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post #21 of 23 Old 04-13-2013, 07:08 AM
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Originally Posted by Stay Puft View Post

I hope intel decides to give the i3s that igpu sherlock. It would be nice to play around with

No LGA cpu will get the "high end" graphics. Only BGA and PGA (mobile) sku's.

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post #22 of 23 Old 04-13-2013, 12:36 PM
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Originally Posted by TheBlademaster01 View Post

You're not completely right here, there's a difference between TDP and max/typical power consumption.

Also, what you're saying in the emboldened part is a little bit weird as well. What does the energy get converted to otherwise? Half of the electrical power gets dissipated into heat while charging the transistors and the other half when discharging. There is no other energy convertion.

The point is that TDP is a cooling/application design concept for a typical load. The load is variable, and TDP is only applicable to cooling purposes and all derivatives there from. If you write code that charges all transistors you can easily go over your TDP. If you have a leakier chip that deviates from the norm you're going to see higher power consumption than one that has less leaky transistors. A 2500K and 2600K have the same TDP, yet power consumption typically is not the same. An E5-2680 has the same TDP as an i7 3930K yet it has lower typical power consumption than the 3930K.
For the emboldened part, there is a minuscule amount energy that is stored instead of dumped as heat, and TDP is a platform design target, not necessarily a chip by chip target, a 2600k uses more power(under load) than a 2500k because more of it's circuitry is active simultaneously, even though it's the same die. If the 2500k weren't targeting the same platform (stock heatsink, and same set of moththerboards) as a 2600k, it would probably be listed at a lower TDP. Similarly for ivy bridge, it's listed TDP is much higher than typical load because Intel wanted the new motherboards to implement a VRM capable of powering the less efficient sandy bridge parts.

Bottom line, TDP isn't enough to know how big a heatsink you need, but it IS enough to know how big a VRM you need. In order to determine how big a heatsink you need, you also need information on the allowable temperature and thermal conductivity of the CPU package.

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post #23 of 23 Old 04-13-2013, 01:04 PM
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For the emboldened part, there is a minuscule amount energy that is stored instead of dumped as heat, and TDP is a platform design target, not necessarily a chip by chip target, a 2600k uses more power(under load) than a 2500k because more of it's circuitry is active simultaneously, even though it's the same die. If the 2500k weren't targeting the same platform (stock heatsink, and same set of moththerboards) as a 2600k, it would probably be listed at a lower TDP. Similarly for ivy bridge, it's listed TDP is much higher than typical load because Intel wanted the new motherboards to implement a VRM capable of powering the less efficient sandy bridge parts.

Bottom line, TDP isn't enough to know how big a heatsink you need, but it IS enough to know how big a VRM you need. In order to determine how big a heatsink you need, you also need information on the allowable temperature and thermal conductivity of the CPU package.

 

I never spoke about heatsinks. I spoke of cooling purposes and derivatives. Meaning everything involved with the thermals of the chip. The term stands for Thermal Design Power, you don't have to redefine the term lol. And TDP also has to do with binning, better binned chips have a lower TDP (meaning SKUs, not individual chips).

 

And no, energy does not get stored in the IC. The transistor works as an integrator, it integrates the current with respect to the time until it is fully charged and stores the energy in the electromagnetic field. Upon switching the pull down network, the field gets completely destroyed. Even if you state that the field is not completely destroyed, the integrator will be charged faster upon the next clock event. There's also no AC component in the semiconductor so all power is active and load resistive. All electrical power that goes in the chip will be converted to heat. I know how ICs work wink.gif

 

Electrical Energy -> Thermal Energy that's all. 


 


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