# So you want PWM control of your new CPU fan?



## Lazzer408 (Oct 29, 2009)

*So you want 4pin PWM control of your new 3pin CPU fan?*

I don't know if anyone can use this but I drew it up because I might need it in my ITX build. Feel free to try it.

*What it is:*
So you went out and bought a high quality aftermarket heatsink/fan but most of them only come with a 3-pin fan. Or maybe you want to use the cpu's fan controller to power additional fans. Or perhapse you want to take advantage of a high speed fan but you don't want to hear it or fork out the cash or deal with adjusting a manual fan controller. Most motherboards have cpu fan throttling adjustments in the bios but you can't take advantage of them with your new 3pin fan! This schematic shows you how to use only 3 components to allow the motherboard's 4-pin CPU fan connector to control 2 or 3-pin fans.

*The motherboard's 4-pin fan header:*
The 4-pin fan connector has power, ground, tach, and PWM connections. The PWM signal from the motherboard is open or grounded. When the signal is open the fan opperates at full speed. When the signal is grounded the fan is at it's lowest speed. The motherboard controls the fan by switching between these states.

*How it works:*
The transistor is held in the 'on' state by applying a current to it's base via. the 10k resistor from +12v to the base. The PWM signal from the motherboard sinks (grounds) the base to turn off the transistor. This resistor is needed to both turn the device on and limit the amount of current the motherboard has to sink to 1.2ma. The second resistor, also 10k, limits how low the pwm signal can sink it (and subsequently reduces the current current the motherboard's pwm circuit to .6ma). This keeps the fan opperating at a minimum RPM.

The NPN transistor was chosen because they are generally easier to find.

The 10k resistor in series with the PWM signal can be substituted with a 10k pot to give you 0-50% adjustability of the MINIMUM fan RPM.


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## Cuzza (Oct 29, 2009)

Nifty idea.
I had to think about that for a while but I get it. Most of it - what does the 10k do? (sometimes I wish I'd taken up electronics instead of chemistry, grrrr)


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## King Wookie (Oct 29, 2009)

Any specific transistors or mosfets you recommend?


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## Lazzer408 (Oct 29, 2009)

Cuzza said:


> Nifty idea.
> I had to think about that for a while but I get it. Most of it - what does the 10k do? (sometimes I wish I'd taken up electronics instead of chemistry, grrrr)



The 10k provides base drive to keep the transistor in the on state. It also limits the amount of current required to pull the base to 0v (during PWM low state) to 1.2ma. 



King Wookie said:


> Any specific transistors or mosfets you recommend?



As long as the transistor's power rating is higher then your fan it'll work. Nothing to specific about it.

EDIT - The circuit was changed to just a transistor to make it easier to maintain a minium fan rpm.


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## <<Onafets>> (Oct 29, 2009)

Sticky?
Good job...I'll get my dad to help me try this!


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## MKmods (Oct 29, 2009)

thanks for posting this, well done.


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## ERazer (Oct 29, 2009)

sweet im soo gonna try this when i get home, ty bud


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## Lazzer408 (Oct 29, 2009)

EDIT - Circuit tested. Image updated. Videos posted.

The 3rd pic is how your circuit should look when assembled. I just used any old NPN transistor I had with the same pinouts. It works as planned and the RPM will shift from 100% to 50% when the PWM lead is grounded or anywhere in between based on the PWM signal from the motherboard. This works exactly the same way as the stock Intel cooler. The transistor is only noticably warm and will not require any heatsink unless your using a very powerfull fan or multiple fans. The TIP120 is rated at 5 amps and could probably control 6-10 fans with a proper heatsink.

The fan tested is a Magic MGT8012XR-A25. I couldn't find any info on it other then it's 80mm, 12v .39a and a medium-speed fan.

Here's some videos to show how the circuit functions vs. a generic 4-pin soc775 replacement fan (with a bad bearing) 

PWM grounded = 50% speed
PWM open = 100% speed

Intel replacement 4-pin fan.

4-pin to 3-pin adaptor circuit.


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## Lazzer408 (Oct 29, 2009)

Here it is potted in hot glue and ready to "safely dangle" inside a case. Potting in hot glue is easy. Get yourself a piece of flat aluminum or copper. The base of an old heatsink would work great. Lightly coat the base with oil and use your finger to spread it around. Put some glue on the parts to be potted then press it against the heatsink and wait a few seconds. This will leave a nice flat surface and cool the glue. Spread the oil every time you flatten the glue. When you do the next side it will squeeze some glue over the side you just did. Just trim that off with a razor blade. Repeat for all 4 sides and wa-la.


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## tw3akm@ster (Oct 30, 2009)

Thumbs up man. Nice mod.


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## Lazzer408 (Oct 30, 2009)

tw3akm@ster said:


> Thumbs up man. Nice mod.



Thanks man.


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## Lazzer408 (Oct 31, 2009)

Has anyone else made this yet? I wan't to hear some good news.


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## Cuzza (Oct 31, 2009)

I totally would just for kicks, but I don't have a board with PWM. *Sigh*


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## Lazzer408 (Nov 3, 2009)

Just a quick update. I swapped out the 10k resistor (from the 4-pin's PWM signal) with a 10k pot. This gives you some adjustment for the fan.
This is how mine turned out.


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## bobosoft (Nov 22, 2009)

Oh man, where do I start. Doing all of the math required to predict that circuit would make my head hurt. What I can tell you is that it is not what you wanted and the only reason it works at all for you is because you are using a TIP120 which is a darlington transistor which is an integrated circuit with 2 NPN transistors to create especially high gain. Anyone who used a normal transistor would just end up with a slow running fan. I can also tell you that won't work well when you plug it a motherboard. The fan will  vary from slow (at 100% PWM) to slightly slower (at 0%). That is because a PWM signal is not grounded or open it is HI or LOW. In other words 0v or 5V, both of which would pull your circuit low. It also violated intels specified maxium 5V at the PWM by delivering 12V there although I doubt it would actually damage the MOBO.

The PWM signal varies the fan speed by changing its duty signal which is the percentage of time is at 5V vs 0V. For example, a 30% duty signal at 25Khz(intel specified frequency) would be a continuous pulse of 12 microseconds 5V followed by 28 microseconds 0V.

The circuit I would suggest would be to interrupt the GROUND wire (Not power) with an N-MOSFET. connect the source pin to the motherboard side and the drain pin to the fan side. Connect the gate pin to the PWM signal through a 100 Ohm resistor. It is that simple. The IRF510 available at radioshack would work great. Set the minimum fan speed in your bios.


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## sneekypeet (Nov 22, 2009)

It's been stuck, nice work with the mod! I hope a lot of members can benefit from this short but sweet tutorial.


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## Lazzer408 (Nov 22, 2009)

bobosoft said:


> Oh man, where do I start. Doing all of the math required to predict that circuit would make my head hurt. What I can tell you is that it is not what you wanted and the only reason it works at all for you is because you are using a TIP120 which is a darlington transistor which is an integrated circuit with 2 NPN transistors to create especially high gain. Anyone who used a normal transistor would just end up with a slow running fan. I can also tell you that won't work well when you plug it a motherboard. The fan will  vary from slow (at 100% PWM) to slightly slower (at 0%). That is because a PWM signal is not grounded or open it is HI or LOW. In other words 0v or 5V, both of which would pull your circuit low. It also violated intels specified maxium 5V at the PWM by delivering 12V there although I doubt it would actually damage the MOBO.
> 
> The PWM signal varies the fan speed by changing its duty signal which is the percentage of time is at 5V vs 0V. For example, a 30% duty signal at 25Khz(intel specified frequency) would be a continuous pulse of 12 microseconds 5V followed by 28 microseconds 0V.
> 
> The circuit I would suggest would be to interrupt the GROUND wire (Not power) with an N-MOSFET. connect the source pin to the motherboard side and the drain pin to the fan side. Connect the gate pin to the PWM signal through a 100 Ohm resistor. It is that simple. The IRF510 available at radioshack would work great. Set the minimum fan speed in your bios.



I had posted a low-side switched mosfet version at one time but for simplicity sake I just left the transistor version. TIP120 was chose because RadioShack has it. Other then the voltage drop across the transistor it's more then sufficient for a single fan. As I stated, the resistors limit the current that the motherboard has to sink. It's cheap, it works. I've tested it with a varity of transistors. 2SD1273 (for example) is not a darlington (see pics) and functions the same as the TIP120 in circuit. My motherboard shows open/grounded on the PWM pin.

If the math for just 3 components makes your head hurt then perhapse electronics is not your strong point. Your more then welcome to post your own solution for a 4 to 3pin adaptor.


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## bobosoft (Nov 23, 2009)

Id rather not argue about it, my concern was preventing frustration on the part people trying to implement your circuit who might think they had done something wrong.


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## Lazzer408 (Nov 23, 2009)

bobosoft said:


> Doing all of the math required to predict that circuit would make my head hurt.QUOTE]
> 
> To help you understand the math I'll leave the transistor's gain, voltage drop, and fan current out of it. A resistor divider made of two equal values will give you 1/2 the -difference- in the voltage across the divider. If that voltage is 0 and 12v then the base receives 6v. The fan will run at aprox. 50%. Since the 'low side' resistor is adjustable, the base can recieve 0-6v so the fan can run between 0-50% when PWM is 0v. When the PWM is open, the base recieves 12v and the fan is ~100%.
> 
> ...


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## bobosoft (Nov 23, 2009)

That's not quite how a bipolar transistor works. The current flowing from the base to the emitter determines the amount of current that can flow from the collector to the emitter. That proportion is called the gain so if you had 1ma through BE and a gain of 1000 (as is typical of darlington) you would allow 1AMP through CE. Calculating the current through BE after your voltage divider and taking into consideration the inductor it has to go through before ground is the math I was referring to.


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## Lazzer408 (Nov 23, 2009)

bobosoft said:


> That's not quite how a bipolar transistor works. The current flowing from the base to the emitter determines the amount of current that can flow from the collector to the emitter. That proportion is called the gain so if you had 1ma through BE and a gain of 1000 (as is typical of darlington) you would allow 1AMP through CE. Calculating the current through BE after your voltage divider and taking into consideration the inductor it has to go through before ground is the math I was referring to.



I wasn't explaining the function of the transistor. I know it's not a voltage device. That's why I said I left the transistor's characteristics out of it. I was focusing on the point you made about the PWM's high state being 5v as opposed to an open state. I see how that could cause an issue and explained why. My board's PWM pin is going from 0 to infinity not 0-5v. If the circuit is connected, then PWM goes from 0-12v. Do you have a link handy to the Intel spec on the 4pin? I can't find it now. I'll look into later. I have to run out for awhile.


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## bobosoft (Nov 23, 2009)

Here is the spec:

http://www.formfactors.org/developer%5Cspecs%5C4_Wire_PWM_Spec.pdf

You are right about it being open, not 5v, Intel says it has to be pulled up in the fan.

"This signal must be pulled up to a maximum of 5.25V within the fan."

the 2SD1273 is not darlington but it might as well be since it also offers veyr high gain (min 500, max 2500)


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## Lazzer408 (Nov 23, 2009)

It's being pulled anywhere from 0-6v with what I built. I'm sure it's lower then 6v if you consider the base current dropping the voltage across R1 with the fan load on the transistor. I doubt it's to 5.25v though. I tried to keep this as simple as I could. It worked good enough for what I needed so I thought I'd post it. Is it perfect? no, but it's dirt cheap and it works. Beats listening to a 120mm fan running at full speed. =)

I never looked up the 1273 datasheet. It was grabbed from my drawer of NPNs. The fan isn't a huge load so most to220 NPNs will work for this circuit. Just don't mount it with the tab.


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## Wile E (Nov 23, 2009)

How in the world did I know this thread was started by Lazzer? lol


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## Lazzer408 (Nov 23, 2009)




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## pvillegeek (May 16, 2010)

*Better late than never*

It looks like I'm a little late to this party, but seeing your schematic made be cringe!

I guess I should start off by mentioning that I am an electrical engineer, so I know what I'm talking about.

What's right:
On the positive side, you did keep ground as ground, and you chopped the 12v line (I've seen several attempts messing this up).  If you chop the ground wire it will mess up the tach output.  Other than that I'm surprised this circuit works at all.

What's wrong:
The PWM pin on the motherboard is designed to sink up to 5.25v, in your circuit it gets up to 12v! (though due to the resisters it can't draw much current, so probably won't damage anything)

The voltage to the base of the transistor is switching between 12v (on) to 6v (off) so it SHOULDN'T ever be turning off!  If it is in fact working it's only because the transistor you chose has a very high base current.

Oh, and varying the resistor will NOT change the minimum speed of the fan.  It will have no positive effect on the circuit, and might make the circuit (if it works now?) stop working completely.

How to fix it:
The PWM input should be connected directly to the base of the transistor between the resistor pair.  And the bottom of the resistor pair should be connected to ground (not PWM).  
Finally you should change the resistor values to:  3.9k between pwr and the base of the transistor, and 2.4k between the base of the transistor to ground.  This will allow a higher current to the base of the transistor, making it more likely that you'll be in the switching region of the transistor, rather than the analog region.

Hope someone finds this useful


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## mastrdrver (Jan 29, 2011)

Are the wires and signals the same coming off a video card?

Thinking about doing what Zalman should have done and make my VF3000a a pwm version.


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## Lazzer408 (Mar 11, 2011)

Hope this isn't too old to post.

I made another version. It provides 5v pull-up on the PWM (It was suggested. I don't know why this is needed. Motherboard detection of PWM fan installed maybe.) and kick start on power-up to get stopped fans moving. It also has a minimum speed adjustment to compensate a wider varity of fans. Roughly 4~10v or something like that. I used a PNP as a high side switch. Seemed more efficient to use the PNP plus the ground is maintained so the tach works right.

It works in the emulator but hasn't been bench tested. Lets pick it apart!





Here is a scope shot of the power-up kick and switching between full-low and full-high speed.








pvillegeek said:


> It looks like I'm a little late to this party, but seeing your schematic made be cringe!
> 
> I guess I should start off by mentioning that I am an electrical engineer, so I know what I'm talking about.
> 
> ...



I forgot to comment on this. The reason I used 2 10k resistors was so the motherboard could NOT stop the fan. The base voltage will alternate between roughly 12 and 6 making full/half speed. This is also why adjusting the resistor will effect the minimum speed. I'm not very informed on Intel's PWM specs but I do know that grounding the PWM pin of a PWM fan will NOT stop the fan. It only reduces the speed.

In your revision, should the motherboard sink the base 100% when it calls for low speed, *your modified version will allow the fan to COMPLETELY stop. We don't want this and I don't recomend your revision to anyone for that reason alone.* My 10k off 12v is only 1.2ma of current for the board to sink. I don't see a problem here. With your 3.9k the board has to sink about 3.1ma. Almost 3x more then my original design. I'm sure a board can sink 3.1ma anyways but it's not necessary. Just use a transistor with high gain and a 1.2ma is plenty of base drive. 
My original schematic works fine if you see my intentions. It's cheap, easy, safe, and more people could handle soldering a few components directly to themselfs vs. building a more advanced circuit on a PCB.


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## mastrdrver (Mar 12, 2011)

The design pvillegeek gives lets you basically make any 3 pin fan a 4 pin pwm fan. Since you always receive a signal to the base even when it sinks it 100%, the transistor will always be switching between 12v and 0v and never go to a constant 0.0v being supplied.

You don't need a pcb either to build it. I should have some free time this weekend (hopefully) so I should be able to get one put together.


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## Lazzer408 (Mar 12, 2011)

mastrdrver said:


> The design pvillegeek gives lets you basically make any 3 pin fan a 4 pin pwm fan. Since you always receive a signal to the base even when it sinks it 100%, the transistor will always be switching between 12v and 0v and never go to a constant 0.0v being supplied.
> 
> You don't need a pcb either to build it. I should have some free time this weekend (hopefully) so I should be able to get one put together.



My point was you don't want the fan to go to zero and I designed it that way by letting the PWM switch the low side of a voltage divider. Is anyone reading?  pvillegeek's design will work much better (as a 12-0 switch) if you get rid of the 2.4k. If not, your fan isn't going to be spinning as fast as it could be so heads up. Most fan's are going to have a hard time starting with pvillegeek's changes so WATCH THE FAN the first time you use it. It may hardly spin at all. His circuit only give the fan 4~5v MAX. DO NOT BUILD IT! I had it right the first time and an "electrical engineer" he should have seen what my intention was with the circuit. If you wan't, swap out both 10k resistors (in my design) with 4.7k if your transistor is low gain. At least the fan won't stall with my original design.


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## mastrdrver (Mar 13, 2011)

Intel spec for the 4 pin pwm circuit states that at startup there be a 100% duty cycle signal sent on the pwm pin. As such I don't see how there could be a problem with a fan not being able to spin up.

As for the electrical engineering part, I've got pvillegeek and another one (other then myself) saying pvillegeek's design will work. So I'll take my chances.


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## NGinuity (Jan 2, 2012)

Just brainstorming for fun here, found this thread via Google....

I realize this thread is long since dead but I was wondering why are you fussing with the BJT when something like an N-Channel FET or similar could drive this without the complication.  Is there a particular reason why you went this way?  Couldn't you just send the PWM signal straight to the gate of a FET?  The PWM signal is likely quite adequate to saturate the FET to full on (most decent ones I see are full on above 4 volts), which is what you want during the high of the duty cycle anyway.  The IRFZ46N immediately comes to mind, albeit extremely overkill, and the switching frequency of 25kHz, no sweat.  Maybe the IRFZ24N, cheaper, but still a bit overkill.  Another thing it could possibly buy you is controlling more than one fan off of the same source pin.....the only problem is that you'd only get tach from one fan in that configuration.


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## Lazzer408 (Jan 2, 2012)

No problem from me with bumping an old thread. Knowledge is power?

Using an N-ch mosfet, you will likely be switching the low side (neg) of the fan. This pin is also used for the tach circuit (return path) and switching it could (will) cause problems with the tach readings. If you are not using the tach you could switch the low side but Intel PWM spec is an on/off "PWM" signal. (It's not actually PWM until you plug the fan in.**) Most N-ch mosfets require at least 8v to the gate. You could use a logic-level mosfet which will accept a 5v signal on the gate but this still leaves the tach issue.

Intel's PWM spec states a minimum fan speed of 30%. This is easy to do with a transistor by feeding a current to the transistors base while the PWM is in it's min. state (0v). Intel spec also calls for a start-up pulse to kick the fan into rotation from a stop. You can't simply apply 30% voltage to a stopped fan and expect it to start 100% of the time. Some motherboards will power-up with 100% PWM to start the fan but my understanding is this isn't a requirement of the board manufacturer.

The circuit was simple and cheap but may need tweaking to work with your particular choice of transistor and fan. It was provided as an example of how you could take advantage of the PWM function without the complex circuitry needed to be 100% "Intel compliant".

Hope this helps.

mastrdrver - I didn't see your post. Your wrong. Pvillegeek's modification will not work correctly. The PWM signal toggles between open and closed. It is not providing 0 and 5v or anything else. It's only opening and closing a path to ground. Imagine the PWM signal is in it's is open state (100%). He has 3.9k and 2.4k dividing 12v to the base of the transistor which is 4.565v. This will not give you full power to the transistor or the fan. When the PWM signal is closed, the base is grounded giving 0v to the fan. He circuit will toggle the fan voltage between 0 and (guessing) 4v. It may not spin the fan AT ALL.

In my original circuit, if the PWM is open (100%), the transistor's base recieves power (12v) via. the 10k from pwr to base (call it R1). When the PWM is grounded, the other 10k (R2) create a resistor divider that esentially gets you ~6v at the base. This toggles the base between ~6v and 12v. Any exact calculation require knowing the specs of the transistor being used. I do NOT recomend using Pvillegeek's modifications.

Transistors are current devices and there's alot more to calculating "in vs. out" voltages, or rather currents, like knowing the transistor's specs and the current of the fan. To visually understand how the transistor works in a perfect world, you can imagine 5v at the base gives you 5v out. It's more like 5v in gets you 4-4.5v out but you get the idea. With Pvillegeek's modification, the base will -never- see more then 4.565v which can NOT drive the transistor into full conduction even in a perfect world.

See post #28.


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## animal007uk (Jan 2, 2012)

Lazzer408 said:


> Some motherboards will power-up with 100% PWM to start the fan but my understanding is this isn't a requirement of the board manufacturer



My motherboard does this, Fan kicks it at full speed then settles down.

Some great info in this thread to its very intresting to read.


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## Lazzer408 (Jan 2, 2012)

Post 33 was edited to add some more information to be found here.

**A PWM signal would be a signal that alternates between high and low states and the duration of both states totals 100%. 

If it's on 1/2 the time, and off the other half, it is a 50% duty cycle or 50% PWM. Since the PWM pin does not actually supply 

a voltage there is no "signal" present on the pin until you plug in the fan. The control circuitry in the fan itself provides a 

very low current 5v signal which is grounded out by the motherboard via the PWM pin. The fan circuitry needs to read the 

voltage alternating between 0 and 5v and adjust the fan speed accordingly. The PWM frequency (or width) to the fan's motor 

itself could be completely different then the PWM frequency the motherboard is switching at. When the board calls for 50% fan 

speed, it sets the PWM duty cycle to 50%. The actual motor's speed in the fan is seldom linearly proportionate to duty cycle. 

50% duty cycle to a motor may not reduce the speed by 50%. It might reduce the speed by say 30%. How do you fix this? The fan 

circuitry gives the motor whatever duty cycle it needs to adjust it's speed to be linearly proportinate to the motherboard's 

request. Placing all of this control circuitry in the fan itself eliminated the motherboard manufactures from having to deal 

with it. Since fan motors vary greatly in their power requirements and designs, it's neither possibile nor cost effective to 

create a "one size fits all" control circuitry on the motherboard. It's up to the fan designer to decide what their particular 

fan may need as far as the electronics are concerned.


Even more about PWM...

I mentioned the frequency to a (quality) fan motor (stator) can be higher then the frequency at the PWM pin. This is to reduce 

or eliminate noise emitted from the motor. The motor can act just like a speaker and may vibrate at the same rate as the duty 

cycle. Because of this, the switching frequency is set high enough that it can not be heard. Every put a cheap fan on a Zotac 

motherboard and heard it buzzing? 


A little more about how PWM works...

Imagine a standard tungsten filament light bulb in a fixture on the ceiling. There is a switch on the wall to control the 

light. If you slowly turn the switch on and off the light will simply turn on and off. There is actually a slight delay in the 

time it takes the light to turn on or off because the filament takes time to heat and cool. If you turn the switch on and off 

faster then the filament can -fully- heat or cool, it will look dim and not flicker at all. This speed is the switching 

frequency of the pulses. If the switch is on half of the time and off half the time the light will dim to 50%. This is the duty 

cycle. The duty cycle is a length of time also called pulse width. To adjust this time is to modulate it. Pulse Width 

Modulation aka. PWM.


And even more about switching and motor control...

Like the filament in a light, a motor takes time to spin up or down. If the switching frequency is too low, the motor will have 

time to slow down, or spin up, faster then we want it to. If we choose a switching frequency that is faster then the time it 

takes for the motor to spin up or down, the motor's speed will stay smooth. This frequency still needs to be higher then just 

smoothing the motor's speed. The magnetic field can still fluctuate greatly and produce noise. Ever hear the whine in a 

cordless drill when you pull the trigger slightly? This is because the switching frequency is not high enough to be inaudable. 

Well why not? Mosfets are wonderful little switches because when they are completely on, their resistance is VERY low. This 

makes them a great choice where high currents need to be controlled by a small device. Mosfets used in applications like this 

typically have 3 leads. Gate, Drain, and Source. Without going into great detail, think of the 3 leads as in, out, and control. 

The gate is the control and I'll put a little focus on that. The Mosfet has a delay when it turns on. Say the gate is a knob on 

a faucet. Turn the knob on and the water flows and visa-versa. Turn the knob quickly on and off and you are modulating the flow 

of water, but, it takes time for you to open and close the valve. A mosfet has the same problem. When the mosfet is changing 

states on or off it has higher resistance. This creates heat. Not good. So, the switching frequency has to be fast enough that 

the magnetic fields don't have time to fully build or colapse (to reduce noise) but not to high or the mosfet will be 

opperating in it's "linear region" where resistance is higher. This is called swithcing loss. In the cordless drill, a single 

mosfet can control 100 amps of current. Impressive. To help keep it cool, a low switching frequency was chose to minimize 

switching losses and heat generated by those losses yet high enough to keep the motor running smoothly. Noise is not an issue with a cordless drill so there's no need to keep it quiet.


Switching and a computer's processor supply.

This information also applys to a switching power supply or switching regulator. Have you seen the newer boards that have a v-reg switching frequency option? Typically 250khz or so. A higher frequency gives better, smoother regulation, as explained above, but at a cost. More heat is generated by the mosfets due to switching losses. If you can keep your mosfets cooler, higher is better. If your not overclocking and your system is stable at a lower switching frequency, then reduce it to lower the amount of heat generated. See those little inductors around your CPU socket? Also called chokes or toroids. These help smooth and average the pulses of the mosfets and do so using a small magnetic field. It's a little more complicated then that but that's the just of it. If you really need to know how they work, just ask.

Hopefully this gives a better understanding to how PWM and mosfets work. 

EDIT- I have no idea why everythying is double spaced. Sorry about that.


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## NGinuity (Jan 3, 2012)

Lazzer408 said:


> Using an N-ch mosfet, you will likely be switching the low side (neg) of the fan. This pin is also used for the tach circuit (return path) and switching it could (will) cause problems with the tach readings.


Couldn't this be easily mitigated by running it through a RC filter and a comparator as a Schmitt Trigger to "rebuild" the pulses?  The RC filter would take out the PWM trash relative to correct ground, leave the tach pulses, and then feed those into + on the Schmitt Trigger, 5VDC on - of the Schmitt Trigger (non-inverting), and clean them up for return to the computer.  This is not needed if the tach circuit on the mobo has a Schmitt trigger input, but there's no harm in using it if it doesn't.



Lazzer408 said:


> If you are not using the tach you could switch the low side but Intel PWM spec is an on/off "PWM" signal. (It's not actually PWM until you plug the fan in.**) Most N-ch mosfets require at least 8v to the gate. You could use a logic-level mosfet which will accept a 5v signal on the gate but this still leaves the tach issue.



The MOSFET models I described both require only 4 volts to the gate to saturate I believe.



> Intel's PWM spec states a minimum fan speed of 30%. This is easy to do with a transistor by feeding a current to the transistors base while the PWM is in it's min. state (0v). Intel spec also calls for a start-up pulse to kick the fan into rotation from a stop. You can't simply apply 30% voltage to a stopped fan and expect it to start 100% of the time. Some motherboards will power-up with 100% PWM to start the fan but my understanding is this isn't a requirement of the board manufacturer.


But you're not talking about 30% voltage.  You're talking about 100% voltage at a 30% duty cycle for every pulse (big difference), give or take a little to compensate for rise/fall slew.  30% voltage would mean you're only powering the fan with 4 volts, but when the PWM signal pulses the gate, you're applying 12 volts from source as long as the gate is open all the way (it would be).  Each pulse will fully saturate the gate of the FET regardless of it's duty cycle (Current flow is also not restricted because the gate is saturated).  The real question is, can it spin up at a 30% duty cycle on it's own (not sure why not unless the fan bearings are causing too much friction)?  Would be an interesting experiment to play with. Perhaps I can breadboard some flavor of PWM generator with the stuff in my junkbox.  We're only talking about a very loose 25KHz generator here, and I can probably use the old 555 timer with 2 diode and potentiometer trick to modulate the PWM duty cycle.  Although to accurately simulate the problem, I'd need to sink the signal instead of generate it.  Not a big deal though.  Would also give me some time to look at the tach wire output problem.



> mastrdrver - I didn't see your post. Your wrong. Pvillegeek's modification will not work correctly. The PWM signal toggles between open and closed. It is not providing 0 and 5v or anything else. It's only opening and closing a path to ground. Imagine the PWM signal is in it's is open state (100%). He has 3.9k and 2.4k dividing 12v to the base of the transistor which is 4.565v. This will not give you full power to the transistor or the fan. When the PWM signal is closed, the base is grounded giving 0v to the fan. He circuit will toggle the fan voltage between 0 and (guessing) 4v. It may not spin the fan AT ALL.



Well, that's actually only part of it.  The 4 volts to the base would be just fine if there's enough current supplied to it as well to bias it.  This is provided that the PWM sink on the MOBO finds 4 volts valid which shouldn't be an issue, but with a MAXIMUM capability of 5.65 volts, I'd find it hard to believe that 4 wouldn't work considering modern day TTL logic is around 3.3 volts.  

The missing part of the equation is the load current on the collector side of the transmitter.  As long as the transistors base is in saturation with respect to the load, the fan voltage will be around 12 volts, not 4.

Keep in mind the following equation:  
RB = (VS * hFE)/(5 * lc)
Where RB is Base resistance, VS is base supply voltage, hFE is transistor beta (ideal), and lc is load current in amps.  The "5" constant is derating the load current for tolerance considerations (assuring trouble free operation by overating the load).  I'm also haphazardly assuming 0 volts between Base and Emitter, but the design constraints aren't that tight and that voltage drop is negligible. Should be ok.  The load current derating should more than make up for this.

If we use a cockroach 2N3904 (not a good choice for PWM but will satisfy the explanation and some may choose it who don't know about the downfalls of them), it has a generally excepted ideal beta of 100 in accordance with prophecy.

The tricky part is sizing the actual load current consumption.  Each fan is different (which is also where using a BJT becomes kind of iffy).  My Nexus fans each run .15 amp according to spec.  If we take that and do a little bit of math (!!!!!), we get the resulting equation:
RB = (4 volts * 100)/(5 * .15)
RB = 400 / .75
RB ~ 533 ohms (ideal theoretical full saturation)

So now we know what it "should" be to drive it into saturation, we see using Ohms law that we end up needing a current of 7 milliamps (4 volts / 533 ohms = .007 amps).  In pvillegeek's schematic, he's supplying a current of only 1 milliamp (4 volts /2400 ohms) to the base, which may cause the transistor to not fully bias, meaning the load current of .15 in my fan will not be 100% no matter the duty cycle, even if the derated constant were taken out.  But that's why the fan theoretically won't spin, not because of the voltage.  For what it's worth though, it would likely work in the transistors analog region pretty well.  He was on the right track when he suggested using the 3.9k/2.4k pair instead of the 10k to properly bias the base into the switching region and bring down the voltage on the base to a safe level for the MOBO per Intel specs.  Replacing the BJT with a FET in his schematic might work just fine to drive the fan, but I'd like to see a little more wiggle room over 4 volts to bias the gate over the FET, just in case.



> In my original circuit, if the PWM is open (100%), the transistor's base recieves power (12v) via. the 10k from pwr to base (call it R1). When the PWM is grounded, the other 10k (R2) create a resistor divider that esentially gets you ~6v at the base. This toggles the base between ~6v and 12v. Any exact calculation require knowing the specs of the transistor being used. I do NOT recomend using Pvillegeek's modifications.



Voltage does not bias the base of a BJT, current does.  (Well, actually, that's not totally true from a physics standpoint but generally accepted that way by EE's, more accurately voltage ALONE doesn't because it's actually the causal relationship of voltage to current :-D). Toggling between 6 volts and 12 volts on your base is immaterial if your resistor is sized improperly and all that is really happening is a current fluctuation relative to that (I = V/R with constant resistance).  You're still getting 12 volts on the fan nonetheless.  The current to the fan, not the voltage, is what fluctuates, causing the fan to slow, but your voltage difference on the base can be whatever you want as long as it's within spec of the transistor, and the resistor that derives the current to the base is proper.  In the case of your schematic, the spec says that the sink is only rated to 5.65 volts MAXIMUM.  You're overvolting the PWM signal back to the computer.  Your SPICE simulation confirms that as well.  That line at the bottom of the fall should be dead on zero with respect to ground, not around 6 because the computer should sink every bit of it.



> Transistors are current devices and there's alot more to calculating "in vs. out" voltages, or rather currents, like knowing the transistor's specs and the current of the fan. To visually understand how the transistor works in a perfect world, you can imagine 5v at the base gives you 5v out. It's more like 5v in gets you 4-4.5v out but you get the idea. With Pvillegeek's modification, the base will -never- see more then 4.565v which can NOT drive the transistor into full conduction even in a perfect world.



Actually 5 volts at your base doesn't mean you get 5 volts out (taking out the voltage drop).  This is only true if your source voltage is 5 volts as well, which it isn't.  Source voltage to the collector is what does that.  If this were not true, for example, you could not use a BJT on a 5 volt microcontroller output to drive a 12 volt load that actuates a 12 volt relay ('cause the relay no workie under 12 volts).  The problem with using BJT's (as you have hinted to in your response, and I've shown above) is that you have to recalculate your base resistor value for every additional load you place on the collector side.  If you undersize the base resistor to accommodate a "maximum load" scenario for multiple fans, running only one fan will generate excess current (although you prevent damage to the transistor by staying within design specs of it).  If you oversize the base resistor, you run the risk of not supplying enough current to the gate to bias it open.  I'm not disagreeing that it could be done this way at all, I'm just saying using a FET seems a little more efficient from a consumption perspective, are pretty much agnostic to current load considerations, and they seem to be a lot more "switching tolerant" than BJT's as well.

I know a lot of what I mentioned would add slightly to the cost, but the reliability would be great if made to work. Either way, great exercise in all regards to get me away from just doing circuit board rework on the bench.  I currently have 3 fans running at full tilt and it's getting into my mic audio.  It's a real shame I have to touch the noise gate console JUST for that.  Cheers!


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## Lazzer408 (Jan 3, 2012)

You need to READ my words sir. Your arguing alot of things as if I stated them as solid facts. I'm not trying to explain things to electrical engineers.

I said "It's more like 5v in gets you 4-4.5v out but you get the idea."

You say "Actually 5 volts at your base doesn't mean you get 5 volts out"

I said "Any exact calculation require knowing the specs of the transistor being used."  and also  "Transistors are current devices and there's alot more to calculating "in vs. out" voltages"

I thought I cleared everything up didn't I?  Stop trolling for an intelectual argument. That's beyond the scope of this thread. My intention was to provide the simplest most inexpensive solution to control a 3pin fan off a 4pin header. It works, I use it, and it won't harm anything. It can also be soldered together lead to lead and shrink-wraped inline with the wires. Is it perfect? Of course not. That would require using predetermined components and more of them. This adds to the cost and complexity. Your more then welcome to offer your own circuit.


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## NGinuity (Jan 3, 2012)

Lazzer408 said:


> You need to READ my words sir. Your arguing alot of things as if I stated them as solid facts. I'm not trying to explain things to electrical engineers.


I mean no ill intention towards you, but I did respond because you didn't clear everything up.  I am simply pointing out the problems as I see in your posted schematic that I see as an unsafe design.  On the contrary, I think my response was more than adequate in the civility department.  My advice costs you the same as your advice costs me.  Take it for what it is.



> I said "It's more like 5v in gets you 4-4.5v out but you get the idea."
> 
> You say "Actually 5 volts at your base doesn't mean you get 5 volts out"


With all due respect, you really should reread what I wrote before you call me out on it.  It wasn't an intellectual reply.  5 volts at the base does not mean you get 5 volts output when you've got a 12 volt supply on the collector.  It doesn't mean you get "4-4.5v" either.  It means you get around 12 on the fan when the gate is saturated.



> I said "Any exact calculation require knowing the specs of the transistor being used."  and also  "Transistors are current devices and there's alot more to calculating "in vs. out" voltages"



My calculations were hardly "exact".  More "ballpark", but these are things you need to know when you want to do these designs properly.  I gave you the extremely stripped down, condensed version at that.



> I thought I cleared everything up didn't I?  Stop trolling for an intelectual argument. That wasn't the purpose of this thread.



Who's trolling?  I understand that you are very defensive of your design (believe me, I understand any circuit designer doing that), but my intention was not to just tell you "you're wrong" like others have done, but to show you, and others with come to this page with no previous experience WHY some of the things you did are an unsafe design (such as overvolting your PWM pin, for instance) so that you can actually apply some of it and not make the same mistake again. Additionally, if you feel I'm wrong, tell me why, but don't attribute valid discussion as "trolling".  I have ZERO problem with you finding flaw in something I have said and up until this point you did an _excellent_ job citing your design reasons of why you chose the BJT over the FET.  I have no desire to get into a virtual slap fight with you and I think some of your criticism is a little harsh toward others, especially when you had an initial outlook of "Knowledge is power" toward me.  It's a two way street you know.  If you're not going to be receptive to feedback, don't put yourself out there.

So, that being said, if you are not going to take feedback on your design, what IS the purpose of this thread?  You've driven off two other people who obviously are very competent in this area because they've been trying to explain to you why things in your design are unsafe, things I agree with and can explain mathematically.  I don't believe I'm the first one to bring that to your attention, either, so chill man, let's just have some fun doing this.  I'm not keeping score or looking down on you.


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## Lazzer408 (Jan 3, 2012)

I take offence when someone like pivllegeek come in and break my design and people like mastrdrver thank him for giving them the circuit? To quote mastrdrver: "The design pvillegeek gives lets you basically make any 3 pin fan a 4 pin pwm fan."

It really gets under my skin when I'm out here trying to help someone just to have some tool come along, break my design, then get credited for it. I've build MUCH more complicated electronics then this little circuit. Such as a 220kw Class-d amplifier.

You on the other hand. I have no problem with you. I completely understand what your saying and why your saying it. But understand this isn't an electronics course.

One thing you just said...

"5 volts at the base does not mean you get 5 volts output when you've got a 12 volt supply on the collector. It doesn't mean you get "4-4.5v" either. It means you get around 12 on the fan when the gate is saturated."

I'd love to see a transistor, that has 5v present on the base, give 12v on the emitter with a 12v collector."

That would revolutionize the electronics industry!   I'll get you some screen shots in a second to show the various voltages and currents in my original circuit.


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## Lazzer408 (Jan 3, 2012)

Here are two screen shots. I don't have a very high gain transistor, or the TIP120, in my emulator so I used a TIP100 which is close. You get the idea anyway.

The first image shows the switch in it's open state. The second image in it's closed state. This is to show PWM 0% and PWM 100%. The fan will never actually see 12v due to the voltage drop across the transistor. Solution? Use a faster fan.









More to come...


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## Lazzer408 (Jan 3, 2012)

Here is the circuit shown with pvillegeek's mods. You can clearly see how it won't work and SHOULD NOT BE USED nor should he be credited for improving or creating anything.









More to come...


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## Lazzer408 (Jan 3, 2012)

NGinuity said:


> 5 volts at the base does not mean you get 5 volts output when you've got a 12 volt supply on the collector.  It doesn't mean you get "4-4.5v" either.  It means you get around 12 on the fan when the gate is saturated.



Here is the circuit with ~5v at the base and 12v at the collector. I guessed 4-4.5v. We can see in the emulation it's actually ~4.8v (will vary depending on load). Where's the 12v your claiming?


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## Lazzer408 (Jan 3, 2012)

pvillegeek said:


> Oh, and varying the resistor will NOT change the minimum speed of the fan.  It will have no positive effect on the circuit, and might make the circuit (if it works now?) stop working completely.



Well lets prove this guy wrong again. Here are two circuits showing the PWM pulled low (0%). The board is calling for the fan to be at it's lowest speed. I've drawn two identical circuits with the 10k (to PWM) changed to 20k in one of them. Note the voltage at the fan. This is why I say you can put a variable resistor there to set the low speed of the fan. It will have no effect on the fan's top speed when PWM calls for full speed (100%).


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## DriveByEE (Nov 12, 2013)

*more drive-by judgement*

Lazzer408, pulling the PWM pin up past 5 volts violates the Intel PWM spec. You haven't had a problem because presumably whatever your motherboard has hooked up to this line can handle 12v. (And note this has nothing to do with the current it's sinking. that's *another* requirement). But this circuit may very well cause damage to a different motherboard. Other than that, it's good :>

pvillegeek, your circuit may not break the motherboard, but won't drive a fan either. With the base voltage only ever being a maximum of 5v, the emitter voltage will be similarly limited. Designing circuits quickly is a good way to forget about all the requirements 

This problem really wants two transistors - something like an NPN low-side that turns on a PNP on the high side. IMHO more straightforward than making one three-resistor ladder between 12v--base--pwm--ground, getting the values right, and not having full range from the fans.

I'm honestly surprised I'm finding these threads years later instead of finding some off the shelf adapter that's overpriced but saves my tinkering energy for something else.. I just got an X9SRA that has 5 PWM fan connectors, but I'm left wondering how popular they are on consumer motherboards. I'm starting to get that design urge to solve this problem once and for all...


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## Lazzer408 (Nov 18, 2013)

DriveByEE said:


> I'm honestly surprised I'm finding these threads years later instead of finding some off the shelf adapter



http://www.camera2000.com/en/speed-controller-by-pwm-for-computer-pc-cooling-fan-4pin-3pin.html


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## jyavenard (Dec 3, 2013)

I bought a few of the devices mentioned above http://www.camera2000.com/en/speed-controller-by-pwm-for-computer-pc-cooling-fan-4pin-3pin.html

Well, they do not work...

My motherboard has 5 PWM connectors, I have 3 PWM fans in my chassis (4 pins)... As they run way too fast, they are incredibly noisy, so I wanted to lower their speed.. And found this thread.

I connected the 3 pins (tac, power, ground) to the connector.

Fans speed go like this: full for 5s; stop for 5s, repeat...

so I'm back to searching ways to lower the fans speed, but not lose the ability to control the speed of the fans by the BIOS.

Any ideas?
Thanks


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## sttubs (Dec 3, 2013)

The best thing to do is replace the loud fans with lower RPM fans. These are quiet & keep the air moving: http://www.newegg.com/Product/Product.aspx?Item=N82E16835103052
Or you could use something like this: http://www.newegg.com/Product/Product.aspx?Item=N82E16835103052 it uses different voltages depending on which connection you use.
Also it would be better that you create a new thread with your inquiry, you may get more responses that way.


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## unnamet (Dec 10, 2013)

#28

Would it work if i pulled the PWM signal to 5v instead of 12v?


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## Nukemaster (Jan 10, 2014)

I just want to point out that post #28 does work quite well. I needed to modify the resistor values for my fan, but it gave a decent range of speed, but it does climb rather fast.

I had changed the resistor to the base to on the npn to 100k and the resistor from base to ground on the pnp to 200k. adding a cap to the output also seems to give you lower slow speeds without effecting the top end speed. I got upto about 11.6-11.7 volts at 100% and down to 4-5 volts depending on resistor values,





I also built pvillegeek's version for the hell of it and guess what full speed JUST turns the fan as the OP suggested it would


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## maximkk (Feb 9, 2014)

Hey I built this on a protoboard but am seeing weird results (damn I did not find the new thread before this one 

So with PWM connected the rpm goes: 20% duty cycle 640rpm - 100% duty 700rpm on a 12v 1150rpm GentleTyphoon. When I unlug the PWM cable the was runs at approx the specced 1150rpm. Using a normal PWM the mobo connector works just fine.

I think it should've been 20% 600 / 100% ~1050rpm (with TIP120 dropoff)

Any ideas if there would be an easy fix?


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## Nukemaster (Feb 9, 2014)

That have several versions of the design here, What one are you using?

I did not have great luck with the one on the first page(my re-claimed transistors just do not have the gain), but the one with the zener diode did as shown above.

Another one to try would be this. It needs a decent gain on the first transistor, but seems to work quite well.
*PWM to Analog converter circuit @ overclockers.co.uk*


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## maximkk (Feb 9, 2014)

Nukemaster said:


> That have several versions of the design here, What one are you using?
> .....



I made the one in the first post - the simplicity of the design made me want to test it. maybe i will pull it apart and try another design. though the simpler the better.

I put it together with the parts mentioned in the post. What I was thinking would there be any change to play with the resistor values to widen the PWM control marginal - or is it that the 5v is just not enough for TIP120


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## Nukemaster (Feb 9, 2014)

If you lower the values(both at the same time), you should be able to get more current from it(if you think current is the issue). You have to ensure you do not pull too much from the PWM. it is only suppose to be good for ~5ma I think(but with such a high gain transistor[tip has a current gain of 1000, that should run a 12 watt fan on 1ma at the base on paper at least] I am surprised you would need it.).

Not all fans spin the same ratio at certain speeds either.

Again, the transistors I had salvaged had very low gain and the resistors limitation reduced my range with the first plan. I did run 2 in a darlington configuration and it seemed to work ok.

When I get a chance, I will try it out again to see what range I get(I have an AP15 to try with,).

I quite liked the PWM to analog converter from overclockers.co.uk because it worked quite(I am almost sure it did not climb as fast. I will have it on a breadboard kicking around) well as long as you had a decent gain on the first transistor(I still had to lower the resistor values to allow more current to the second one because of my crappy[ok they have a use still] transistors.).

Please note, I am not engineer here, I have just played with this because I have lots of non pwm fans I would like to control with PWM(and I am cheap so I recover parts from old hardware.).


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## Retrorockit (Jun 5, 2016)

I first would like to thank everyone who posted here for taking the time to do so. It was very interesting and educational. I have a fan problem or two that are different but somewhat related and I thought if I bumped this thread maybe some of the knowledgeable people that posted here might respond who otherwise might not see my questions.

A couple years ago I took on the challenge of overclocking my Dell BTX computer. In BTX one fan cools everything. This requires a big PWM fan. Especially for computers dating back to the Pentium4 and Pentium D era. Typically a Delta AFC1212DE 1.6A. 120mmx38mm and I believe 2800rpm fan. I was (am) running a Core2Extreme QX6800 at speeds up to 4GHz. Usually 3.72GHz for benchmarking.
When attempting 3.72GHz I found that the fan would speed up as temperature rose towards 80*C, then the computer would crash before cooling down. I suspected that the fan was pulling down the CPU voltage. Also the fan noise due to the high RPM was objectionable. The fan had not reached maximum speed yet so I think it had the capacity to do the cooling.
 To solve this I installed a delta GFB1212VHG two motor fan 120mmx50mm. The 2 motors draw a total of 3.4A. Both motors are PWM although they turn different RPMs (and directions). I ran the 4 power wires to a Molex connector, and paired the PWM signal wires to the one from the MB, and chose one of the tach leads to connect also.
 This actually worked very well. My  voltage was much lower, and my full load (Prime95)  temperature improved also. I now had enough voltage available  to attempt 4GHz for a validation. It crashed fairly quickly and I had to interrupt the PWM wire to send the fan to full speed so I could reboot and have time to return the settings to 3.72Ghz. This mod got me from 3.45GHz. up to 4GHz. ( My BIOS is locked so all settings are set and saved in Windows. I actually had to wait for it to boot into Windows to reset it. Thank God for SSDs.)

 There were 2 issues.
1- the GFB1212VHG was still pretty loud under load. Not as loud as the AFC at high RPM but still louder than I would like. It did move more air at lower PWM% and kept temps from rising as high in the first place.
2- Dell BIOS checks for fan function at boot and sets an error that requires me to hit F1 to get past it. No way to disable this setting. I tried to wire a small .3A exhaust blower into the fan header but the BIOS was not impressed. I don't know what the MB is looking for from the fan. I've heard rumors that it must be a Dell fan. The GFB isn't Dell. The only fix I've seen posted is to wire a 10 Ohm resistor in, which  = 1.2A. load. This kind of defeats the purpose. 1.2A.-1.8A. is the range of fans I've seen used by Dell in this application.

What I would like to do is have a device that allows the fan to run off of the MB header for a few seconds, then switch to Molex power to provide the voltage improvement I noted.

There is another fan I've started working with. The AFC1512DG, 150x50mm 1.8A, and I suspect it doesn't go  over 2000 rpm. Lots of air at low speed and designed to be quiet in workstation environments. I've plugged it into the MB header and so far so good. So I'm back to the big fan on the MB problem again. Also I'm planning a couple of Dell builds that will require more than one of these fans.
So If I could run one fan, or motor, off of the MB header long enough to satisfy the BIOS, then switch  everything over to Molex power. I could have it  all. Quiet, cool, stable, and no BIOS errors.


I understand only basic electricity, and have the skills and tools to wire something up. But I posses no electronic knowledge or experience. I did read a book on it to learn some vocabulary and a little theory so maybe I can understand enough to follow instructions. I might be running up to 4 motors (2-GFBs) each motor in the 1.7-1.8A. range. It IS an assumption on my part that the draw on the 2 motor fan is equally split between the 2 motors, but it might not be so. I think 2 A. draw for each motor would be a good design limit.

I'm thinking a 555 with a capacitor and rely would provide the delay,and the signal. But what sort of relay to use I have no clue. At least the 5V. control power is available from the Molex connector.


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## JayCan73 (Jun 5, 2016)

No offence but TLDR(It is Saturday night), maybe the simplist solution would be to run some daisy chaind fans at 100% from a molex connection, might be a little noisy but a few cheap fans could solve your problems.


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## Retrorockit (Jun 5, 2016)

Don't know what TLDR means? I'm in USA. The 150mmX50mm Delta fan is $15 or less from surplus sources. So adding more cheapness won't help. I understand very few people have bothered with modding BTX so not very many understand it. The big fan is "built into" the CPU heatsink housing. All air from the fan is forced over the VRM, then through the heatpipe cooler, and out the back of the case. The coolers I use can be had for $10-$12, the fans $12-$15 ($25 for the 2 motor fans) That and some heatsinks on the MOSFETs will get the result shown in the link in my sig. I don't have a cooling problem. I have a voltage supply problem. Thanks for taking the time to reply though.


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## revin (Jun 5, 2016)

Retrorockit said:


> Don't know what TLDR means?


*T*o *L*ong *D*idn't *R*ead


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## Retrorockit (Jun 5, 2016)

We call that ADHD here.


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## P!nkpanther (Sep 18, 2017)

I developed a tested solution for this problem and published it here in a new thread:
https://www.techpowerup.com/forums/...ump-using-4-pin-pwm-control-from-mobo.237118/


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