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Fitting own LEDs in Fan. Resistors + Pots help

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December 30, 2009 8:05:20 PM

Hey Toms Hardware

I'm in need of help, I want to add in my own LEDs to my fan like this -





My fan has its own holes all ready for me to do it but I need help with which Resistor I need and which Potentiometers I need as I'm wanting to add some RGB LEDs into my Fan, which means I can adjust to any colour I wish.

They share the same cathode but each have separate anodes.

The Red section has a max of 30mA and a normal forward voltage of 1.9V
The Blue section has a max of 30mA and a normal forward voltage of 3.2V
The Green section has a max of 30mA and a normal forward voltage of 3.2V


I would like to know what Resistor I would need for each of these, bearing in mind I am using 4 LEDs connecting to one Molex connector. Also I would like to know what Potentiometer I would need to be able to dim the LED from 0 to 100% brightness.

I am thinking of going Power > Resistor > Potentiometer > 4 LEDs in Parallel which would work for me for each the red, blue and green sections, although I was then wondering if there was a way to be able to add in LEDs running off the same power supply and being controlled by the same potentiometer (I am going to have three, to control red blue and green sections).

Help asap would be helpful, I will be online lots to check for updates from people.
a b B Homebuilt system
December 30, 2009 8:21:21 PM

See: http://forums.bit-tech.net/showpost.php?p=650636&postco...
For calculating resistor value. Remember: V = I*R.
http://www.electro-tech-online.com/electronic-theory/37...

More info: http://forums.bit-tech.net/showpost.php?p=927009&postco...

Quote:
I am thinking of going Power > Resistor > Potentiometer >

That will do, just remember to take in to consideration of what voltage the resistor needs drop if the pot is set to "zero" resistance so the LED doesn't burn out. Also, are you dropping +12v or +5v?
a b B Homebuilt system
December 30, 2009 8:24:34 PM

Please provide a link to a website that provides technical specifications for the exact LEDs that you have.
Related resources
December 30, 2009 8:30:16 PM

http://www.rapidonline.com/Electronic-Components/Optoel... is the LED I wish to use. The specs are there.

And Shadow, thanks for that, I should be able to work it all out.


Main question I want to know is is there anyway to be able to add more LEDs into the parallel circuit without having to change the resistors, only add in more.
a b B Homebuilt system
December 30, 2009 9:41:51 PM

Quote:

Main question I want to know is is there anyway to be able to add more LEDs into the parallel circuit without having to change the resistors, only add in more.

There probably is, but I can't think of any way with out using a Voltage regulator
a b B Homebuilt system
December 30, 2009 10:31:02 PM

I did the math on the Kingbright LEDs, and across 12 volts, you need a 1/2 watt 360 ohm @10% for each of the green and blue, and a 1/2 watt 300 ohm @10% for the red to keep the current under the max. To limit current to the typical values represented by 1.9 and 3.2 volts, use 390 and 330. Each leg of the parallel circuit, that is to say, each LED, will have to have its own resistor. With those current limiting resistors in place, one pot between the 12 volt supply and the LEDs will dim them all together. Try a 1k (maybe as much as 10k) linear pot at maybe 2 watts, but be careful not to let the smoke out, and shield your eyes.

What you're trying to do is possible, but inadvisable. The pot will get quite warm, maybe hot, and you'll be wasting power that could be better used to fly well-dressed college graduates around the world to global warming conferences. Google on "LED dimmer", and see whether you wouldn't rather use something like that to dim your LEDs.
a b B Homebuilt system
December 30, 2009 10:38:35 PM

^ I think the reason OP is using the pots are to change color (as in vary the RGB level), not so much as to change brightness if I understand right. ;) 
a b B Homebuilt system
December 30, 2009 10:47:47 PM

You would be changing color by varying the relative brightness of the three colors, so you would need three dimmers, of course.
a b B Homebuilt system
December 30, 2009 10:50:05 PM

^ True that
December 30, 2009 11:29:40 PM

Petrofsky said:
I did the math on the Kingbright LEDs, and across 12 volts, you need a 1/2 watt 360 ohm @10% for each of the green and blue, and a 1/2 watt 300 ohm @10% for the red to keep the current under the max. To limit current to the typical values represented by 1.9 and 3.2 volts, use 390 and 330. Each leg of the parallel circuit, that is to say, each LED, will have to have its own resistor. With those current limiting resistors in place, one pot between the 12 volt supply and the LEDs will dim them all together. Try a 1k (maybe as much as 10k) linear pot at maybe 2 watts, but be careful not to let the smoke out, and shield your eyes.

What you're trying to do is possible, but inadvisable. The pot will get quite warm, maybe hot, and you'll be wasting power that could be better used to fly well-dressed college graduates around the world to global warming conferences. Google on "LED dimmer", and see whether you wouldn't rather use something like that to dim your LEDs.



I've looked up that LED Dimmer but the thing is, the cost of it, like £9.99 for the cheapest ones, and I don't understand how they work properly. The main aim of it all is to have three pots coming from my PC case and control the colour of my LEDs.

Also with this pot inbetween the 12 volt supply and LEDs, does this mean I have Power > Pot > Resistor > LED. I'm guessing this would mean I could just load up LEDs & Resistors in parallel after the pot (Giving Power > Pot > Resistor > LED is correct) and not have to change the pot that is in there.
a b B Homebuilt system
December 30, 2009 11:48:48 PM

Each parallel leg would consist of an LED and its current-limiting resistor. You could have as many legs in parallel as you like, the whole parallel circuit controlled by a pot, something like this (http://www.chowfookcheong.com/Electronics/LED%20Circuit...) with a pot right above the battery. You would need three setups of this kind, one for each color, obviously, but they could all be connected to the same 12 volt supply as long as it could provide enough current. Just add up the 30 milliamps for each LED.

Again, though, this is not a good way to do it. Much better is some sort of pulsed controller. You could build three of those, instead. By the way, it is of course possible to have one resistor limit current for several LEDs in series, but I've given you the simpler setup, one that will, for example, mainly keep working if one of the LEDs opens.

I was serious about shielding your eyes, too.
a b B Homebuilt system
December 30, 2009 11:56:22 PM

^ Just reposting image:
a b B Homebuilt system
December 31, 2009 12:44:33 AM

Thinking about it some more, the more legs you have, the lower the resistance of the pot will have to be to allow good control, and the higher, obviously, its wattage will have to be. Experiment with fixed resistors to arrive at a good value for the pot. But, one more time, it's not good to use a pot as a rheostat like that, burning off voltage drop in heat.
December 31, 2009 7:50:05 AM

As you keep saying it's a bad idea to use a pot what else could I use to make this all work? Without spending out a fortune.
a b B Homebuilt system
December 31, 2009 9:45:52 AM

Here is a circuit that ought to work for like three LEDs. The 7805 costs 20 pence here. This is the kind of heatsink you need for it.

You are now entering the realm of the electronics hobbyist/experimenter. Welcome. Shield Your Eyes (our mantra). Keep One Hand In Your Pocket (our high-voltage mantra).
December 31, 2009 11:35:05 AM

Thanks for welcoming me xD And I'm guessing one hand in pocket is so I don't die from electrocution. Shield my eyes, from bright light?

Anyway, back to the subject at hand. This circuit, I'm taking the thyristor (I'm thinking thats right) feeds into the 10k pot, but the question is, what wattage power should this pot be, along with the 18k resistor and the 82's in there (The 82's would vary in wattage power and ohmage depending on the voltage and amperage of the colours I'm guessing).
a b B Homebuilt system
December 31, 2009 12:25:24 PM

With one hand in your pocket, the current can't flow directly through your heart, yes, but you're shielding your eyes from exploding capacitors and semiconductors and flying molten copper, mostly, though, you're right, light can be a problem if you get a good, snappy arc going.

Not to type a discouraging word, but if you aren't up to speed on Ohm's law, you're asking for trouble monkeying around with this stuff. That said, the voltage divider on the base is in the ratio 18:10. This reduces to 9:5, giving the equation 9/14=x/12 for the voltage drop across the (larger) 18k resistor, or about 8 volts. Power is E^2/R, so 64/18000=about 3.6 milliwatts. Practically nothing, and the same goes for the pot. The more LEDs you stack on each current-limiting resistor, the less will be the load on the voltage regulator, so that would be better. The value of the resistor will depend on how many and which type, yes, but 1/2 watt will be plenty no matter how many LEDs are there. He shows four, but you can't have that many because your forward drop is like 3 volts, and the voltage to the leg will be no more than 10 volts. You are limited to three per leg.

If you take the maximum of 3.2 v at 30 ma, three LEDs will drop 3 X 3.2 = 9.6 volts. This leaves 0.4 volts to be dropped by the current-limiting resistor if indeed we have 10 volts. That's cutting it fine, but 0.40 volts at 0.030 amps gives the equation (by R=E/I) 0.40/0.030 = 13.3 ohms. That's cutting it too fine, because you'd need very tight tolerances. It would be better to use two LEDs at 3.2 volts. I leave the solution as an exercise for the reader, unless you want me to do that.
December 31, 2009 1:46:49 PM

Well, I do know Ohm's law and etc. from Science and Electronics, but this is slightly more advanced than I know about.

I don't understand this section -

Quote:
That said, the voltage divider on the base is in the ratio 18:10. This reduces to 9:5, giving the equation 9/14=x/12 for the voltage drop across the (larger) 18k resistor, or about 8 volts. Power is E^2/R, so 64/18000=about 3.6 milliwatts. Practically nothing, and the same goes for the pot.


The rest I seem to know what you're talking about, I understand the voltage drops with the LEDs and the left over leaving things close from the resistor tolerances. I will be using 12V from molex cables inside my PC btw.


Just a few questions :p 

So I'd need 0.1A +12V Voltage Resistor?

1/2 watt LEDs will be more than enough for what I need

I'd be best off setting two sets of 2 for 4 LEDs

I need a 10K pot, something like - http://www.rapidonline.com/productinfo.aspx?tier1=Elect... would be suitable?
December 31, 2009 1:50:14 PM

1/2 watt resistors will be more than enough for what I need*



Sorry for all these questions, being 15, interested in electronics and computers, I don't know everything.
a b B Homebuilt system
December 31, 2009 3:13:54 PM

No prob, I love to show off. If you want to study the matter, I recommend two books: Electricity 1-7 and Electronics 1-7. They are pretty much the bible for this stuff.

Half a watt is kind of the default value when you go to buy cheap resistors.

I don't know what this means, "So I'd need 0.1A +12V Voltage Resistor?", and neither do you.

If you want four of each color, then 2 and 2 would be good.

That pot is OK, but the shaft is longer than usual.

About the divider, the 18k resistor and the 10k pot (also a resistor) constitute a path from supply (12v) to ground. The current through that path is wholly determined by the sum of the resistances, namely 28 kohms, and the current is the same throughout the path because it would have to be when you think about it (I'm ignoring the very small current through the wiper and the base of the transistor). Ohm's law will give you that in amps. If you take a voltage reading from the top of the 18k resistor to ground, you will get 12v, obviously. Equally obviously, a reading from the bottom of the pot to ground will be zero, because the bottom of the pot is connected to ground. But if you measure the voltage from the junction of the resistor and the pot to ground, you will read a voltage that will be in the same proportion to the total voltage across the path (12v) as the 10k of the pot is to the total resistance of the path. The voltage gets divided up among the resistances in the path, hence the name "voltage divider". (This is a simple concept in math--the ratio. You should sue your school system if I'm not just reminding you of this.)

Current through a resistance makes heat. That heat is energy lost from the circuit. The energy loss shows up as voltage drop. This phenomenon is both glaringly straightforward and sublimely mysterious. This is where the deep nature of reality rises almost to the surface where we can see it and touch it. I love electronics.
December 31, 2009 3:20:58 PM

Lol, okay, I might look into that.

And okay, I can get like 2W resistors from that rapid online but if half a watt is fine then that's good.

Here - http://www.rapidonline.com/productinfo.aspx?tier1=Elect... has many different types of Voltage Regulators, which I'm unsure which type I need to purchase, i'm thinking it's the 0.1A +12V version.

I've been taught about potential dividers in electronics :p  Not "Voltage Dividers"

So to reiterate which I need

My LEDs
330(Depending on all my LED details) Ohm 1/2 Watt Resistors
18K 1/2 Resistor
10k Pot
The Transistor in the circuit


The pot shaft size doesn't really bother me much, I can cut it to the size I want later on.
a b B Homebuilt system
December 31, 2009 4:24:29 PM

Ah. "Potential" is the more proper term for "voltage", "potential difference", to be super correct, or "electromotive force".

The lower power resistors are physically smaller, which is desirable.

The regulator you link to is the one I was looking at. I think that will be fine, but I'd slap a heat sink on it just for laughs. It's rated for 100 ma, and you need like 60.

Shame on you. You didn't do the math assignment I gave you. Two LEDs in series.... Me, I'd probably get an extra 1k pot and use it in series with a 100-ohm resistor to determine empirically the best values for the current-limiters, then permanently install that value fixed resistor. You want to go from black to dazzling and no farther. The math gets you in the ballpark, but you have to fudge once you're in there sometimes.

Get extras of your components if you can afford to. If you make a mistake and damage one, you won't want to wait a week to try again, not to mention the shipping charge.

!