.4 GB/sec of bandwidth.
This statement from MacCentral is poorly worded. A processor does not "consume" the voltage it is operating at. Voltage is like an electrical force, and the power supply keeps it constant. Power dissipation follows the law P = VI, where "I" is the electrical current drawn by the device. At higher clock rates the processor draws more current to do the work, and hence power goes up. To reduce power, therefore, the CPU can be operated at a lower clock rate so it draws less current. At lower clock rates the operating voltage can also be set lower, so power goes down even more. Both V and I are less. At 1.1 Volts and 1.2 GHz, the 970 will draw about 17.27 Amperes and dissipate 19 Watts.
Originally posted by snoopy
"At 1.8GHz, the PowerPC 970 will consume 1.3-volts and dissipate 42-Watts. At 1.2 GHz, the PowerPC 970 will consume 1.1-volts and dissipate only 19-Watts. For comparison, a 1GHz G4 consumes 1.6-volts and dissipates 21.3-Watts."
This statement from MacCentral is poorly worded. A processor does not "consume" the voltage it is operating at. Voltage is like an electrical force, and the power supply keeps it constant. Power dissipation follows the law P = VI, where "I" is the electrical current drawn by the device. At higher clock rates the processor draws more current to do the work, and hence power goes up. To reduce power, therefore, the CPU can be operated at a lower clock rate so it draws less current. At lower clock rates the operating voltage can also be set lower, so power goes down even more. Both V and I are less. At 1.1 Volts and 1.2 GHz, the 970 will draw about 17.27 Amperes and dissipate 19 Watts.
Could you run that by me again? It was my understanding that there would be no math.
Originally posted by snoopy
"... At 1.1 Volts and 1.2 GHz, the 970 will draw about 17.27 Amperes and dissipate 19 Watts.
17.27 amps seems high to me, just asking. I'm mean that will drive a pretty hefty sized electric motor in our plant. Color me confused.
Originally posted by rickag
17.27 amps seems high to me, just asking. I'm mean that will drive a pretty hefty sized electric motor in our plant. Color me confused.
That's just what the equations show. It's worse at 1.8 GHz and 1.3 Volts, where it draws 32.3 Amperes and dissipates 42 Watts. It is surprising to me too, as I never did the math til now. They will need large gauge wire from the power supply for that, like number 10 or bigger. I haven't verified the 1.3 Volts from another source, but if it is incorrect the current will be a different value. I know the 42 Watts is correct.
The big difference between motors and CPUs is the operating voltage. At 120 Volts and 17 Amperes for example, the power would be about 2000 Watts. On AC sources there is something called power factor, which makes the math more messy. Power is not exactly current times voltage.
'Course, having a combination laptop and arc welder would be something completely different. Let's see Dell compete with that!
(Edited to get *my* units correct. Jeez, now everyone's doing it!)
Originally posted by Voxapps
Um, the typical full-sized household vacuum cleaner draws 11 amps. A 32-amp laptop would probably have about 30 seconds of battery life before it melted through whatever it was sitting on. . .
The laptop would be 17 Amperes, and 19 Watts. Your 11 Ampere vacuum cleaner is about 1300 Watts. It is the Watts, the power dissipation, that determines the amount of heat generated, not the current. High current requires good size conductors however. If a 1.1 Volt DC source supplies 1 Ampere to the load, that is only 1.1 Watts. The 970 is said to dissipate 19 Watts at 1.1 Volts.
Originally posted by snoopy
That's just what the equations show. It's worse at 1.8 GHz and 1.3 Volts, where it draws 32.3 Amperes and dissipates 42 Watts. It is surprising to me too, as I never did the math til now. They will need large gauge wire from the power supply for that, like number 10 or bigger. I haven't verified the 1.3 Volts from another source, but if it is incorrect the current will be a different value. I know the 42 Watts is correct.
Huh?
The current G4's electrical specs call for 6.4A max current. You're trying to tell us that the PPC970 will more than triple the maximum current (at the high end)? I don't think so. I don't know what formula you're using, but I doubt your figures.
don´t get me wrong, I am a newbie !!!!!
faxe
from macrumors message board:
http://www.spymac.com/forums/showthr...5&pagenumber=3
(Is Loop Rumors Wrong)
Originally posted by keyboardf12
I.L.R.W.* ?
from macrumors message board:
http://www.spymac.com/forums/showth...15&pagenumber=3
(Is Loop Rumors Wrong)
just get an error message when i try to access this link
faxe
Originally posted by Tomb of the Unknown
Huh?
The current G4's electrical specs call for 6.4A max current. You're trying to tell us that the PPC970 will more than triple the maximum current (at the high end)? I don't think so. I don't know what formula you're using, but I doubt your figures.
The G4 CPU at 1 GHz actually draws 13.3 Amperes, since it is said to operate at 1.6 Volts and dissipates 21.3 Watts. The 6.4 Amperes Maximum on the back of your Mac refers to the most line current it will draw from a 120 VAC power source. That would seem like the Mac dissipates over 700 Watts but it really does not. The max amps on a product is for UL listing. First, it draws the highest current only at low line voltage conditions. Second, there is a power factor, which means the actual power is less than the number you get by multiplying RMS current by the line voltage. Third, the product makers almost always state the number a little high. The number of amps is meant to guide people on what load there will be on the building's electrical circuits.
Originally posted by faxe_tv
can anybody provide a link ( for newbies ) for this electric things
don´t get me wrong, I am a newbie !!!!!
faxe
I don't have a link, but what we are talking about is how to calculate power, using the formula:
power = current X voltage
If you look at that formula you will notice something. If the voltage is low, it takes more current for the same power you would get at a higher voltage. If you compare the currents of household devices, which operate at 120 V, to currents of automotive devices, which operate at 12 V, you will see a big difference. Now, when we get down to something like 1.3 V or 1.1 V in the 970 CPU, the difference is even bigger.
Originally posted by keyboardf12
I.L.R.W.* ?
from macrumors message board:
http://www.spymac.com/forums/showthr...5&pagenumber=3
(Is Loop Rumors Wrong)
Very interesting thread. I for one believe the Powerjack (MacWhispers guy) to be legit. He just sounds too serious to be BS-ing. Besides, he should know if he screws around with us with lies no one will buy his prodcuts to be released later this year.
So it looks like 970 around August, so maybe announced at MWNY and shipping immediately or within weeks.
Macintouch reader reports
I thought this might be helpful to those blinded by rumors.
He's nailed a couple of things and we should know by WWDC what his long term prospects as a rumor monger are.
Originally posted by KidRed
Very interesting thread. I for one believe the Powerjack (MacWhispers guy) to be legit. He just sounds too serious to be BS-ing. Besides, he should know if he screws around with us with lies no one will buy his prodcuts to be released later this year.
So it looks like 970 around August, so maybe announced at MWNY and shipping immediately or within weeks.
Well sounds like were back to the original schedule August-September release and maybe pre-order in July or at WWDC
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