Fan Speed Controller - Build your Own

Silvion said:

Hey thanks for the reply!
Yeah, i understand that. The resistor on the LED won't affect the potentiometer at all?
Or did you mean have a "Y" section? splits to resistor/led and to potentiometer?


I can make two seperate circuits to have two 12v inputs, no worries there..
Also, would you know where i went wrong with these calculations?
12 = 12(85.71/(Rr+85.71))
1 = 85.71/(Rr+85.71)
1(Rr+85.71) = 85.71
Rr + 85.71 = 85.71
Rr = 1???
My fan is drawing 0.14amps on a 12v line:/

I just realized that I defined Vf in 2 different ways.

You want Vf to be 5v in this equation. I will go back and clarify that.

And yes, you want a Y connector off the switch. The reason is that if you put the LED inline with the pot, you won't have enough to power the fan as the fan needs 12v, not 2v.

~Lordbob

ZaLiTH said:

Silvion said:

Hey thanks for the reply!
Yeah, i understand that. The resistor on the LED won't affect the potentiometer at all?
Or did you mean have a "Y" section? splits to resistor/led and to potentiometer?

To be on the safe side, you could connect the LED directly to the ground. So you'd have two lines going from the switch to the ground, such as:

Code:

/ = switch
o = potentiometer
- = LED
~ = resistor
* = fan
=================

  +12v
    |
    /
 |    |
 o    ~
 |    |
 *    -
 |    |
(ground)

That way, the fan will only be connected to the potentiometer and the LEd will only be connected to the resistor, while both the resistor and potentiometer will be controlled simultaneously by the switch.

Exactly.

~Lordbob

ZaLiTH said:

Lordbob75 said:

Exactly.

~Lordbob

Lol.. I think I'm coming dangerously close to hijacking your thread here... Oops...

Don't worry about it.

I am reworking the equation to be SOLVED for Rr, so it will involve a lot less algebra for those reading.
However, I keep getting 2 completely different resists, so I am still working on that.

~Lordbob

Silvion said:

Hey thanks for the reply!
Yeah, i understand that. The resistor on the LED won't affect the potentiometer at all?
Or did you mean have a "Y" section? splits to resistor/led and to potentiometer?


I can make two seperate circuits to have two 12v inputs, no worries there..
Also, would you know where i went wrong with these calculations?
12 = 12(85.71/(Rr+85.71))
1 = 85.71/(Rr+85.71)
1(Rr+85.71) = 85.71
Rr + 85.71 = 85.71
Rr = 1???
My fan is drawing 0.14amps on a 12v line:/

Here is the new equation:

Rr = ((Vs*Rf)/Vf)-Rf

Rr = ((12*85.7)/5)-85.7
Rr = 119.98 Ohms, assuming you want a 5v minumum.

Sorry about the confusion.

~Lordbob

Silvion said:

Lordbob75 said:

Here is the new equation:

Rr = ((Vs*Rf)/Vf)-Rf

Rr = ((12*85.7)/5)-85.7
Rr = 119.98 Ohms, assuming you want a 5v minumum.

Sorry about the confusion.

~Lordbob

Ah thanks for clearing that up, so i need a potentionmeter's around 120 ohms? (can't go below 120, but can go higher? Just means that they won't go as slow when spinning?)
So fans will run at a minimum of 5v, max of 12v?

It is variable.

I have found that the fan needs about 10v minimum to start, but will keep running down to around 3 or 4 volts.

Now, if you INCREASE the size on the potentiometer (i.e. 150 ohms), then the fan will spin slower. If you decrease the pot (i.e.100 ohms), then it will spin faster.

It is up to how slow you want the fan to go. I picked 5v as a low because that is the lowest it is really rated at, and it just turned off after that.

~Lordbob