Tamariniak

Tamariniak OP t1_ja3izr0 wrote

So please let me know if I'm understanding this right:

  • The stated voltage (12V in my example) is the maximum voltage
  • The stated current (0.3A in my example) is the maximum current
  • The current going through the fan can be calculated using Ohm's law, I = V / R, where R = Vmax / Imax (40 ohm in my example)
  • I just need to make sure the sum of the currents going through each of my fans is less than what the PSU is rated for

I saw some people elsewhere suggest that the current flowing through the fan is constant and not dependent on the supplied voltage, but that doesn't really make sense in my head since I don't think the fans could just magically reduce their resistance.

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Tamariniak OP t1_ja34giq wrote

So the voltage on a power supply is a constant supplied voltage and the current rating of the PSU is just its maximum current? Meaning I just have to match the voltage and get a higher (maximum supplied) current than the sum of the fans?

Also, would you happen to know how I could regulate the speed? Is 12V the slowest or the fastest? Is it safe to run the fans at a lower voltage than what their rating is?

EDIT: one more thing, at a lower voltage, would the current flowing through the fans be lower than the rating (via Ohm's law) or does the current somehow stay the same regardless of the voltage?

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Tamariniak OP t1_ja30ptt wrote

What about the voltage and current ratings though? What does it mean when a fan says "12V/0.3A"? Is that the maximum, the minimum, or something else? It makes me a bit scared to use lower voltages.

I also saw people saying that the fan "will always draw 0.3A regardless of the voltage," but I find that a bit weird. Surely a 12V/0.3A rating would mean that the fan has a resistance of 40 Ohms and the current would just be I = V / R.

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Tamariniak t1_isa5z6e wrote

That's what the article says it was. Also supposedly exploded with an unaccounted-for amount of force, shredding the metal canister and flinging the shreds out at the person's face.

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