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[Various] AMD’s Zen Eight Core CPU Delivers Double the Performance of the FX 8350 (Update 2) - Page 48

post #471 of 564
Quote:
Originally Posted by Blameless View Post

Quote:
Originally Posted by cssorkinman View Post

What is causing the extra heat load when you operate a cpu at the same voltages , under the same load, but increase the clockspeed?

It takes more current to do more work.
Quote:
Originally Posted by Fyrwulf View Post

You can't increase clock speed without increasing voltage somewhere.

Yes, you can. Indeed, voltages will often be lower as temperatures and resistance rises, even in the absence of things like vdroop, if you keep supply voltage constant but increase clock speeds and/or the load on the CPU.
Quote:
Originally Posted by looncraz View Post

A modern CPU, though, should make this less likely - an electron entering a CPU without doing some work is a worthless electron.

Less power does work in modern CPUs than past CPUs. Leakage power as a percentage of total power has been rising exponentially.

Go back to 1995 and 90%+ of the power dissipated by a CPU was used to switch transistors. It's about 10-30% now, in high-performance parts. Of course, transistor switching power requirements have been decreasing even faster, so performance per watt keeps going up, but there have had to be many innovations to prevent leakage current from outpacing those gains.

I appreciate the answer. That would be reflected in wattage values, wouldn't it?
Something I've observed in the workplace has puzzled me. We have 4 large volume fans , each powered by 500hp electric motors producing 1.5 million CFM. If you set up our facility in a manner where they are generating little to no pressure ( the fans are basically free-wheeling) , the amp gauges will read higher than if you pressurize the building to 7 + inches of water column. Can you explain why this happens?
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post #472 of 564
Quote:
Originally Posted by cssorkinman View Post

What is causing the extra heat load when you operate a cpu at the same voltages , under the same load, but increase the clockspeed? Resistance? Just curious , sorry so far O.T.

Averaged resistance over time goes down such that you'll note the processor drawing more current and thus more power overall. However, instantaneous resistance isn't changing much. Reactance in a logic die is going to be minimal though technically present. Mostly, the answer to your question stems from the charge and discharge of FET gates. The higher the frequency, the more times this happens per unit of time. A small number of electrons are trapped on each gate each time they are charged to potential. These are wasted as heat when discharged.
post #473 of 564
Quote:
Originally Posted by Fyrwulf View Post

Bristol Ridge uses Excavator...

Yes. So 40% over Excavator (Bristol Ridge).
post #474 of 564
Quote:
Originally Posted by AmericanLoco View Post

That is literally not possible. What your describing there is a 0-resistance superconducting path from core voltage to ground, if such a path existed, you would have a direct short the moment you turned your computer on and all CPU v-core power would be shunted directly to ground.

Please read up on resistance, super conductors and semi-conductor leakage.

I think once you realize that a CPU drops into another partial circuit and is, therefore, only part of a large circuit, that you will realize what I am saying.

Some of the current flows right through the CPU without doing work or inducing heating. The more this happens, the less efficient the CPU is. Now, yes, most of the electrons flowing through the CPU will encounter resistance and result in an increase in kinetic energy within the CPU, however this is not 100%.

Current flowing from the CPU in the form of I/O will not induce heat within the CPU, that heat will be created elsewhere - on the board, in the PSU, or on the building leads. I/O power was much higher in the P3 era, which is the entire extent of my real world experience with this. It could have been 3~5W... for a 35W CPU (we were using low end P3s and Celerons - some of which I got to keep thumb.gif).

From there, some of the current undoubtedly went back to the VRM without inducing heat within the CPU (instead inducing it in the VRM).

Some of the current will have found itself on the ground-plane of the motherboard, which was not directly grounded. The heating would not occur within the CPU here, but within the entirety of the ground plane. The entire path to the chassis, the chassis itself, and the leads into the wall.

It's not that you are wrong - at all - beyond where the heat is produced. It just isn't actually well contained within the CPU. That does change with temperature and load, though, but we were testing the ability to radiate certain heat-loads from a point source into a vacuum for a commercial satellite - my mentor wanted to make a lot of money, which he didn't do :mad - so we were just running max-loads with some software in OS/2.
post #475 of 564
Quote:
Originally Posted by flippin_waffles View Post

Yes. So 40% over Excavator (Bristol Ridge).

Bristol Ridge uses a tweaked version of Excavator, so it's more like Excavator+.

Carrizo uses Excavator.
post #476 of 564
Quote:
Originally Posted by looncraz View Post

Bristol Ridge uses a tweaked version of Excavator, so it's more like Excavator+.

Carrizo uses Excavator.

And it may not be large, but it appears the IPC has increased with Bristol Ridge Ex over Carrizo Ex. A 40% increase over that is almost certainly going to change the entire x86 CPU landscape.
post #477 of 564
Quote:
Originally Posted by cssorkinman View Post

That would be reflected in wattage values, wouldn't it?

Yes.
Quote:
Originally Posted by cssorkinman View Post

Something I've observed in the workplace has puzzled me. We have 4 large volume fans , each powered by 500hp electric motors producing 1.5 million CFM. If you set up our facility in a manner where they are generating little to no pressure ( the fans are basically free-wheeling) , the amp gauges will read higher than if you pressurize the building to 7 + inches of water column. Can you explain why this happens?

Assuming these fans are cooling things and that feathering the blades to make them free wheeling causes them to move less air, and other equipment to run warmer, that is probably responsible for the power differential you see.

Cooler running electronics, all other things being equal, use less power because of lower resistance.
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post #478 of 564
Quote:
Originally Posted by cssorkinman View Post

I appreciate the answer. That would be reflected in wattage values, wouldn't it?
Something I've observed in the workplace has puzzled me. We have 4 large volume fans , each powered by 500hp electric motors producing 1.5 million CFM. If you set up our facility in a manner where they are generating little to no pressure ( the fans are basically free-wheeling) , the amp gauges will read higher than if you pressurize the building to 7 + inches of water column. Can you explain why this happens?
Could be your motors have a low power factor at low torques, so while you might be drawing more current, you're actually using less power because that current draw is out of phase with the voltage.
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post #479 of 564
I seriously cannot wait for these to hit the shelves! There is good reason to go AMD again instead of just being the fanboi that I am for them. Also, it would be a great upgrade from my phenom biggrin.gif
post #480 of 564
Quote:
Originally Posted by TranquilTempest View Post

Could be your motors have a low power factor at low torques, so while you might be drawing more current, you're actually using less power because that current draw is out of phase with the voltage.

if voltage were going down and current going up then wattage would stay the same and thus power stays the same. Volts*Amps=Watts. A low power factor would also cause the customer to be charged more by the electric company, as the electric company "sees" the load higher that the true wattage being drawn, so even though wattage would be the same if the power factor is dropping then they are being charged like it is drawing more power.
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