"Dave D" wrote in message
...

"Andre" wrote in message
m...
Hi group .

I recently scored a pile of 5V 1W fans . Problem is, for what I want
them for they are too fast, and suck too much current . (200 mA)

I am left with :-

1) Adding a constant current to limit current (and therefore speed)-
easy enough to do . How come there isn't a 100mA "constant current"
diode ?!?!?!
Also I'd need the regulator to drop less than a volt for this to work
.

Put two silicon diodes in series in the 5V supply, works perfectly.


2) Using a lower supply voltage (this would make things worse !)


How? That's exactly what you want to achieve!

3) Using a different fan . (too expensive)

4) Modifying the fan somehow to limit the speed (possible, adding a
capacitor across the Hall sensor output to limit the speed would
likely work but wouldn't solve the current problem)


Are you saying these are speed regulated fans? If so removing the hall
effect sensor might be the first step as they will always attempt to make
up
for your efforts to slow them down. Sometimes (often in PC fans) the hall
effect sensor is there simply to report the rpm to the device it is
connected to.

5) Pulsing the fan voltage so it gets 4V mainly with x % 5V pulses
(this would keep the speed reasonably constant) - anyone done this ?
The idea is to allow the logic to work so fan won't stall, but
limiting the current enough so that it keeps the speed down .


Rather like a sledgehammer to crack a nut! Try the diode idea first.

Dave


In my experience with 12V fans, a series resistor works nicely, as does
running them from a stiff 6V supply.
Unless you have something unusual, these fans do vary RPM in response to
power supply voltage.

You will have to determine the resistor by trial and error, since the fans
do not respond linearly to voltage.

"Robert Morein" wrote in message
...

snipped

In my experience with 12V fans,

They're 5V fans.

a series resistor works nicely, as does
running them from a stiff 6V supply.

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.

Unless you have something unusual, these fans do vary RPM in response to
power supply voltage.

You will have to determine the resistor by trial and error, since the fans
do not respond linearly to voltage.



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

"Dave D" writes:

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.

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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"Sam Goldwasser" wrote in message
...
"Dave D" writes:

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.

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.


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

Dave D wrote:

"A E" wrote in message
...
Dave D wrote:

"Robert Morein" wrote in message
...

snipped

In my experience with 12V fans,

They're 5V fans.

a series resistor works nicely, as does
running them from a stiff 6V supply.

A series resistor generates heat, diodes drop voltage and run cooler..
They

So, you're saying with the same VI, a diode generates less heat than a
resistor???????????


Nope, see Sam's post. It will always drop a relatively fixed voltage,
regardless of current draw, so it is more efficient than a resistor in this
application.

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?

Dave D wrote:

"Sam Goldwasser" wrote in message
...
"Dave D" writes:

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.

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.


Exactly, so the diode will run cooler, like I said :-) Naturally, the losses

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.

"Dave D" wrote in message ...
"Robert Morein" wrote in message
...

snipped

In my experience with 12V fans,

They're 5V fans.

a series resistor works nicely, as does
running them from a stiff 6V supply.

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.

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


Unless you have something unusual, these fans do vary RPM in response to
power supply voltage.

You will have to determine the resistor by trial and error, since the fans
do not respond linearly to voltage.



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

Interesting idea - thanks


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

Dave

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).


"A E" wrote in message
...


Dave D wrote:

"A E" wrote in message
...
Dave D wrote:

"Robert Morein" wrote in message
...

snipped

In my experience with 12V fans,

They're 5V fans.

a series resistor works nicely, as does
running them from a stiff 6V supply.

A series resistor generates heat, diodes drop voltage and run
cooler..
They

So, you're saying with the same VI, a diode generates less heat than a
resistor???????????


Nope, see Sam's post. It will always drop a relatively fixed voltage,
regardless of current draw, so it is more efficient than a resistor in
this
application.

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?

A E wrote:
Dave D wrote:

a series resistor works nicely, as does
running them from a stiff 6V supply.

A series resistor generates heat, diodes drop voltage and run
cooler.. They

So, you're saying with the same VI, a diode generates less heat
than a resistor???????????


Nope, see Sam's post. It will always drop a relatively fixed
voltage, regardless of current draw, so it is more efficient than a
resistor in this application.

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?

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.

"A E" wrote in message
...
Andre wrote:


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

Interesting idea - thanks

Except it does nothing more than a resistor, and a 0.6V drop at a supply
of 5V gives you 12%
increments, hardly small.
I'd stick with a resistor, or go the PWM way, that's the right way to do
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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"A E" wrote in message
...

dave's theory was that a diode was more 'efficient'
than a resistor, which doesn't compute.

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
...
Andre wrote:


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

Interesting idea - thanks

Except it does nothing more than a resistor, and a 0.6V drop at a supply
of 5V gives you 12%
increments, hardly small.
I'd stick with a resistor, or go the PWM way, that's the right way to do
it.

Dave D wrote:


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 cap is cheap.


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.


Hm, true.


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

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...

greater effect on startup torque and running torque than a diode.


Hm, yes yes I see now.


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.

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...

Sam Goldwasser wrote in message ...
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.

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 .


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"Andre" wrote in message
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

On Wed, 30 Jul 2003 13:19:08 +0000 (UTC), "Dave D"
wrote:

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!

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

--

+ Ken +

"Dave D" wrote in message ...
"Andre" wrote in message
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!

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


Dave

Andre,

There are current regulator diodes.
http://www.vishay.com/docs/70711/70711.pdf
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

(Tom) wrote in message . com...
Andre,

There are current regulator diodes.
http://www.vishay.com/docs/70711/70711.pdf
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.

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


Tom

"Andre" wrote in message
om...
(Tom) wrote in message
. com...
Andre,

There are current regulator diodes.
http://www.vishay.com/docs/70711/70711.pdf
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.

Interestingly, you cannot parallel LM317x regulators to get more
current .

Try using the LM317 to drive a pass transistor capable of passing the
required current. Mind, the LM317 is more than capable of supplying the
200mA you say your fan requires!

Dave

"cpemma" wrote in message
...
A E wrote:
Dave D wrote:


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.

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...

Snag with 317 is that 1.5-2v loss; better method with a 0.75v loss and 3-4
parts ;-)

http://www.cpemma.co.uk/ef.html



Alternatively the TL431 adjustable zener driving a pass transistor. (The TL
431- aren't they a great little device ;-))

Dave