# Pulse Width Modulation Help

Discussion in 'Electronic Design' started by Mike, Oct 17, 2004.

1. ### MikeGuest

Hello everybody!

I'm seeking help in building a little device that will allow me to
control speed on high current fans from a low current fan header found
on motherboards.

Preface: The fan header on motherboards was designed to run small
fans, typically around 5W. Most motherboards nowadays support fan
speed control via pulse width modulation. Speed is controlled by
software which reads various thermistors (ex. CPU, hard drives,
ambient…)

What I basically want to do is have a high current 12v input from the
power supply, pulse width modulated low current 12v input from the
motherboard header, and a high current pulse width modulated output.

(Preferably with parts available from the neighborhood RatShack.)

2. ### Al BorowskiGuest

Since its just a fan, I suppose a simple transistor would work. It
depends on what the motherboard actually modulates - the 12V supply, or
the fan earth return. If the 12V is modulated an NPN transistor should
do the trick

12V
|
|
|
FAN +

FAN -
|
|
|
|
|
|
___ |/
+12V from MB------|___|-----|
|>
|
|
|
|

0V

created by Andy´s ASCII-Circuit v1.22.310103 Beta www.tech-chat.de

cheers,

Al

4. ### MikeGuest

If the 12V is modulated an NPN transistor should
Al, yes it is 12V modulated on 3 different motherboards I tested. I
also did find one interesting bit of information while testing this
out.

It seems there is a capacitor before the header, as when I was
measuring voltage on the header after turning it off via software, the
voltage took a good 3 minutes to settle at around 0.3V.

So I have 3 questions:
(please be patient as I have no formal education in electronics
engineering or prior experience :-] )

1.) I'm assuming that the capacitor before the header will make the
single transistor approach impossible, unless a load is presented on
the header. Resistor? Is that a good approach? And how would I
calculate watts from ohms? Given voltage.

2.) I found an old NPN TIP120 transistor laying around and tested it.
I'm getting 420 ohms between the emitter ad the collector and 1.6k
ohms between collector and emitter. Is it dead?

3.) If it is, would a new TIP120 be good for what I'm trying to do?
How many watts (ballpark) would it be able to handle on a small
heatsink inside a sealed project box?

5. ### Al BorowskiGuest

Hello,
If you are just measuring with a multimeter, you might just be measuring
the stray capacitance?
Theres a fair chance the loading from the transistor will be enough.

Resistor? Is that a good approach? And how would I
I'm tired, so theres likely to be a mistake in this, but here's what I
would do (note: The following are _very_ rough 'ballpark' figures that
should be OK for a one-off).

1) Assume that the 'capacitor' is just capacitance, and thus has a very
small value. Assume that 100ma current flow is enough to discharge the
capacitence quickly enough

2) Since I = .1, and V = 12, use ohm's law to get a resistor to use. V =
I*R, R = V / I, R = 120 ohms.

3) check the resistors power rating. P = V * I = 12 * .1 = 1.2W.
However, considering that the PWM probably won't always be at 100%, you
can probably get away with 1W. Or play it safe and use 2 1W resistors.

4) hook up the circuit as follows

12V
|
|
|
FAN +

FAN -
|
|
|
|
|
120 ohms | Collector
___ |/
+12V from MB------|___|-----|
Base |>
| Emitter
|
|
|

0V
No idea sorry.
It looks like it should be OK.
What kind of box? Metal?
No problem, but as I said I'm very tired right now, and may be speaking
crap!

cheers,

Al

6. ### Tim WescottGuest

Is the circuit one that drives the motor straight off of a PWM
transistor, or does it make a switching amplifier out of it with a
filter coil and capacitor to spare the motor? If it's filtered (which
the presence of a cap would indicate) then the simple transistor circuit
we're giving you won't suffice; you'll need something to turn the analog
voltage back into a PWM duty cycle.

You should check: load the output (with a resistor or a real fan). If
you have an o-scope then look at the voltage to the motor -- this will
be good because the simple circuit given is also sensitive to frequency;
above 1kHz you need to actually worry about the transistor speed. If
all you have is a DVM then all you can do is check to see if there's
significant AC on the pin, and hope that it's slow enough for your
transistor circuit.

7. ### MikeGuest

It seems there is a capacitor before the header, as when I was
I thought it would be impossible to find the actual capacitor, but the
are fairly large near the header, and can be traced with the naked
eye. There
is indeed one sitting right before each header, and it is 47uf @ 25v.
Unfortunately no. I tried another small NPN transistor that I found
and the
fan went into full power as soon as the header was turned on. It made
no
difference in fan RPMs if the header was set to 5% or 100%.
And given that the capacitance is now known ([email protected][email protected] here
right?)
based on that how would one calculate the speed at which it discharges
with
a given resistor value @ 12v?

And in this scenario, I guess we also need to know how fast the signal
is
pulsated? As I'm assuming if the capacitor is not discharged fast
enough,
and the transistor has a certain threshold before it switches off, the
pulses coming off the transistor will be longer. Logic correct?
Plastic, 1" x 1" x 2".

Please bear with me for a second, and this might sound silly, but
shouldn’t the resistor connect with one end to the 12v MB and
the other end to 0V? (I think 0v is common ground in computers, that
is the case and every other component shares that ground.)

Power Supply
| 12v+
|
|
|
|
|
12v+ ----------------------| NPN
| |> Transistor
.-. |
| | |
| | |
'-' |
Resistor | |
| |
| | +
| |
| .-.
| (FAN)
| '-'
| | -
| |
| |
___ ___
_ Ground _
| |

Would this schematic work? (Modulating (+) instead of ground?)

Thanks,
Mike

8. ### Al BorowskiGuest

Hi,

[...]
Rats. That makes thing harder.

[...]
It doesn't work like that I'm afraid. The '25V' on the capacitor means
the capacitor is rated for a maximum of 25V. It still has the valur of 47uf.
"rc discharge equation". Seeing there is a real capacitor there, the
calculations become a bit trickier. I'd crunch the numbers myself but i
don't have time right now.
Yes.

As I'm assuming if the capacitor is not discharged fast
Yes. One danger is that the transistor may operate in the linear region,
where it is neither fully on nor fully off. This means it will get much
hotter then if it was switched fully on or fully off. In a sealed
plastic box, this could cause problems.
I'm not expert, but it sounds like that could be asking for trouble.
Maybe a cheap metal box would be the way to go?

It effectivly does!

There is a (small) vltage drop accross the Base-Emitter pins, but
otherwise its connected to 0V through the transistor.

Possibly, but its asking for trouble:

1) Too much base current may flow. The resistor in the first circuit
served to limit the bse current, but also to discharge the capacitor.

2) since you just want to use the NPN transistor as a switch, you are
better off placing the fan between 12V and the collector. I don't have
time to explain why right now, but google "common emitter" "emitter
follower" and "transistor switch".

I can give more details later if needed.

cheers,

Al

9. ### MikeGuest

That I do not know, and slowly trying to figure out. The chances of me
getting to an engineer at any motherboard manufacturer are slim to
none.

From what I can see on the board with my own two eyes, there are only
two transistors, a few resistors and a capacitor in the final stage of
the PWD fan control.

The first transistor is small, with the base connected to the central
microprocessor, and the base gets a 0 to 5.38v+ DC, linearly
proportional to how many % the fan header is set to. The second
transistor is larger, with the base feeding from the smaller
transistor's emitter, collector a constant 12v+ DC and emitter going
directly to the capacitor and then the actual fan header.

23.61V AC. The strange part is, if I set the multi-meter to AC, use
the positive probe on the DC (+) I get the 23.61V but if I use the
negative probe on the DC (+) I get 0V AC. Should it not be showing
same voltage either way?

10. ### MikeGuest

That's all very good stuff. I was just looking to be pointed in the
right direction. With some google, and maybe *gasp* a book that is not
too big I'll figure this out, hopefully without burning up all fan