Custom Fan Controller question

[yiannisp]

Γιωργο καλησπερα!

I stumbled accross your site while looking for a solution to my "problem".

I have bought a computer case which has a built-in fan controller which I am not satisfied with.

First of all, the progress of the voltage regulation is not linear at all. I am providing you with a picture so you can see for yourself. I think the potentiometer used in the circuit must be a logarithmic rather than linear potentiometer,

After some reading and searching, I came accross a post of someone who has the exact same case and modified his PCB layout with new components and claims that he can now control the voltage from 5v to 12v. I am attaching you a picture of the original schematic, components and values (top) along with his updated schematic.


So far what I've done is change the SOT89 transistors with better ones (originally were D882, now I have ZXTN19055DZ). From what you can understand I need some help trying to figure out what component values I need for my controller. Ideally I would like to control from ~6v to ~12V. My 3 fans use very little current (0.10-0.20A @ 12V) and I will be using 1 fan per channel (one will be empty, since I only use 3 fans)

[kam]

Γειά σου φίλε μου,

First of, you can't reach 12V with an emitter follower. There is always gonna be some 0.6V less at the output due to the base-emitter diode voltage drop. So do not worry if you do not measure 12V.

Second, i really do not see why you need 2 extra resistors. What you want is a voltage divider with Vin 12V and Vout 12 to 5. Follow this connection:
http://www.pcbheaven.com/circuitpages/Simple_Linear_Fan_Controller/
Remove R2 completely - As you said, the fans draw just a little current so the base current will not blow your transistor (hopefully).
Now for the values:

When the potentiometer "looks" at the +12V, the base voltage is 12v so the output is some 11.4V

What you want is this: When the pot looks at the resistor (in my circuit this is R3), the potentiometer resistance and the resistor must perform a voltage divider to provide about 5.6V at the base, so that the motor gets 5V. Here is the calculator:
http://www.pcbheaven.com/drcalculus/index.php?calc=voltagediv

With a 5K potentiometer and a 4700 resistor the voltage divider output is 5.8V.

Try this circuit, it maybe help.

[yiannisp]

So during the weekend I picked up some SMG resistors and tried to experiment.

I changed R1 from 3kΩ to 4.7kΩ without removing R2 and the voltage output at the fans was 9.8V. And today I changed the R2 value from 15kΩ to 10kΩ and the voltage output was 11.85V.

This tells me that I'm going the wrong way. I will be keeping the 10K resistor to R2 and I will lower the R1 value. What do you reckon is a good value for R1? Do you think I should reincrease R2 to 15K? Keep in mind this is a 10K potentiometer (I cannot find a substitute 5K potentiometer because it won't sit in my case properly).

Thank you very much.

ps. I was taking a look at the pcb, and removing the R2 completely didn't seem like a good idea, hence I didn't try removing it altogether.

[cheerio]

Are you satisfied with those potentiometer fan controls? i always ditched them and made a self ragulated temperature reading fan control.

[kam]

Here is what you want to keep in mind: Say that the transistor hfe is 200. Say also that each fan draws 300mA So, you need a base current AT LEAST 300/200 = 1.5mA. This is true for max power. For minimum power you want to do some tests and measure how much current the fan draws at the requires voltage. Lets assume that the current is 1/10 of the max current, so it is 30mA, or 0.15mA base current. For 4 fans, you need at least 4x0.3=0.6mA current.

As a general rule of the thumb, the voltage divider must allow at least 5 times the current required for the base, that is 3mA.

I suppose you're talking about your circuit??? Its kinda hard to calculate this circuit - its a funny connection. But anyway. When the potentiometer is at max, then R1 becomes 0 and R2 becomes P//R2 (// means parallel). Same, at min, R1 becomes P//R1 and R2 becomes 0. So you do the math and find the values.

Using my circuit its much easier. The current through the voltage divider is always constant, its the sum of R1+R2+R3 under voltage. Roughly, its 12V/15KOhm = 0.8mA. As you see this is too small for your application. So you want to have a smaller potentiometer in order to further decrease R3. So, since the V_D calculation is linear, we get a ratio of 1/5, the potentiometer becomes 1KOhm and R3 becomes 2KOhm, total resistance is now 3KOhm, so the current is 12/3K = 4mA.

This is how you want to do your calculations.