Re: Reduce power draw on a fan ? (or, speed control) ?

They're 5V fans.

A series resistor generates heat, diodes drop voltage and run cooler.. They can be soldered into the fan leads then covered with heatshrink sleeving. A

6V supply will speed these fans up.

Silicon diodes in series allow small increments and an easier way to select the required speed.

BTW, you replied to my post, I did not post the query :-) Maybe your news service isn't showing all the posts?

Dave

Not quite. :) If you have a voltage drop and current flow, you have IV=P generates the same heat. Doesn't matter if it is a diode or a resistor. The difference is that the power dissipation is roughly proportional to current rather than a square function because the voltage drop is more or less constant.

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They

sleeving. A

Exactly, so the diode will run cooler, like I said :-) Naturally, the losses are to heat, I didn't mean that diodes magically dump the energy into a black hole!

Dave

Makes no sense. You will simply not draw more current through that diode than with a resistor, because the load is the same... If you put a resistor in series with the fan, and the resistor happens to drop .6V (simple to figure out, if you can draw a load line, otherwise a bit of experimenting), the same amount of current will go through that resistor as through a diode... Same heat dissipation in both cases. Don't you think? :)

It'll only run cooler than a resistor in the case of increasing the input voltage to the fan-diode combination, but then, the fan starts to run faster, which is goes against what the OP wants, so it's no longer doing its job! Therefore, it's not efficient. So you add another diode to drop the extra voltage, and guess what, you're back to dissipating the same as a resistor... "Efficient" would be if the diode did the same thing as a resistor dropping .6V, while dissipating less heat, which it doesn't. Given an input voltage, a given fan, and a given target of reduced RPM, and you only have diodes and resistors to do it with, you'll dissipate the *same* amount of heat in either case. There's no 'efficiency' to using a diode... You *will* have to drop the *same* voltage with either a resistor or a chain of diodes, and the current will be the same in both cases. There's no magic about this. P=VI.

I tried running the unit off 4.5 V and it works - but almost no torque .

Interesting idea - thanks :)

I'm with you AE on this one. Same voltage drop * same current = same power.

If that's too much power (unlikely) then then the only way to beat this is to use a digital speed controller (eg PWM).

cooler..

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The major drawback with resistors is that the v drop depends on current, and at start-up current is high, so v drop is high - the supply is anything but "stiff". Aiming for say 7v on a 12v fan, the fan may not even get going. Diodes drop a fairly steady voltage over the range of fan currents likely.

select

of 5V gives you 12%

it.

I don't think PWM is the right way to drive an electronic brushless motor, unless the pwm waveform is converted to a dc level with a capacitor. PWM is a great way to drive a brush motor, I don't think it would have a desirable effect on a brushless one, it may even damage the circuitry in the fan.

Your claim that a diode does nothing more than a resistor is not true. It provides a (approx) 0.6V drop without reducing the available current significantly, not the same as a resistor at all. A resistor will have a far greater effect on startup torque and running torque than a diode.

Anyway, the best way to control one of these brushless motors IMO is a LM317 or similar regulator with a pot so the OP can set the exact speed he wants. Not the simplest way but a sure way of getting the speed right.

Dave

One thing to note: In order to guarantee that the fan starts with a series resistor, a large electrolytic cap (e.g., 500 uF) should be placed across the resistor or diode(s). Then, power will initially be applied at full voltage long enough to get it going.

A typical 12 V 3-1/2" fan will run just above stall speed with a 200 ohm resistor in series to 12 V. I've built regulated temperature controllers by using the speed range from this to full speed in a feedback loop.

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OK, cheerfully withdrawn! Maybe it was a bad choice of words. Let's just say IMO a diode is a cleaner way of dropping voltage because a 0.6V drop (approx) can be predicted, and all available current is still available, unlike a resistor. I'm well aware of the maths regarding dissipation, on reflection I didn't really say what I meant.

Dave

I wouldn't mix diodes and resistors when it's about voltage drop... You're right about P=V*I, what you forgot is that P also equals R*I2.... Now: go and measure the diode's resistance and do this calculation for the same voltage drop and the same current as your resistance. You'll then see the difference. Apply any voltage acros the resistor the current is goind to be V/R. Do the same thing with the diode: it won't last long, first of all, but if you keep the voltage in reasonable limits so you won't exceed the maximum current of the diode you'll see that there's a differnt relationship between voltage and current and it's definitely non linear. The resistor is a component with high (I mean from a bit more than a copper wire to something that one might consider an insulator) resistance while the diode drops voltage because of the P-N junction polarization voltage. Think of the P-N junction as an insulator (which it really is, even if it's called "semiconductor") which starts leaking electrons in a direction if properly polarized. If the voltage across the diode is less than 0.6V for Si and 0.3V for Ge the diode acts like an insulator. It won't open at all because the electric field acros the P-N junction is not strong enough to free electrons. If the voltage goes over the above mentioned values some elctrons are freed (remember that the main, only and single difference between conductors and insulators is the status of electrons inside it...) and it turns into a conductor. That 0.6 to 0.7 volts remains constant no matter the current passing through the diode. The ideal diode won't drop more than 0.6 volts, the real diode drops a bit more because other than the semiconductor effect the P-N junctions has it's own internal resistance (imagine it as an ideal diode series with a tiny resistor). Anyway, the internal resistance is much lower than that of a resistor you would use for the same purpose, if you don't exceed the rated current of the diode you can consider it's resistance "zero". We can now return to our problem: first of all I never tried to reduce speed of a brushless motor... I must admit I never thought of it. The PC power supply fans are sometimes quite noisy, the processor fans are really good, I don't remember one noisy enough to make me want to lower the rpm. I also believe that 12V motors are easier to control than 5V anyway, you can drop the voltage quite a lot until you reach the point where the transistors won't polarize any more. What would a capacitor do in parallel with a hall sensor? In a brushless motor a hall sensor is used to determine the position of magnetic poles of the rotor and trigger the energization of the corresponding coil to create the spinning magnetic field. This scheme simulates the comutator and brushes in a nomal motor motor. What you can try is to insert a little diode series with each coil inside the motor. You'll keep the same polarization voltage for the control circuits so you'll still have the transistors properly polarized, but the voltage on each coil is going to be 0.7V lower which will result in a lower torque. Or you may try a resistor... If you have enough room inside, of course (I don't think so). But I think that'll do the trick. Good luck!

SM

"A E" wrote in message news: snipped-for-privacy@videotron.ca...

select

of 5V gives you 12%

it.

>

A cap is cheap.

Hm, true.

True for transient state, but I think that in the steady state, a resistor will do fine, esp. in this case where the resistor will be small. There's still the question of the only having 0.6V increments to play with. Unless you put in a bunch of tunnel diodes, Schottky and Ge diodes for extra increments... :)

Hm, yes yes I see now.

Simple enough, though. 5 parts, and we can all agree on that! (input cap, LM317, output cap, resistor, pot) This is more complex than required. Maybe the OP can just remove half the fan's blades instead...

Hi Sam,

That's interesting, thanks ! :)

Putting a capacitor across a constant current diode (LM317 + resistor) might work also, for the same reason .

BTW, thanks everyone for your suggestions so far, most useful ! :)

-A

Next problem . How to delay turn-on long enough to charge a fairly hefty capacitor to boost start the fan ?

I'm thinking some type of MOSFET, and a capacitor/resistor combination to get the 5 second or so delay . Possibly also a zener diode .

e.g. it waits until the voltage is more than say 3.6V and start charging the capacitor . When the voltage hits the MOSFET's turn on voltage it starts, dumping the charge stored in the main reservoir capacitor into the motor .

No need for that. The cap is in _series_ with the motor, and in _parallel_ with the resistor or diodes or LM317. On startup, the cap acts as a very low resistance shunting the resistor to apply full voltage to the fan. When it charges up its resistance goes high and it is effectively no longer there, so the resistor does its job. When the fan is powered off, the resistor discharges the cap so it's ready to go again. Great idea, so simple like the best of them!

Dave

It's the best annd it works perfect! I use it since many years ago.

--
 
 + Ken +

Yeah, should work well . With a few tweaks of my own (to take into account a slowly rising supply voltage) . Email direct for more information .

-A

Andre,

There are current regulator diodes.

and there are others. These range from .43 ma to 4.7 ma. May not be enough to keep your fan going, but you could put a few in parallel to adjust.

Tom

Interestingly, you cannot parallel LM317x regulators to get more current . Even if you use diodes to separate them it will not work .

2*100mA regulators give about 143 mA or so .

Any ideas why ?

-A