AMD Unleashes First-Ever 5 GHz Processor

[ZetZet]

FX-9590: Eight "Piledriver" cores, 5 GHz Max Turbo
FX-9370: Eight "Piledriver" cores, 4.7 GHz Max Turbo


max turbo...

MAX... it is NOT going to be ALL cores, MAX turbo works when its a single thread load !!!!

my god, amazing no one knows this already... :shadedshu

But it will overclock to 5Ghz so easy. ^^

 

[Huguito]

But it will overclock to 5Ghz so easy. ^^

1 core @ 5ghz its fairly easy to achive.. 8 cores its a totally diferent history

 

[Mathragh]

Short circuits generate massive amounts of heat. Stuff tends to melt when there is a short circuit...

Yeah, but only if the wire the short circuit is running through cant handle the current. The real world examples you connect with do no necessarily apply to everything.

 

[Aquinus]

Edit: just think of a short circuit: no all energy is used(transformed in heat), but there is still a lot more energy running through than without a short circuit

...yes, electricity flows through it but it doesn't do any work except for the work done by heating the wire. AC doesn't work that way. Remember that current changes directions and un-used electrons that went from hot to neutral go in reverse back from neutral to hot. So any electricity that leaves the circuit and isn't converted into energy will get returned to the PSU, and if you consider electron drift you will see that very slowly, despite electrons moving in both directions, that the overall flow of current is going into the device and not leaving it as electricity, but as heat.

Yeah, but only if the wire the short circuit is running through cant handle the current. The real world examples you connect with do no necessarily apply to everything.

Wattage is work done which describes that transfer of energy from one form to another. You're thinking of current which is the flow of electrons, not work done.

Also consider Ohm's law. I = V / R

Unless you have a super conductor, there is always resistance, which means work is done and since electrons can flow freely, you'll heat up that wire really fast off the mains. Either that or you will fry whatever is generating that electricity (or trip a fuse.)

 

[newtekie1]

max turbo...

MAX... it is NOT going to be ALL cores, MAX turbo works when its a single thread load !!!!

my god, amazing no one knows this already... :shadedshu

Which works out very well since applications that work better with high frequencies tend to be single threaded, and applications that don't need high frequencies tend to use multiple cores.

I'm surprised no one knows this already...:shadedshu

Yeah, but only if the wire the short circuit is running through cant handle the current. The real world examples you connect with do no necessarily apply to everything.

No, heat will still be produced regardless of the wire used.

 

[Sempron Guy]

It would be interesting to know how AMD will price these chips and how far the FX9590 can be pushed further maybe 5.5 on a good AIO cooler? ( I wish )

 

[silkstone]

Yeah, but only if the wire the short circuit is running through cant handle the current. The real world examples you connect with do no necessarily apply to everything.

...yes, electricity flows through it but it doesn't do any work except for the work done by heating the wire. AC doesn't work that way. Remember that current changes directions and un-used electrons that went from hot to neutral go in reverse back from neutral to hot. So any electricity that leaves the circuit and isn't converted into energy will get returned to the PSU, and if you consider electron drift you will see that very slowly, despite electrons moving in both directions, that the overall flow of current is going into the device and not leaving it as electricity, but as heat.



Wattage is work done which describes that transfer of energy from one form to another. You're thinking of current which is the flow of electrons, not work done.

I teach this stuff up to grade 12. It fairly easy, but confusing to visualize. Go with the water example I added to my edited post above.

Wattage is power is work done over time. Mathragh is visualising it wrong.

Also the CPU takes DC current, which is quite different (and less complicated) than AC.
But, you are right that he is thinking of current. Where current in = current out (always)
Energy in =/= Energy out. But, you don;t look at the energy coming out of a system to determine the amount of energy used by that system.

You need to look at the voltage drop across the component to determine how much energy has been used (Voltage is joules per coulomb)

 

[Mathragh]

Energy in =/= Energy out and is quite different.

This is what I meant, thanks. I do know current stays exactly the same, but apparently I'm not always as clear as I think I am when I'm explaining what I'm thinking.
Lol, sorry for all the pants I got in a twist people .

Anyway, back on topic, and restating my former question:

Does this chip involve some new mojo(trinity->richland mojo?(sort of)) or,

Just an aggressive binning of vishera?

I personally think it's just binning, but that would mean that they could potentially also implement the new turbo to current lower-end vishera chips.

Edit:

You need to look at the voltage drop across the component to determine how much energy has been used (Voltage is joules per coulomb)

Thanks again, this was the point I was trying to make, but apparently didn't have the words for. When current leaves the chip, its not at 0V, but its also not useable anymore, so its effectively wasted energy. Ofc all appliances work like this, but in this case, it is the difference between the power consumption, and heat generated(someone again correct me if i'm wrong)

 

[Aquinus]

When current leaves the chip, its not at 0V, but its also not useable anymore, so its effectively wasted energy.

Well, that depends. You have your voltage going in and you have the current flowing towards ground. Ground is 0v, yes, but if the CPU is the only resistive load between the voltage source and the ground, then the voltage between the ground and the CPU is 0v. If there is voltage between ground the CPU, that means there is a second resistive load between the CPU and ground which uses a portion of that voltage, which would result in less voltage for the CPU. That would indicate something other than CPU is doing work (like the VRMs, they get release heat too,) but even then, that's before the voltage going into the CPU. The loss is inside the VRM itself and that makes the CPU the second load.

So I disagree. I think the CPU is the only major resistive load there and that after the CPU, the voltage is very much so 0v. The current may be several amps though, varying with load.

I'm done talking about circuits though, we digress way too much for our own good. Hopefully this chip will be able to squeeze more than stock speeds out of it with something other than phase change.

 

[silkstone]

Not exactly.

Power consumption = Heat generated. It's by definition. Basically, if power is used then energy has changed forms.

The potential drop across a component will only be different from the potential difference across the terminals of the supply if you have other elements in series.

Voltage is measured in Joules per Coulmb. It is the amount to energy stored in the electrons' electric fields due to their proximity.

Think about having rubber balls (electrons) all squashed together at one side and rubber balls far apart next to them, but separated by a 1-way barrier.

When you create a path for the rubber balls to flow (creating a circuit) the balls are going to push away from each other at the one end, go down the path (the cpu) be slowed down and push into the rubber balls at the other end which are further apart.

When they are pushed into those balls at the other end, with some extra help, they are going to push those rubber balls through the barrier in to the rest of the squashed rubber balls.

The rubber balls themselves are not used up, they don't disappear or go anywhere, they just lose energy as they travel down the path.

I believe the potential difference across a CPU is controlled very tightly by the NB. If you increase the voltage (potential difference) across the CPU, it starts using more energy and overheats.

The CPU will have a power rating at a certain number of volts. Increasing the voltage will cause the power used to go up as you are causing the CPU to use more voltage.

The rate at which electrons can carry energy out of the CPU is limited and would be very very minimal. But again, that energy they carry is "re-used."

To get into details would be too involved, and I have to admit, I don't understand it 100% as it's been a long time since studying it at Uni.

Edit - I found a water analogy illustration:

 

[Ravenas]

When current leaves the chip, its not at 0V, but its also not useable anymore, so its effectively wasted energy.

Current is in Amperes not Volts. Additionally you are using the term energy too loosely. Energy in terms of circuits is in Joules. A Joule can be called a W*s (Watt * second). A watt is a unit of power and you're using that term in the same sentence as current as if current and energy are interchangeable terms...

 

[Aquinus]

The CPU will have a power rating at a certain number of volts. Increasing the voltage will cause the power used to go up as you are causing the CPU to use more voltage.

I want to work off this phrase to make it more clear.

The CPU will consume so much power with a certain number of volts some some load that is consistent. Increasing the voltage will cause the power used to go up because more voltage allows more current to flow with the same amount of resistance in the circuit. Heat is generated by the flow of electrons (current) through a resistive medium. The heat energy released is directly proportional to the resistance multiplied by the square of the current. So as current increases and resistance remains the same, for any increase in current will cause heat to increase exponentially.

The only reason voltage impacts the amount energy being released as heat is because, generally speaking, resistance isn't changing a whole lot under a constant unchanging load and higher voltages make it easier for electrons to flow through a medium. The only other way to increase current without increasing voltage is to reduce the amount of resistance and the best way to do this is to make the CPU very, very cold (hence LN2 and phase change).

 

[Norton]

Does this chip involve some new mojo(trinity->richland mojo?(sort of)) or,

Just an aggressive binning of vishera?

I personally think it's just binning, but that would mean that they could potentially also implement the new turbo to current lower-end vishera chips.

My thought is that this may be a little of both- agressive binning is the most likely source but I would not be surprised if there are some elements of Steamroller baked into these chips.

AMD went with a more modular design since Bulldozer so it may be possible that certain design elements that are ready to go from Steamroller have been added into the chip.... it may be as simple as a cut and paste on the chip layout prior to making the wafer.

or not

Anyone see any reason into the numbering scheme for these?

 

[Prima.Vera]

you guys need a life, a girlfriend, something...

 

[Norton]

you guys need a life, a girlfriend, something...

My wife won't let me have a girlfriend nvm

Why?

 

[de.das.dude]

i think the 220W was for the other extreme chip.

plus 220W is the max TDP,. no matter how hard you overclock it wont go past 220W!

 

[silkstone]

i think the 220W was for the other extreme chip.

plus 220W is the max TDP,. no matter how hard you overclock it wont go past 220W!

Not true.

"The thermal design power (TDP), sometimes called thermal design point, refers to the maximum amount of power the cooling system in a computer is required to dissipate. The TDP is typically not the most power the chip could ever draw, such as by a power virus, but rather the maximum power that it would draw when running "real applications". This ensures the computer will be able to handle essentially all applications without exceeding its thermal envelope, or requiring a cooling system for the maximum theoretical power (which would cost more but in favor of extra headroom for processing power)" - Wikipedia

 

[bacan]

It looks like these CPUs are OEM models only. (retail maybe later)
http://www.pcgameshardware.de/CPU-Hardware-154106/Specials/FX-9590-FX-9370-AMD-Centurion-1073412/

 

[librin.so.1]

you guys need a life, a girlfriend, something...

But getting a girlfriend would mean spending a lot of money on her [both directly and indirectly]. Meanwhile, that money could be spent on much more important things - like these kind of chips, for example.

 

[drdeathx]

Yeah, but only if the wire the short circuit is running through cant handle the current. The real world examples you connect with do no necessarily apply to everything.

These are binned chips and don't use as many volts as 8350's at 5GHz. I would assume around 1.5 volts which any board can handle.

 

[Dent1]

This is disappointing. I had thought that AMD learned the lessons of Bulldozer (more GHz and more cores don't matter if single-thread performance sucks), but this raises some doubts. Up until now I was sure this was a rumor, or perhaps a pre-Rory Read holdover that had been nixed.

Firstly, AMD couldn't give a rats ass if you liked Bulldozer, it made them money.

Actually, more GHz does increase single threaded performance otherwise us enthusiasts wouldn't bother overclocking.

If AMD have the technology to release a 5GHz CPU on air, then why not? Surely going down in the record books as the first is a big achievement. Think of all the positive press they'll get which translates into free marketing and increased brand awareness and sales.

 

[cadaveca]

I would assume around 1.5 volts which any board can handle.

My 8350 already does 5 GHz @ 1.5V. I expect better than that. Seems pretty simple for AMD to do this, to me. They could have been binning for this 5 GHz chip since they started making Vishera, a long long time ago, in a distant land, with fairies.

 

[de.das.dude]

Not true.

"The thermal design power (TDP), sometimes called thermal design point, refers to the maximum amount of power the cooling system in a computer is required to dissipate. The TDP is typically not the most power the chip could ever draw, such as by a power virus, but rather the maximum power that it would draw when running "real applications". This ensures the computer will be able to handle essentially all applications without exceeding its thermal envelope, or requiring a cooling system for the maximum theoretical power (which would cost more but in favor of extra headroom for processing power)" - Wikipedia

exactly what i said. its the max power draw.
however i have heard that intel does things a bit differently. they take the RMS or something

 

[librin.so.1]

they take the RMS or something

Wha? Stallman?

 

[silkstone]

exactly what i said. its the max power draw.
however i have heard that intel does things a bit differently. they take the RMS or something

You said the opposite. 220 W is the standard application maximum power draw. It other situations, it will easily exceed that. It can easily exceed that number. I think that it is given so that companies will know what they should set as a minimum for their heatsink design.

RMS is for AC where you have a wave-like potential difference. But, yes, they calculate it differently from AMD and Nvidia.