Rheostat Question

[Mobile]

Will a 12.5W 50 ohm Rheostat work to control two 120mm Panaflo fans?...Each Panaflo produces 4W.

I know what the 12.5w means, but I have no clue about the 50 ohm, what does a higher or lower ohm mean?

I am ordering the Rheostat online, and need to know what I should get, 25ohm 10ohm, 7ohm.

Thx.

 

[rlemieux]

I had one controlling a 120x120x38mm delta and it worked fine. You should be able to put two of those on there. Oh btw, check your post in the casifieds.

 

[macklin01]

Ohms are a measure of resistance. The higher the ohms, the greater the resistance. (And higher resistance will slow your fans down more.) Watts are a measure of how much power the resistor can handle (max capacity). Watts = current * volts. At 12V, your resistor can handle at most 12.5 W / 12V = (approx) 1 amp. Your 120 mm panaflow is 4W, or 4W / 12V = .3 A = 300 mA. So, you're just fine there. -- Paul

 

[Mobile]

So I should have no problem running 2 Panaflo (4w each) fans off of a 12.5w, 35ohm rheostat?

 

[packratbob]

the lower the ohms the less u will be able to adjust the voltage. the more ohms the more u can adjust.

 

[macklin01]

So I should have no problem running 2 Panaflo (4w each) fans off of a 12.5w, 35ohm rheostat?

You should have no trouble running 8W of power through the rheostat. I'd just double-check to see whether or not a heatsink is recommended for that particular rheostat. (Many fanbuses based on rheostats have heatsinks on the rheostats.) Personally, I use a different configuration which doesn't require heatsinks (and I only use 1/4 W rheostats), so I can't really advise further there.

And I think that the value of the resistor (35 Ohms) is enough. One fan running at full speed uses .333 A. So, you can estimate an "internal resistance" (this is very crude, but okay) of 36 Ohms. (12 / .333 ). So, if you turn the rheostat up all the way on one fan, you'll roughly be running 6V across the fan. (Careful -- it may stop turning at that speed.) With two fans in parallel, the "internal resistance" is 1/(1 / 36 + 1/36) + 1 / ( 2/36) = 36/2 = 18 Ohms. So, cranking up to the full 35 Ohms gives a little under 8V drop across the rheostat and 4V drop across the fans. It'll definitely be enough to slow them (to a halt, in this case).

These are guesstimates, and you should certainly experiment with them when you get them before you solder them all together.

I hope this helps -- Paul

 

[Mobile]

Ok, thanks. I doubt I will ever have to crank the rheostat all the way, I'm am just going to use to to slow the fans a bit when needed.

One more question though,

How do I wire it up, I understand you have the positive wire passthrough the rheostat; but a rheostat has 3 termials...

So If I am looking at the rheostat from the bottom, with the terminals facing up, which wire goes where...and what happens to the 3rd(extra) terminal?

 

[rlemieux]

This is how they should be wired. The third does not need anything and the yellow wire is split between one side and middle post. Make sure you use some heat shrik tape to prevent grounding.

http://forum.oc-forums.com/vb/showthread.php?s=&threadid=95377

 

[Mobile]

Great, thx again for the help!

 

[rlemieux]

No problem, figured you got one from me, I might as well show you how to hook it up.

 

[macklin01]

The way I would think of the 3 leads is like this: There's a stretch of coiled wire (or conductor) between the two outer leads. The middle lead attaches to a contact that can slide anywhere along that coil. Turn the knob to the right, and the contact moves to teh right along the coil. In that case, there's now decreasing resistance between the middle lead and the right side of the coil, and there's increasing resistance between the left side of the coil and the contact. Turning the knob left has the opposite effect.

So, you do have to take some care to make sure you use the correct outer lead. (Assuming you want to use the common convention that turning the knob to the right makes the fan faster, and turning the knob to the left makes the fan slower.)

rlemieux was spot on.

Best of luck! -- Paul