Motherboard fan header boost circuit

[dangar]

Hallo to all ,

I was wondering if anyone have built a circuit that allows you to add more than one fans on a single fan header , exceeding the permittable motherboard capacity . I can understand that this may be built with transistors or mosfets wired in series with the fan header  , but how exactly?

[kam]

As you said, adding more than 1 fan may exceed the mb/s capacity. Moreover, any functionality that this header would have-such as to measure the RPM- will not work. So, why don't you add the fan directly to the PSU 12v? You can add as much as you want there.

[dangar]

The idea is to control the fans according to system temps , not  only to power them. I believe using the motherbord header is the only way to do it. As for the tacho signal , I will connect only one fan rpm wire to the header , as the fans will all be the same.

[kam]

o i see. one more question: you have 4 wire fans or 3 wire fans?

[dangar]

Three wire fans . I have tried in the past with 4wire fans and  managed to control 3 fans with a LM358 based circuit , but never with 3wire.

[kam]

what control does your motherboard have over a 3-wire fan? The only thing that it can do is to measure the rotation speed. There is no speed control, am i right?

[dangar]

The mobo is MSI neo2 FR and I can control two 3pin case fan headers except the 4pin cpu fan header via speedfan. It uses as superI/O chip the Fintek F71882 and I believe MSI uses the recommended method from chip's datasheet , as I can see next to each fan header a P06P03LCG P-mosfet and a 16V/100μf cap. And between them is a lm358 chip. This method is described into the F71882 datasheet  . I can control a 120mm zalman fan from its full rotation speed - 1900rpm-  down to 500rpm , so I believe it's using a kind of SMPS technique with smoothing output cap .

[kam]

i see. so, this is what i would test first. From the pin, i would get the red wire (12V) and connect it to the base of a transistor through a small limiting resistor (such as 220 ohms for example). The collector of the transistor goes to the 12V of the PSU. The fan red wire (12V) is then connected to the emitter of the transistor. Each fan should have a transistor of its own.

[dangar]

Is it safe to for the mobo to do this? And if I understand correct this would be like an emmitter-follower topology and assuming the driving signal is up to 12V the output would be 0,7V lesser  , more or less?And , lastly , why not to chose one transistor capable enough to drive 2-3A fans?

[kam]

Yes it is safe. You will only draw a few milliamps for the base of the transistor. The topology is indeed emitter follower thus the output is 0.7 volts less. If you know exactly what method the mobo uses to control the fan, then you can use CE instead and provide the full 12V to the fan. If the mobo uses linear control over the fan, then you must use emitter follower. If it uses PWM then you can use CE.
A transistor capable to drive all the fans is also a solution, but this would increase the emitter current which means more power dissipation per transistor. 2Watts of power dissipation may not sound much, but for the small area of the transistor means a lot of heat.

[dangar]

Thank you friend for your time , I will try this and let you know. And keep walking friend , great job with this site.

[dangar]

Yesterday I found the time to try something and I think it will do!. I used this circuit basicly http://www.redcircuits.com//Page109.htm ommiting the voltage divider formed by thermistorR1-R2 and instead I put a limiting resistor as recommended in series to the base of the NPN transistor - in my case 2N3904 . At this resistor I imported the drive signal ( + of the mobo fan header ) and with this I managed to control fans up to 1,3A - Vdrop 0,23V without even heating the output transistor ( I used a BD176 with a small heatsink ). This on the testboard. Next is to place the circuit and I will post the results.   

[kam]

you can calculate the heat of the transistor (power dissipation) with the following formula:

P = Ic x Vce

the transistor will only heat up of the speed of the fan is low. At that point, the Vce is at maximum. If the fan runs full speed, the Vce is minimum and so does the power dissipation. So, if you plan to use the circuit at full speed, you may not need to use a heatsink at all.

if you drop 4 volts from the 12 to power the fan with 8 volts, this means that Vce is 4 volts. If the fan draws 100mA at that voltage, the power dissipation on the transistor is 0.1x4= 400mW.

[dangar]

The cirquit is finally at its place and working like a champ! The heatsink isn't heating at all with 3 fan running  - total consumption is 1,15A , Vdrop only 0,13V. The think is that the cirquit is also working as linearly as expected to be.Quite happy with this. Thanks again!

[kam]

just for educational reasons this voltage drop is converted into power dissipation therefore:

Pdis=1.15 * 0.13 = 150mWatts... You do not need a heatsink at all