Ivy Bridge review (By tweaktown)

[mjw21a]

^^^Ah... I think I understand. Basically we may find that from the wall these chips are using less power than the old Sandy Bridge chips, on the other hand the heat being generated is still rated at 95W TDP due to the increased resistance..... Thats what you're getting at, am I correct?

 

[lilg_06]

^^^Ah... I think I understand. Basically we may find that from the wall these chips are using less power than the old Sandy Bridge chips, on the other hand the heat being generated is still rated at 95W TDP due to the increased resistance..... Thats what you're getting at, am I correct?

Yes, that is exactly what I am saying! lol

 

[Bobnova]

You cannot leave E and I static and raise R. It violates ohm's law.
Raising R drops I. Really, really, really basic electrical theory here.

If at max load it has a constant current and constant voltage it has a constant resistance as well.

Please read about the relationship between resistance and current. Start here: http://www.allaboutcircuits.com/vol_1/chpt_2/1.html


Try it.
Raising resistance lowers current. To be really specific, it's a linear relationship.
Twice the resistance = half the current.
You're continuing to calculate as if current were fixed with a varying resistance. It is not, it varies with resistance.


As far as IB goes, I read it as it leaks like crazy and draws more current for a given voltage level.
That is the behavior of lower resistance. Lower resistance with static voltage always equals more current and more wattage used.

Watts = power.
They don't use less power and generate the same power, heh.

They use the same power, and run hotter.
That is very, very, very different.
Heat is related to power, but different.
For example, 95w into a penny will result in a very hot penny very quickly.
95w into a 200lb lump of the same material will take an awfully long time to get hot.

The IB die is smaller, it has less surface area to get rid of the power. As such it runs hotter for a given amount of wattage consumed. That fits with thermodynamic and electrical laws.

 

[Archer0915]

The IB die is smaller, it has less surface area to get rid of the power. As such it runs hotter for a given amount of wattage consumed. That fits with thermodynamic and electrical laws.

I think this would have been enough.

Another thing is we are talking internal temps. So in the end even though it may be rated 95w TDP (just a number mind you) it may actually may be easier to cool at OC than SB is. Within limits, that is, because the smaller core is less area to wick off of.

 

[Bobnova]

I thought so too, that's why I stated it in post #27.

 

[lilg_06]

Once again, we can't change the variables one at a time, I was simply trying to tell you that increasing R leads to a higher power output, general physics, forget the chip for a minute.

The following equation: P=I^2 * R
The above equation tells you how current and resistance change power consumption, if you have a simple circuit, and replace the resistor with one of a higher rating, to maintain that same CURRENT, you will increase the power output, P increase.

Because the current was constant, another variable had to have changed to lead to the increase in P:
P=VI (You can calculate the new voltage required to maintain the same current)
Use your new power value, you already know what current you had set and what resistor.

All I'm saying, is increasing the resistance means that the power consumption increases, more heat output, of course a real life application to this scenario is difficult, because we don't know exactly what current is going through the chip, the only thing I was trying to get across is we know the chips are smaller, so physically should have a higher resistance, to get electrons flowing through them we need more power, more power = more heat, if vcore is set manual, then the cpu will pull MORE current, as V is not a variable....
Its very complicated, but simply, Resistance increases power output, so does current.....

The point of all this.

Compared to SB, IB has a higher resistance, therefore pulls more current to draw its power requirements, the resistance of the chip is a very complex thing to look at, because it changes with the number of transistors being on/off... but from a logical perspective overall it is higher, do smaller diameter wires not have a higher resistance?

NOTE: When we talk about the cpu's resistance, to picture the scenario more simply, think of it as the resistance of just ONE transistor, not all of them it will definitely be more resistive than the same transistor in SB due to it being smaller, you cannot adjust the resistance of this one resistor, it is FIXED. more current is drawn to fulfil power requirements....


And yes, surface area may very well be the issue here..... But surface area does not explain why its sucking more power from the wall.

 

[Bobnova]

To maintain the same current while raising the resistance you must INCREASE the voltage. That is all.

Are we increasing the voltage here? Or is it maintaining a given level? Please think carefully before you answer.

 

[lilg_06]

To maintain the same current while raising the resistance you must INCREASE the voltage. That is all.

Are we increasing the voltage here? Or is it maintaining a given level? Please think carefully before you answer.

In other words, this law is wrong: P=I^2 * R
lol

 

[lilg_06]

I have another question, are there more transistors in IB, because that can also lead to an increase in current draw??

 

[Bobnova]

No, that law assumes voltage is whatever is required to push I through R.
You can find the voltage required easily enough by looking at I and R, as the amount required is very specific for a given I and R.

 

[Archer0915]

I think there are two thoughts here and three things to look at.

Power in (wall watts) and Power out (heat watts) are the thoughts and what we need to look at are:

In, out and actual measured temps as it relates to actual thermal output.

Remembering that 95 watt TDP is not necessarily 95 watts of heat energy.

 

[Audioaficionado]

I'm waiting on the massive reviews once this puppy is released from NDA.

I'm more interested in the mobile performance as I want a nice laptop in the near future.

 

[mjw21a]

Yeah, I think we need to wait for more sources of information. We've only seen one relating to heat output so far, the other is just confirmation (via pictures) of TDP etc.

 

[lilg_06]

But, at say 1.3V, a motherboard is capable of supplying a range of current to the cpu, it doesn't constantly supply all the transistors at once, it draws current when required, SB may not have been using the max current at a certain vcore, you want proof, I have my cpu clocked at 4.7GHz 1.42V, when I'm not using my system, the power consumption in AI suite drops, my vcore does not drop, it is manually set in the bios, however my power draw drops, P=VI, if power draw drops, less current is delivered, why? because the cpu is not doing much, less current is delivered to it, this explains why more current is delivered at the same vcore....
First we need to determine what the given current range is at a set voltage, it isn't fixed, the above law is 100% correct, you can deny it all you like, doesn't change anything. lol

 

[lilg_06]

Yep, lets leave it be and wait for the reviews....

 

[Brolloks]

Certainly the reviews will get us the detail....that said it will only confirm what we tell you

 

[MattNo5ss]

It's self-explanatory that the equation V = I * R shows a linear relationship between V and R with a slope of I (or vice versa). Using the simplest example of I = 1 for a constant, then V and R would have to be the same number at all times. Basically, V and R have to be multiples of each other when assuming constant I.

However, the whole discussion about it is getting a bit off topic really, since it's now general discussion about the relationships between voltage, current, resistance, and power.

I'm curious about how fast Ivy Bridge power consumption and temps will rise based on Vcore and MHz as variables in actual testing.

 

[mjw21a]

Lol, perhaps. I've never known Bobnova to be wrong with anything I've looked at yet so I trust him on this. He is still human though so it IS possible.

There's too many unknowns for us at this point. Easiest thing is to wait for the reviews (much like with new CPU architectures).

 

[Archer0915]

Lol, perhaps. I've never known Bobnova to be wrong with anything I've looked at yet so I trust him on this. He is still human though so it IS possible.

There's too many unknowns for us at this point. Easiest thing is to wait for the reviews (much like with new CPU architectures).

There is really not a right or wrong here as I feel there are two different aspects here. Hell I might just get my geek on and go greek

 

[Brolloks]

At least it wont get as hot as a Q6600 and it will certainly clock easier.