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I am hoping for some advice on powering a 12v computer fan from the mains. I'm not an electrician, just doing a humble diy project.

The manufacturer states the fans actual input power is 1.6 W, with a maximum rated input current of 0.20 A. So, I believe I ideally need a current of 0.13A (1.6/12).

I'm hoping to achieve this with a 12v DC led driver. My issue is most of these provide far greater than 0.13A. I assume excess current above the 0.2A max will kill the fan.

Can this be solved with a simple resistor? I have a led driver rated at 12V and 1.66A. What resistor is necessary to meet the correct ampage and dissipate this much power?

Many thanks for any help
Ed
 
Ignore all this waffle about shunts and ohm's law etc. The fan requires 12V - connect it to a regulated 12V power supply with rated output of an amp or so. Job done. E.g.
PowerPax UK SW4010G 12VDC 1 AMP Mini Plugtop Sw Mode PSU - https://www.rapidonline.com/Electrical-Power/PowerPax-UK-SW4010G-12VDC-1-AMP-Mini-Plugtop-Sw-Mode-PSU-85-3070

Your LED driver might do the job if it is a constant voltage type and does not require a minimum load. If it doesn't like such a low load as the fan, it's easier to get a regular power supply than to faff about providing that extra load to keep it happy.

BTW motors don't obey ohm's law; the windings do have resistance but that is not what determines how much current they take.
 
The key points are:
You need a fixed voltage of 12V DC
The power supply needs to be able to deliver the required power or more and be able to deliver the voltage at an output current as needed for the fan (as mentioned some PSUs need a minimum load to deliver an output)

You can't really use Ohms law on a fan as its not just resistance but impedance that's in play. The fan is made up of coils and the generation of and collapse of magnetic fields will alter the power factor of the circuit so the "resistance" will vary.
But the fan is small and the current low so its unlikely to be an issue.
 
the generation of and collapse of magnetic fields will alter the power factor of the circuit
There's no power factor... the supply is DC!

Winding inductance plays a significant part in the commutation behaviour of a motor but that is purely an internal parameter. From the outside, any permanent-magnet motor, whether brushless like a computer fan or with a copper commutator, obeys an approximate proportional relationship between torque and current, also between voltage and speed. On a fixed voltage supply, the voltage determines the speed, the speed of the impeller determines the load torque, the torque determines the current. So the relationship between voltage and current in theory is that of the fan curve, the motor simply introduces a further constant of proportionality which is the flux constant. However, in a small motor like a brushless fan, the motor departs very far from the ideal and the proportionality is lost, so the motor plays a part in defining the V/I curve as well as the impeller.

OP: Re. ignoring waffle, my waffle is better because it's technically correct, but you can ignore it all the same.
 

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