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4 phase psu

post #1 of 8
Thread Starter 
Hi guys
im wondering, since idk about it, would 4 phase psu work on 8 phase mobos, like new ones that are out...
i have 2 psu, and both of them are 4 phase... my mobo does support 8 phase but never used... thx in advance
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post #2 of 8
The motherboard VRM doesn't care how the PSU is built so long as it meets spec for voltage/current/ripple etc.
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post #3 of 8
Are you talking about power phases or the PSU power cables? If it's the latter, 4-pin cables usually work in 8-pin sockets.
post #4 of 8
I've never heard of "4 phase" PSU. Do you mean "4 rail"?
If so, the number of rails of the PSU is not related to the phases of the mobo.
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post #5 of 8
Phase?

Do you mean as in 4 phase AC? In which case, why do you have that and why are you asking the question?

Or do you mean like phases in the VRM on the motherboard? In which case, it doesn't matter, and ATX PSUs don't have phases.


Or do you mean "pins". As in 4-pin ATX12V connector and 8-pin EPS12V connector. In which case, 4 pin can usually be used on an 8 pin motherboard as long as you're using a CPU with a TDP under ~100W.
post #6 of 8
1. Yes it will work

2. it's not "phases", it's 4-pin or 8-pin. The 8-pin is just a double 4 pin for reduced resistance.
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post #7 of 8
Quote:
Originally Posted by railfan844 View Post
1. Yes it will work

2. it's not "phases", it's 4-pin or 8-pin. The 8-pin is just a double 4 pin for reduced resistance.
I don't get it. If you deliver the same current over 2 identical paths (the 4-pin connectors) instead of 1, you should encounter more resistance (twice), not less...
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post #8 of 8
Quote:
Originally Posted by ronnin426850 View Post
I don't get it. If you deliver the same current over 2 identical paths (the 4-pin connectors) instead of 1, you should encounter more resistance (twice), not less...
With more pathways to choose from, the flow of charge encounters less resistance. Think about what happens if you have one low resistance path and you add a high resistance path. Most of the current still goes through the low resistance path--you're not increasing the total resistance by adding a second line. Otherwise, you can always think about free space as being a path with lots of resistance. If all the paths added together, then...

Or also think about what happens to the cross-sectional area if you increase the number of wires.

Anyway, for two resistors in parallel:
R_eq = (R1 * R2) / (R1 + R2)
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