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Power Supply Capacity Formula

I'm adding the TDPs (Thermal design power) for the GPU, CPU and HDDs and adding 20% of that subtotal for the motherboard, fans, etc...

Is this a valid approach? If I buy an 850W power supply but only use 600W, will my power supply use the same amount of power as an 750W would assuming similar efficiency ratings?

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marked as duplicate by Troggy Feb 24 '11 at 23:01

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Any of these answer your question? superuser.com/questions/16889/power-supply-capacity-formula , superuser.com/questions/86144/how-much-power-is-enough , superuser.com/questions/83569/power-supply-too-big Please flag for moderator attention if you feel this is not a duplicate and why it is different than what has been asked previously. Thank you. –  Troggy Feb 24 '11 at 22:59
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Power supplies are rated for their maximum output. So if you've calculated all the components and what they use at their maximum, then as long as your power supply can feed that needed power then it'll work. It's important to make sure you don't go over the rating of the PS though, as it'll most likely result in a voltage drop and it certainly isn't good for many of the components.

But to answer your other question, no if the PS only needs to put out 50W and it's a 1000W power supply (to use extremes), it won't suck 1000W out of your outlet. It'll suck near 50W. More efficient power supplies are more efficient at delivering power and use less themselves. Often bigger ones may "eat a little more themselves" for the fan and circuitry, but it shouldn't be huge.

Adding 20% is a safe option since you definitely don't want to hit the limit. But also keep in mind that you may add components to the system in the future, which is admittedly hard to predict.

Generally, one of the most expensive times is actually at boot time. That's when all the components fire up and are eating power near their maximum. Typically after that the power drawn is actually quite a bit less.

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