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Question regarding power supply heat output

ut you where implying that heat is purposely dissipated through the wires. This is not the case. The wires generally get warm from carrying the current
purposely really? So wires get warm because they generate cold? :banghead: maybe i should have worded my post differently.....
 
I wholeheartedly disagree when someone says a higher rating is not worth the cost over a lower rating. Talking about ROI for PSUs is plain silly. If you consider the saying, "you get what you pay for", then count how many times people post on TPU in regards to PSU failures, you should 2nd guess those that are saying Gold+ is better than Platinum+. As long as the buy does the research and compares the benefits and minds the review ratings, finding a PSU within the budget should not be an issue.

I'd also like to add that electronic heat is expelled in cabling, its a small amount for sure, but that also depends on the size of the wire.

Yes perhaps different wording would bring a different meaning.
But general rule of thumb is to avoid dissipating heat through cables.
And I stand firm, you dont want hot cables ever.
 
So, let's assume my current power supply is 80% efficient, and my rig draws 350w, pulling ~440w from the wall. Let's also assume this is a constant load, all day every day. Does this mean that 90w is lost as heat? Now, let's assume I replace it with a unit that is 90% efficient. This means the rig now pulls ~390w, roughly 50w less than it currently does. Would this mean I'm generating 50w less of heat? Is this any meaningful amount at all?
Regardless, I would simply buy a unit with a bit more headroom on it. A quality 600W gold is a good sweetspot.

Look at the 80 Plus rating tier (wiki) and you can see the difference in efficiency... it isn't much.

And those talking about PSU efficiency curves..... again, look at the 80 Plus ratings. You'll notice that from the measuring points (20%/50%/100%) the tiers are, at most, 3% different. So its a pretty damn flat curve once a 'real' load is applied wavering, at most, 3%. Notice the graphs don't start at zero............... So while efficiency is the best around 50%, the difference in reality isn't much at all. And certainly NOT worth running 2x the needed power.
 
I wound up getting this one in the end:


It's got a little more power behind it than my current 550w, and it's also Gold rated, rather than "80+" (not even Bronze). I cheaped out a bit from my original plans for going with a Titanium unit, but I had a ton of other crap to buy and this power supply should be a fair bit better than what I have... and I already know FSP to be a decent brand.

...I spent nearly $500, basically building a whole new computer... low end, but modern.
 
If a PSU is 60% efficient, 40% of the load current is being dissipated as heat, in the PSU.

A PSU sending 7A down an 18awg wire is losing 6.385ohms/1000ft x current^2 in watts, or 6.385(r) x 7A(i)= 0.044695V per foot, or .312W/ft.
E=ixr; P=e^2/r or P=i^2xR
A pin (any pin, really) on the power supply is rated for 7A, so multiply/divide for the number of pins, and length of cables.

The cable drop is not visible to the PSU, so losses there are in addition to PSU losses.

All Losses show up as heat. Entropy always increases.

A 650W PSU that's 80 % efficient, fully loaded is dropping 650x0.2=130W as heat, plus any wiring losses.
 
If a PSU is 60% efficient, 40% of the load current is being dissipated as heat, in the PSU.

If a PSU is 60% efficient, that means for every 100 watts the load demands, the PSU must pull from the wall ~167 watts (167 x .6 = 100.2) to compensate and make up the difference. And that extra 67 is indeed wasted in the form of heat. The point is, the load (in this case, the computer or motherboard, CPU, GPU, RAM, drives, etc.) still gets its full 100W.

The PSU does not "drop" any wattage. A quality 650W 80% efficient PSU that is fully loaded will still output 650W if that is the demand. But it will pull from the wall ~812 watts (812 x .8 = 649.6).

an 18awg wire is losing 6.385ohms

The cable drop is not visible to the PSU
Wires don't "lose" ohms (resistance). They add ohms to the circuit. As the length of the wire increases, the resistance (ohms) in the conductor (wire) will "increase" too. And that added resistance is indeed seen by the PSU as a slightly (very slightly) greater load.

So while it is true wires do "add" resistance, the wires/cables used for computer power supplies are not really long enough to be significant. However, the load will still need what the load demands. So if the cable adds a little resistance, the PSU will simply see that as a slightly larger load, and will compensate for it. The PSU does not know that added resistance (demand) is caused by the wires, but it still sees the added demand because that added resistance is just part of the total load put on the PSU.

Frankly, cable connectors present a significantly greater potential for more resistance than the wires themselves. A straight wire is just that, a straight wire. Assuming it is not damaged or kinked, it is going to have the lowest resistance possible through that conductor.

But when you have a connector in the middle of the cable somewhere (say on the PSU's housing) that wire is soldered to one end of the male pin. The male pin must make a solid "mechanical" connection to the female socket. And the another wire is soldered to the other end of the female connector. That's 3 points of potential added resistance (and potential failure) that otherwise would not be there. The solder joints could be "cold", contaminated, or faulty in other ways. The "mechanical" connection (essential for a quality electrical connection) of the male and female parts of the connector could be loose, dirty, or otherwise improper and add resistance.

Those potential issues with cable connectors is one reason why many professionals don't like modular power supplies. They would rather the power connectors be directly connected to the tap or PCB connector point inside the PSU instead of going through a connector in the PSU case.
 
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