[Voidling]

I picked up two PC psu today. It seems like a waste of power to just run 2-4 fans. Can they be used to power a DIY panel?

[T_B_M]

Yes, but you would need more driver circuits because you are limited to 12V DC max. The cheapest driver I built using wire wrap was with the LM317 IC, a capacitor, and a power resistor.

Here is the data sheet: LM317. Page 18 gives the circuit. To figure out what size resistor you need, use the following:

The reference voltage on the Adj. pin is 1.25V. That never changes. So we use Ohms Law to figure out the resistance based on the current we want. For my case I ran 700mA, or .7A. So if I want .7 Amps based on a reference voltage of 1.25V, I would use the formula:

E = I x R (Voltage = Current x Resistance)

We want to solve for R, so:

R = E / I
R = 1.25V / .7A
I = 1.78 Ohms

Pick a value that is close to this one. A larger valued resistor will give less current. A smaller resitance will give more current.
I will go with 2 Ohms for convenience sake.

So 2 Ohms will result in:

E / R = I
1.25V / 2Ohm = .625A or 625mA (This is the actual current using a 2 Ohm resistor)

What power rating will the resistor need?

Power = V x I
Power = 1.25V x .625A = .781W

Shoot for a 1W to 1.5W resistor. You want to go at least 25% more.

Calculating LEDs strings:

Blues/Whites:
12V / 3.3V = 3.6 LEDs. You can bump to 4, they will just run a tad less current. Usually they will only draw 3.2, but I calculate using a slightly higher operating voltage to account for irregularities.

For Reds:
12V / 2.3 = 5.21 LEDs. I would run 5. Running 6 would drop the total power for each LED.

You should now see how you would need multiple driver circuits using 12V. I used 24V with my LM317 to power 12 LEDs. Two parallel strings of 6 each running on them.

So lets say you decide to run 4 white LEDs at 625mA using the 2 ohm resistor.

4LEDs x 3.3V each = 13.2V.

This is higher than 12V supply, so you wind up with (13.2V - 12V) / 4LEDs = .3V/LED.

So there will be a .3V loss per LED unless you can adjust the output of your supply over 12V. Most PC supplies have internal potentiometers that you can set adjust the voltage with. Bump it up as high as you can.

Now onto power loss in the LM317:

P = power
Vs = supply voltage
Vd = voltage drop of LEDs (measure the output pin to ground to get actual value)
I = current

P = ( Vs - Vd ) x I
P = ( 12V - 12.9V) x .625A
P = -.5625W

This isn't the actual power loss, your output will just be the maximum allowed by the LM317, and you will have very little power loss. It is most efficient to use up all the voltage you can to eliminate power losses. Since your output can't be larger than your input...To solve the right way you would use (12V - 12V) x .625A = 0W. This isn't possible and you will always have a little loss.

If you only use 3 white LEDs, it would look like this:

3LEDs x 3.3V each = 9.9V.
P = ( 12V - 9.9V) x .625A
P = 1.31W

Assuming the ambient temperature doesn't exceed 90F / 32C, the maximum allowable temp. rise would be 93C.

The maximum junction to ambient thermal resistance that the LM317 would see is:

θJA = [Tr(MAX) / P]
θJA = 93C / 1.31W
θJA = 70.99 degrees C per Watt

The spec sheet states a max Junction to Ambient rating of 50 degrees C per Watt.

This would require a heat sink. I would try to boost your PSU 12V rail up through the potentiometer if it has one. You would rather not dissipate more than 1W through the LM317, it will get quite hot. I am right around .5W power loss on mine and had to add a small DIY aluminum heat spreader for them. Calculating for the size heat sink you need is another thing all together. I just add a small piece of aluminum until its no longer too hot to touch. Yeah, there are ways to solve for it, but I like to take the shortcut and ball park it.

This is a lot to take in, but there are readings all over the web on constant current sources.

You usually lose a little output voltage from the LM317, so your power loss on the LM317 would be negligible since you are using as much of the voltage on the output as you can. The LEDs won't be as bright, but at 625mA it will most likely not be noticeable.

Real world scenarios will vary, but this is the jist of it. The LM317 is a good place to start for a constant current source. Can't really beat the three components involved. I also feel how hot the IC gets after running. I usually add a small heat sink to them to keep em warm rather than hot.

Wow, this got wordy. Sorry if its confusing, but all the math is here.

[Voidling]

Thanks for all that detail.

I saw on instructables where they combined the 12v rail and -12 volt rail to make a 24 v. I'm not sure how many amps that'd give as +12 is 15a and -12 is 1a. Might not be able to run fans then.

Looks like I'd be better off getting a meanwell or other such driver doesn't it?

[T_B_M]

Originally Posted by Voidling

Thanks for all that detail.

I saw on instructables where they combined the 12v rail and -12 volt rail to make a 24 v. I'm not sure how many amps that'd give as +12 is 15a and -12 is 1a. Might not be able to run fans then.

Looks like I'd be better off getting a meanwell or other such driver doesn't it?

You can run fans on the +12V and GND with no effects on the -12V rail. It does limit you to only a 1 amp string though when using the -12V rail.

Yes, a Meanwell would save you time. Calculate your total voltage you need for your strings to see what size you need. The ELN-60 series would be good for you if you are only planning 50W of LEDs.

Here is some math examples on figuring all this out.

If you want to run red and blue together in the same strings, keep the quantities per string the same.

Assuming I want to run 6 red and 4 white/blue per string at 500mA, this gives me:

(6 x 2.2V) + (4 x 3.2V) = 13.2V + 12.8V = 26V total string voltage.

The ELN-60-27 would be perfect for this. Since it has 2.3A available, you can run multiple strings in parallel. Just use a 700-750mA fuse in line with each string in case something happens to one string, the fuse(s) blow(s), and not the other string(s).

The estimated power for this string would be:
26V x .5A = 13W total string power

Since the driver has 60W available, I could run 4 strings in parallel. The math is:

60W / 13W = 4.6

Since you have some extra power, you could boost them up some if you wanted. The internal Pot. SV2 will let you adjust the current. Start at 500mA and increase a little to use up the extra power. It shouldn't go much over 550mA max with this setup.

[Voidling]

Thanks. I did some sidework 4 months ago with plans to put the money to building a light. So if I ever get paid I'll get to put this to use

[tiltbit]

Originally Posted by T_B_M

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Since you have some extra power, you could boost them up some if you wanted. The internal Pot. SV2 will let you adjust the current. Start at 500mA and increase a little to use up the extra power. It shouldn't go much over 550mA max with this setup.

This seams like an ideal setup for some Osram Dragon Plus's...

I think if you wired up directly to the PSU you'd run into problems with current, as LEDs need a constant one to run at peak efficiency. LBM explained it better with his example using resistors.

It's the reason I just used meanwells in my case. I have the PSU running but it just powers the fans and fan controller/temp sensor.

[Voidling]

Do you need the psu fan to keep it cool just running fans? My psu fan is loud

Yes, I would recommend it. It got pretty toasty without fans drawing air through it somehow. Plus it's keeping my odor control going so i wouldn't want it to fail. My psu fan is not loud, so I guess it depends on the model.*

[Voidling]

These are quite old. I scratched it and using a single wallwart for a single fan at the moment. Not worried about odor as they've popped over the last 4days.