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Trying to get my head around de-lidding effects

So I’ve got an 8700k that I’m currently keeping cool enough @ 5ghz but delidding is on my mind

 

ehat I’m trying to understand is, we all know that delidding these things sees a good drop is cpu temps.

 

but what sort of impact does it have on the cooling system? I assume since the cpu is still generating an identical amount of heat the cooker will run under the exact same load? Or am I missing something and deliddig will put less strain on the cooler?

 

basically If my cpu is already <80 degrees and I don’t want any more performance/voltage/OC would I actually see any benefit with a delid?

 

will my cooler be under less strain or will it be identical?

 

thanks

 

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From what I remember the main performance gain through delidding was the fact that there is better contact between the IHS and the die because the silicone holding down the IHS is removed, so it sits a little lower. Liquid metal of course also helps.

If you're content with what you have now, then no, there's no reason to delid (and risk breaking an expensive part). It can improve temperatures though, so the choice is yours.

Crystal: CPU: i7 7700K | Motherboard: Asus ROG Strix Z270F | RAM: GSkill 16 GB@3200MHz | GPU: Nvidia GTX 1080 Ti FE | Case: Corsair Crystal 570X (black) | PSU: EVGA Supernova G2 1000W | Monitor: Asus VG248QE 24"

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3 minutes ago, Jay Deah said:

currently keeping cool enough @ 5ghz but delidding is on my mind

don't

 

if it's working fine then theres no reason to delid it. 

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the whole point of delidding is improving thermal conductivity, if you are pushing 5ghz, in theory you could do a direct die cooling with a special cpu block and further improve your temperatures and maybe push even higher clock speeds without increasing your cooling.

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You will almost certainly benefit as long as your cooling system isn't already thermally soaked and maxed out. It really depends how many RPMs of fan you are willing to listen to since cranking up your RPMs will reduce the heat soak a bit, but there's only so much heat that can be dissipated from a given mass and surface area with a set fan rpm. Your entire system will be more thermally responsive, for better or worse. (Generally better in the world of desktops.)

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5 minutes ago, Jay Deah said:

will my cooler be under less strain or will it be identical?

From my admittedly very loose theoretical understanding. There will be slightly more strain on your cooler.

 

Yes, the amount of heat produced by the chip is the same. However the rate at which it can transfer to the cooler via the LM/TIM to the IHS to the cooler different. The removable of the stock TIM (or changing the Z height, blah blah, aint going to argue that here) means there is less of a heat blanket around the die.

 

So assuming you are not using pure garbage thermal compound on your IHS--Cooler, the heat from the chip is more efficiently moved away from the die when you use LM. So there should be a higher rate of energy transfer between the die --> cooler, since you are reducing the previous bottleneck, which was the die --> IHS contact.

 

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@For Science! is correct.

 

Your CPUis generating the same heat over time but de-lidded  it will transfer that heat away from itself (and into the cooling system) faster. So the coiling system will be hotter, while the CPU will be cooler.

 

Another way of putting it: With better thermal conductivity between the parts of the system, those parts will be closer to equilibrium. 

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