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I am trying to measure the current of a DC computer cooling fan (Intel E18764-001) at various voltages. Its rated specs are: 12V 0.20A. Check it out here.

So I am suspecting that this is telling me: "give it 12V and you should get 200mA of current" - Am I right?

Anyway, I supplied the fan with various voltages using my power supply starting from 5V all the way to 12V. I measured the current with a multimeter connected in series. Here are the results:

  • @5V --> 50mA (0.05A)
  • @6V --> 30mA
  • @7V --> 30mA
  • @8V --> 30mA
  • @9V --> 40mA
  • @10V --> 40mA
  • @11V --> 50mA
  • @12V --> 53mA

Interesting results. Here are my questions:

  1. Why isn't there a proportionate increase in current with every increase in voltage? In fact, after 10V, the current goes up as voltage goes up! Does it have anything to do with either the fan going all funny because it is not getting its designed voltage or the wires can't handle any current more than 50mA? (Last one unlikely as that still doesn't explain the drop to 40mA.)

Why are the current readings nowhere near the power rating of 0.2A?

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    \$\begingroup\$ perhaps your multimeter is choking the current to ~50mA Max. Could be damaged, with a high resistance? \$\endgroup\$
    – KyranF
    Commented Oct 22, 2014 at 8:58
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    \$\begingroup\$ I once taught my uni's robotics club some basic electronics, using DC computer case fans and transistors etc. The fans were 12V, and showed 120mA on the datasheet. Lo and behold, every single one of them pulled almost exactly 120mA from the bench power supply when given 12V (directly, or when MOSFET used as low side power switch). So I think there is something wrong with your set up, or that fan is special and has built in control circuitry that you have not shown us. \$\endgroup\$
    – KyranF
    Commented Oct 22, 2014 at 9:35
  • \$\begingroup\$ For example that fan seems to have either 3-phase brushless, or it has a tacometer and something else. \$\endgroup\$
    – KyranF
    Commented Oct 22, 2014 at 9:36
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    \$\begingroup\$ And as @CamilStaps said, it's probably a maximum DC average rating, however you have a complex CPU fan there, not an ordinary one, it has 4 pins not 2 or 3 (simpler ones), so I have come to the conclusion that you are not driving it correctly. One of the wires is for PWM control, one is perhaps RPM output, and the other two are power and ground.. play around with it more and see what happens. \$\endgroup\$
    – KyranF
    Commented Oct 22, 2014 at 9:42
  • \$\begingroup\$ Hi @KyranF and Camil Staps thanks for the advice. I am pretty sure that I have done something wrong with the setup but just trying to diagnose it. I think I got the wires correct - ie not using PWM and Tacho. I'll play around with it a bit more. Many thanks \$\endgroup\$
    – Gil
    Commented Oct 22, 2014 at 9:45

5 Answers 5

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Since DC motors are also generators, the reverse voltage raises proportional with the RPM. So the higher the RPM, the higher the reverse voltage -> less current. Also the ambient temperature is most probably not stable, which leads to fluctuating current measurement results.

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  • \$\begingroup\$ I think this is rather speculative. That fan is probably a BLDC. While its controller may let it look like a reverse voltage, it probably isn't. \$\endgroup\$
    – Ariser
    Commented Oct 2, 2019 at 4:13
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Motors are inductive loads and work differently than resistant loads. The resistance of a motors coils may only be a few ohms while stationary, but while spinning its resistance is proportional to its rpm. The faster the motor spins, the more magnetic flux is generated, and the higher the electrical resistance becomes. This causes an equilibrium speed to be reached; the motor's final amperage is equal to input voltage divided by resistance.

Under-voltage, or a heavier load will cause the motor to slow down, a slower rpm means less resistance and increased amperage. This explains why the lower voltages are a higher power use, then it hits a sweet spot, most likely the motors most efficient speed. Further increasing the input voltage increases the amperage, increasing the rpm slightly.

Lastly, about motor rating: just because the motor is rated for 200mA at 12v, doesn't mean the fans load is big enough to reach that rating. This fan seems to be running at roughly 25% rating, but that's probably a design feature to limit current, noise, and increase efficiency and longevity. I suspect you could put a more aggressive or heavier fan blade on the motor and still be within spec.

I know this is almost 6 years old, but I hope someone in the future see this and it helps them. Sorry if my description isn't 100% accurate.

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    \$\begingroup\$ Welcome to EE.SE! "The resistance of a motors coils may only be a few ohms while stationary, but while spinning its resistance is proportional to its rpm" That's not correct. Coil resistance remains the same but back EMF comes into play. Please read up before posing an answer. \$\endgroup\$
    – winny
    Commented Oct 15, 2020 at 9:52
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With motors you actually have different current draws one being the starting current of the motor (LRA- locked rotor amerage) and the other being the running current (FLA- full load amperage). Typically, and it depends on the type and design of the motor (there are a few) for a 3 phase squirrel cage motor the LRA is 5-7 times the FLA. So, if I were to guess, and I am because in the commercial electrical industry we don't use pc fans and the motors we use have very descriptive nameplates as per NEMA (national electrical manufacturers association) I would say that maybe the current rating stated is the starting current.

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Are you using a true RMS multimeter? Remember that the fan is PWM controlled, so the current is not a DC current. Maybe you are actually measuring the DC level of the fan?

Also, as Keelan mentioned, it vary very much depending on external measures. I remember when I was designing a helicopter; If I held the copter down while applying constant throttle (voltage), the current was much higher than when I released it with the same throttle due to the drop in resistive force applied. If the fan has an easy path of air flow, it may not draw as much current as expected. Also, I think the label represents average current consumption.

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  • \$\begingroup\$ Thanks for the clues. I'll have a play around with it a bit more with your advice. Cheers \$\endgroup\$
    – Gil
    Commented Oct 22, 2014 at 9:46
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Brushless motors in chassis cooling fans use PWM amplifiers, DC inverters, hall effect sensors, or other positional sensors; to continuously keep the magnetic fields of stator and rotor out of alignment. In the attempt to align the fields, the motor produces torque, which gives an impetus to rotate. Solid-state switching, microprocessor control, PWM, or inveterter modulation control speed and torque, making the current rating printed on the dc fan meaningless in terms of inline DMM current measurement.

The current spec is an average determined by the manufacturer, at a certain voltage, ambient temperature, and no air resistance. Speed control leads sometimes must be tied to one of the power leads, with a certain value resistor, to run the unit at full speed. Other designs cause full-speed operation with the control lead dangling.

The stated current rating is not to determine cooling performance. It is a design parameter used to determine supply current requirements, overall power consumption, or, where the fan is specified after the fact, by the end user, to supply a specification for maximum current.

Frequently, as was mentioned above, small fan motors use only a fraction of the rated current, to decrease noise and increase service life. However, such designs will increase current draw to maintain speed as resistance to airflow increases due to dust accumulation, air-filter resistance, or placement with inadequate clearance from walls.

CFM is the most pertinent factor in cooling performance, and this has 'almost' nothing to do with current rating, except under conditions determined by the manufacturer.

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