Fans and ohm's law

[1234568]

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

 

[Taylortwocities]

I'm not an electrician, just doing a humble diy project.

Yes, I can see that!
The rating of the power supply is the maximum current that it can deliver. The fan will just take 0.13A and that's what will get provided by the power supply. It's like your car, it is capable of doing 130MPH, but you drive it at 30mph. You do not need a giant flywheel to get rid of the excess 100mph!

Just make sure that the voltage is correct and that your driver is a constant VOLTAGE driver and not a constant CURRENT one.
Sorted?

 

[1234568]

Thanks, a good analogy. So in short I do not need to worry. Out of curiosity, how does the circuit know the desired current? I may decide to drive at 30mph, but what limits the current in this situation?

 

[Taylortwocities]

what limits the current in this situation

The resistance of the fan's windings.

Its Ohms Law. V=IR so I=V/R so the greater the resistance is, the lower the current will be.

Revision here

 

[dmxtothemax]

you will need to shunt the extra current around the fan,
do you know the maths to work it out ?

 

[1234568]

The resistance of the fan's windings.

Its Ohms Law. V=IR so I=V/R so the greater the resistance is, the lower the current will be.

Revision here

Thanks for the link. The bit that throws me is the voltage is fixed at 12v and the resistance of the motor is fixed, based on the number of coils and wire gauge. Wouldn't ohms law imply the current also has to be fixed?

you will need to shunt the extra current around the fan,
do you know the maths to work it out ?

No. Any help would be appreciated!

 

[dmxtothemax]

What is the voltage and power output of the LED driver ?

 

[1234568]

12V 1.66A 20W
https://www.amazon.co.uk/gp/aw/d/B0191JTLZA/ref=yo_ii_img?ie=UTF8&psc=1

 

[Marvo]

If you're going to use an LED power supply I'd suggest a constant voltage supply rather than a constant current type.

I think that you're going to have an opposite problem if your power supply is too large. Electronic power supplies have a minimum load requirement for the output circuitry to give a stable voltage and often this figure isn't stated on the supply itself, you've got to go digging in the manufacturers spec sheets to find it.

If you use a power supply that's too big it won't cause too much current in the circuit, as already mentioned the current that will flow is governed by the impedance (resistance) of the item connected so it won't damage the fan motor but it might mean it won't give 12V and the fan simply won't run.

 

[1234568]

So this may be a better option at constant 12V and 1A? I'll still need to shunt the extra current around the 0.13A fan?

https://www.amazon.co.uk/gp/aw/d/B0...nt+voltage&dpPl=1&dpID=41dc9EQl7UL&ref=plSrch

 

[telectrix]

you need a 2A , 12V power supply. not sure if your fan is A.C. or D.C. though. the plug'in PSU you get with a netgear router will do if it's D.C..

 

[Taylortwocities]

you will need to shunt the extra current around the fan,

I really do not understand this. The fan will draw the amount of current commensurate with its resistance.
Where does this extra current come from???
If you put a shunt across the fan then the resistance decreases and the current drawn from the power supply will increase BUT the amount of current flowing through the fan will remain the same because
1. the voltage will still be 12volts
2. the resistance of the fan will not have changed.

Please, @dmxtothemax explain to me why the elecrtrical networks in your world are different to everybody elses

 

[telectrix]

who the hell mentioned the word shunt? that's a red sardine if i ever saw one. plug the fan into a car battery. clamp the current, get a psu a bit > the current drawn. NEXT?

 

[Taylortwocities]

who the hell mentioned the word shunt?

you will need to shunt the extra current around the fan,

 

[telectrix]

prefer werthers original.

 

[Marvo]

Yeah, I don't get that 'shunt' suggestion either......

 

[Lucien Nunes]

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.

 

[Pat H]

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.

 

[Lucien Nunes]

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.

 

[PEG]

Fluxing as lyrical as ever,Lucien