Ivy Bridge review (By tweaktown)

I.M.O.G. said:

Maybe a bit, but often times Intel doesn't budge prices a whole lot. Look how much a 990X costs currently at newegg. (1K+) But with microcenter selling 2600k for $200 on special now, there very well could be some more deals coming later.

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I forgot about the 990X still costing that much. I seen it back when I was first looking at the 3960X on Newegg. And thought it was kind of dumb for it to still be priced that high.

I wish I had a Microcenter around here. All we have is Compuzone's, and in there a 2600k will cost about as much as a 990X or 3960X. lol

Guess I'll wait one more year to upgrade. Bummer.

I'm still excited to build my IB system. It's been so long since I've upgraded I think I've waited plenty of time.

Hmmmm, beginning to wonder if I should switch back to Intel.... I've not run an Intel chip since my old Pentium 100.....

I'm not too thrilled to see high temps for air/water with IB limiting overclocks. If this is the new normal with these chips then I hope that Intel can come up with a revision down the road that will improve this particular facet of their performance.

I think it makes sense for heat output to increase, the chip is using a smaller manufacturing process, overclocking pushes more current through the chip, so the temps will increase as more electrons flow in a smaller space, think if it this way, these chips are more resistive due to 22nm architecture, more resistance = more heat....
Though I could be wrong, but from a physics perspective, it sounds like a plausible explanation?

More resistance means less current flow and less heat, given a specific voltage.

It makes sense for core temps to go up if the same amount of heat is being generated in a smaller space, that makes it harder to get rid of.

Yep, thats current, theoretically speaking these should be drawing less current at the same vcore, but we don't really know if thats the case....
I guess we'll find out soon enough.
But if they are drawing the same current as SB when overclocked, then they will definitely heat up as resistance increases with a decrease in conductor size.

Just re-read your message, more resistance increases heat output at a given voltage, because with a higher resistance, electrons collide more often in the conductor material increasing temperatures, yes the current flowing through will decrease, but this is as a result of the higher resistance....

Watts of heat produced is identical to watts of electricity consumed.
If you have a 1V supply voltage 0.9ohms of resistance will always generate more heat than 1.1 ohms.
Watts is volts multiplied by amps.
Same voltage, lower resistance means more amps means more watts.

In a circuit the part of the circuit with the highest resistance will be the hottest part of that circuit. In this case that is the CPU, it has a much higher resistance than the solid copper planes and massively paralleled MOSFETs that feed it power, so that is where the heat is (mostly) generated. If you decreased the CPU resistance to the point where the MOSFETs or the copper planes were the highest resistance point then they would be the primary place the heat was generated.

In any case, lower resistance = more current flow = more watts.

From what I've read the TDP of these chips is 95W. The fact the power draw hasn't gone down is what appears to be causing the issue. Same power draw over a smaller surface area = bad news. Much harder to disperse the heat.

Intel will either need to fix their process tech for lower volts, or redesign the chip to spread out thinner over a larger surface area. As I understand though this would decrease the efficiency of the chip electrically.....

Power(W) = Voltage(V) * Current(I) can be written as P=V*I
Voltage(V) = Current(I) * Resistance (R) can be written as V=I*R

Therefore
P = I * I * R
... P = I^2 * R (Increase in resistance means higher power consumption)

This equation tells us that although IB processors require less current, there is also an increase in resistance compared to SB, this does not mean that chip consumes more power overall, it just means that although we are reducing power output for processing power, we are at the same time increasing power output for HEAT.

Also remember the equations above are related to power consumption in a circuit, the amount of watts consumed does not all account for heat. In the case of IB, although less energy is required overall, a larger proportion of that energy consumed is now wasted as heat....

mjw21a said:

From what I've read the TDP of these chips is 95W. The fact the power draw hasn't gone down is what appears to be causing the issue. Same power draw over a smaller surface area = bad news. Much harder to disperse the heat.

Intel will either need to fix their process tech for lower volts, or redesign the chip to spread out thinner over a larger surface area. As I understand though this would decrease the efficiency of the chip electrically.....

Click to expand...

Yep, I personally think its still drawing the same power as SB, hence higher output...

Its all down to the relationship between current and resistance.
P = I^2 * R
Even if the current required is lower, resistance is now higher, it could be the changes are not highly beneficial, and thus, the same power is drawn through, in which case resistance is the biggest problem...

lilg_06 said:

Power(W) = Voltage(V) * Current(I) can be written as P=V*I
Voltage(V) = Current(I) * Resistance (R) can be written as V=I*R

Therefore
P = I * I * R
... P = I^2 * R (Increase in resistance means higher power consumption)

This equation tells us that although IB processors require less current, there is also an increase in resistance compared to SB, this does not mean that chip consumes more power overall, it just means that although we are reducing power output for processing power, we are at the same time increasing power output for HEAT.

Also remember the equations above are related to power consumption in a circuit, the amount of watts consumed does not all account for heat. In the case of IB, although less energy is required overall, a larger proportion of that energy consumed is now wasted as heat....

Click to expand...

Sorry, no.

On question of watts in = watts out: Where, pray do tell, does the other energy go?
Is it being condensed into matter? I ask because that is quite literally the only place it can go that does not result in heat.


On the question of power consumption:

Voltage does not equal current times resistance, sorry.

Voltage is a fixed number set by the voltage controller. It is not a variable.
I high recommend playing with the calculators available here: http://www.the12volt.com/ohm/ohmslawcalculators.asp
Remember that voltage is fixed. Let me know if you can make watts drawn go up by raising resistance with a fixed voltage.


Here's where you went wrong:
P = I^2 * R
This is true, however you must have voltage to push the current through the resistor. If you raise the resistance you will have less current with a given voltage. Example:
E(volts) = 100
I = 10
R = 10
10 ^2 * 10 = 1000 watts.

How do you get that ten amps through a 10 ohm resistor? You need that 100 volts pushing.
Now we'll raise the resistance to 20 ohms.
E = 100
R = 20
To find the current we use I = E / R. The result: 5 = 100 / 20.
Five amps and 20 ohms gives 500 watts.



EDIT:
Please don't say that the energy not dissipated as heat goes into "work done". A processor does no work in that sense and all energy invested in work ends up as heat eventually anyway.

Bobnova said:

Sorry, no.

On question of watts in = watts out: Where, pray do tell, does the other energy go?
Is it being condensed into matter? I ask because that is quite literally the only place it can go that does not result in heat.


On the question of power consumption:

Voltage does not equal current times resistance, sorry.

Voltage is a fixed number set by the voltage controller. It is not a variable.
I high recommend playing with the calculators available here: http://www.the12volt.com/ohm/ohmslawcalculators.asp
Remember that voltage is fixed. Let me know if you can make watts drawn go up by raising resistance with a fixed voltage.


Here's where you went wrong:
P = I^2 * R
This is true, however you must have voltage to push the current through the resistor. If you raise the resistance you will have less current with a given voltage. Example:
E(volts) = 100
I = 10
R = 10
10 ^2 * 10 = 1000 watts.

How do you get that ten amps through a 10 ohm resistor? You need that 100 volts pushing.
Now we'll raise the resistance to 20 ohms.
E = 100
R = 20
To find the current we use I = E / R. The result: 5 = 100 / 20.
Five amps and 20 ohms gives 500 watts.

Click to expand...

Actually not 100% of that energy is wasted as heat, some of it is sound, produced in a very high frequency which you cannot hear as a result of vibrations in the chip.
For simplicity, lets assume that all the power input is converted to heat, and the relationship V=IR is correct, its ohms law, I didn't say we are changing the voltage, I mentioned it to show you the relationship between POWER, CURRENT AND RESISTANCE, these are the variables we were looking at.
P = I^2 * R (Go to any physicist and he will confirm this relationship for you)

You can combine equation v=ir and p= vi to give the equation above.

In this equation P = I^2 * R
The VOLTAGE IS CONSTANT, we only change the current and resistance, we already know the current is lower for IB, but the resistance is higher?

But its producing more heat, which means its pulling MORE power, because R has increased.

Why don't you use the formula above, and see how changing the resistance affects power consumption, google the equation, it should come up, like you said IB uses less current, that's correct lowering the current in the equation above, will reduce the value of P.
Increasing R on the other hand will INCREASE P....

Maybe Intel will license Resonant Clock Mesh process tech from Cyclo's. That would be an easy fix for them, plus once the worked out the kinks with Trigate tech they'd come out massively ahead. According to AMD implementation of that clock mesh is relatively simple with big power/heat benefits (10-15% less power)

Current however is a variable, found via: I = E / R. Raise the resistance with a fixed and you have less current.
Sidenote: Current is not something a CPU can have more or less of, current is defined by voltage and resistance.
In short, you cannot use just P = I^2 * R to calculate wattage with multiple resistances, unless you're raising the voltage AND resistance in your example. Voltage being fixed, you cannot raise resistance and leave current the same.
With static voltage a raised resistance will lower current dramatically, giving a lower wattage.

Please re-read my post above yours and the equations and explanations and/or play with the ohms law calculators I linked.



"Sound" is pressure waves in a fluid. Given the lack of fluids in a CPU I'm assuming you mean vibration.
Vibration in a solid decays rapidly into heat. Where a vibrating solid meets air it produces "sound" waves by pressurizing the air. Pressurizing the air creates heat. The sound waves in the air decay into heat as well.

Sorry, but I'm simply not able to compute that. It's above my mathematical ability. I understand the concept of more resistance = more heat, therefore the same current/voltage with a higher resistance will equate to more heat, though I'm never going to understand the equation.

Bobnova said:

Current however is a variable, found via: I = E / R. Raise the resistance with a fixed and you have less current.
Sidenote: Current is not something a CPU can have more or less of, current is defined by voltage and resistance.
In short, you cannot use just P = I^2 * R to calculate wattage with multiple resistances, unless you're raising the voltage AND resistance in your example. Voltage being fixed, you cannot raise resistance and leave current the same.
With static voltage a raised resistance will lower current dramatically, giving a lower wattage.

Please re-read my post above yours and the equations and explanations and/or play with the ohms law calculators I linked.



"Sound" is pressure waves in a fluid. Given the lack of fluids in a CPU I'm assuming you mean vibration.
Vibration in a solid decays rapidly into heat. Where a vibrating solid meets air it produces "sound" waves by pressurizing the air. Pressurizing the air creates heat. The sound waves in the air decay into heat as well.

Click to expand...

Of course you can't for multiple resistance, but the fact of the matter is, overall IB has a higher resistance because of the smaller components, we are looking at this from a theoretical perspective....

And yes, voltage,current, power and resistance are variables, but at max load, the processor will be fed a constant current, and voltage (or near enough a constant current and voltage), for example lets say at max load, vcore is 1.3V and cpu power consumption is 95W maintained, this is the absolute MAX.

P=VI, therefore I = P/V = 95/1.3 = 73A
P=I^2 * R
P=73^2 * R

Now assuming current flow is the same, NOT that it will be, the only way to increase P to generate more heat is to increase R.

Btw, I'm not saying find a way to increase the power output, I'm simply trying to tell you that increasing the resistance, will LEAD to an increased power output.