12v RGB fan mod project (testing)

[WhitetailAni]

This is an extension of my RGB LED fan mod, but this question is better suited for this part of the forum.

I did study Ohm's Law in October/November, but I can't figure this out:
If I'm supplying 12 volts at 3 amps (I think that's what 12V RGB headers provide?), and I need 3 volts, what size resistor should I use?

[Hackentosher]

1 minute ago, ragnarok0273 said:

This is an extension of my RGB LED fan mod, but this question is better suited for this part of the forum.

I did study Ohm's Law in October/November, but I can't figure this out:
If I'm supplying 12 volts at 3 amps (I think that's what 12V RGB headers provide?), and I need 3 volts, what size resistor should I use?

So you want to drop 9v across a resistor? At 3 amps, P=IV tells us that 27W will be dissipated by that resistor. You'll need a bigass resistor to do that and it's horribly inefficient. Idk what you need 3v for, but look for a step down module on Ebay that can handle the voltage and current you're looking for. Something like this should work https://www.ebay.com/itm/LM2596S-DC-DC-3A-Buck-Adjustable-Step-down-Power-Supply-Converter-Module-Arduino/224156800207?hash=item3430c954cf:g:-agAAOSw-JlfX3he

[WhitetailAni]

1 minute ago, Hackentosher said:

So you want to drop 9v across a resistor? At 3 amps, P=IV tells us that 27W will be dissipated by that resistor. You'll need a bigass resistor to do that and it's horribly inefficient. Idk what you need 3v for, but look for a step down module on Ebay that can handle the voltage and current you're looking for. Something like this should work https://www.ebay.com/itm/LM2596S-DC-DC-3A-Buck-Adjustable-Step-down-Power-Supply-Converter-Module-Arduino/224156800207?hash=item3430c954cf:g:-agAAOSw-JlfX3he

I haven't been able to find single LEDs that take 12v.

[igormp]

4 minutes ago, ragnarok0273 said:

I haven't been able to find single LEDs that take 12v.

Do you really want to use a single LED? Otherwise, having some of those in series could make up for it.

[WhitetailAni]

Just now, igormp said:

Do you really want to use a single LED? Otherwise, having some of those in series could make up for it.

I was planning to run them in parallel, but series makes sense.
Do you know how much it would take up?
Here's the part I'm planning to get:

https://www.digikey.com/en/products/detail/kingbright/WP154A4SEJ3VBDZGW-CA/6569334?s=N4IgjCBcoLQBxVAYygMwIYBsDOBTANCAPZQDaIA7AKwAsIAugL6OEBMZIA6gApi0CCNAMoBRAFIBmAGoAhACIAtAOKcA9AGF%2BDRkA

I'm not sure, though, if a 12V RGB header supplies 12V and has different R, G, and B grounds, or the other way around.

[Hackentosher]

3 minutes ago, ragnarok0273 said:

I was planning to run them in parallel, but series makes sense.
Do you know how much it would take up?
Here's the part I'm planning to get:

https://www.digikey.com/en/products/detail/kingbright/WP154A4SEJ3VBDZGW-CA/6569334?s=N4IgjCBcoLQBxVAYygMwIYBsDOBTANCAPZQDaIA7AKwAsIAugL6OEBMZIA6gApi0CCNAMoBRAFIBmAGoAhACIAtAOKcA9AGF%2BDRkA

I'm not sure, though, if a 12V RGB header supplies 12V and has different R, G, and B grounds, or the other way around.

okay this is different from what I expected. I thought you were just trying to use this for powering something else. Looking at that part, I notice that each LED has a different forward voltage drop. Sooo I guess the way to do it would be something like this. This isn't exactly how it would work, but the diodes and series resistors are correct if you set the buck converter to 4v. I plugged 4v, forward current, and the forward voltage drops of each of the individual diodes in that package. So you'd need to connect a resistor to each annode of the LED and connect the cathode into 4v.

 

What's different here is the step down converter would not be connected like I showed, this is a more classical circuit view. For a step down converter like I linked above, you would connect + and - on your LED header to the + and - IN on the step down converter. Then the + and - OUT on the converter would be connected to your resistors and ground respectively. I don't think LT spice has a symbol for step down converters so I just mocked it up with voltage sources. 

Spoiler

 

[WhitetailAni]

1 minute ago, Hackentosher said:

okay this is different from what I expected. I thought you were just trying to use this for powering something else. Looking at that part, I notice that each LED has a different forward voltage drop. Sooo I guess the way to do it would be something like this. This isn't exactly how it would work, but the diodes and series resistors are correct if you set the buck converter to 4v. I plugged 4v, forward current, and the forward voltage drops of each of the individual diodes in that package. So you'd need to connect a resistor to each annode of the LED and connect the cathode into 4v.

 

What's different here is the step down converter would not be connected like I showed, this is a more classical circuit view. For a step down converter like I linked above, you would connect + and - on your LED header to the + and - IN on the step down converter. Then the + and - OUT on the converter would be connected to your resistors and ground respectively. I don't think LT spice has a symbol for step down converters so I just mocked it up with voltage sources.

Bit confusing to me, so I'm going to try to explain it to myself:

1. I cut, separate, and strip the 12v extender so I have 4 wires - 12v, R, G, and B.

2. I take the voltage converter and step the 12v line down to 4v.

3. I then split the 4v line out into four.

4. I take one of the 4 4v lines and connect it to the common anode of the RGB LEDs.

5. I then take the R, G, and B cathodes and merge the like cathodes together, and connect it to the respective ground pins.

All of this is if the 12v pin on a 12v RGB header supplies 12v, rather than taking 12v.

[Hackentosher]

19 minutes ago, ragnarok0273 said:

Bit confusing to me, so I'm going to try to explain it to myself:

1. I cut, separate, and strip the 12v extender so I have 4 wires - 12v, R, G, and B.

2. I take the voltage converter and step the 12v line down to 4v.

3. I then split the 4v line out into four.

4. I take one of the 4 4v lines and connect it to the common anode of the RGB LEDs.

5. I then take the R, G, and B cathodes and merge the like cathodes together, and connect it to the respective ground pins.

All of this is if the 12v pin on a 12v RGB header supplies 12v, rather than taking 12v.

1. yes
2. yes, but you'll also need a common ground for the converter to work. Electricity needs a way to come in and a way to come out.
3. no. Looking at the datasheet of the LED you linked, we connect 4v to pin 2, then the resistors to the other pins. I guess then you would connect the anodes of each LED to R G and B respectively, that should give you control over each color. You need the resistors in between each anode and RGB pin because the resistors limit the current going through each LED so it'll keep them from blowing up. If you connected 4v directly across an LED it would try to pull as much current as it could before it melts. 
   

perhaps my SPICE schematic wasn't as helpful as I hoped, but I hope this clears it up.

[WhitetailAni]

1 minute ago, Hackentosher said:

1. yes
2. yes, but you'll also need a common ground for the converter to work. Electricity needs a way to come in and a way to come out.
3. no. Looking at the datasheet of the LED you linked, we connect 4v to pin 2, then the resistors to the other pins. I guess then you would connect the anodes of each LED to R G and B respectively, that should give you control over each color. You need the resistors in between each anode and RGB pin because the resistors limit the current going through each LED so it'll keep them from blowing up. If you connected 4v directly across an LED it would try to pull as much current as it could before it melts. 
   

perhaps my SPICE schematic wasn't as helpful as I hoped, but I hope this clears it up.

Ah.

So I connect 4v to the anode, and then resistors to the R, G, and B cathodes.
I then duplicate this by 4 as I'm planning to run 4 LEDs, and do I:
1. Combine the R, G, and B cathodes after the resistors, then connect them to the R, G, and B pins.

2. Combine the R, G, and B cathodes and then put resistors in between the combined cathodes and the R, G, and B pins.

[Hackentosher]

39 minutes ago, ragnarok0273 said:

Ah.

So I connect 4v to the anode, and then resistors to the R, G, and B cathodes.
I then duplicate this by 4 as I'm planning to run 4 LEDs, and do I:
1. Combine the R, G, and B cathodes after the resistors, then connect them to the R, G, and B pins.

2. Combine the R, G, and B cathodes and then put resistors in between the combined cathodes and the R, G, and B pins.

If I'm understanding correctly, the first one. Each R resistor should be connected to the R pin, each G resistor to G pin and B resistor to B pin. Your usage of "combine" is confusing me. Can you draw the circuit you're thinking of?

[WhitetailAni]

Here ya go:

Spoiler

I'm planning to run 4 LEDs total.

Sorry if it's a bit messy.

So:

I take the 12v line and convert it to 4v. I then split it into 4, and connect each 4v line to an anode of the RGB LEDs. After that, I connect resistors to each cathode of each LED, connect the like cathodes, and connect all of that to the respective pin.

[WhitetailAni]

@Hackentosher

Was looking more at the voltage converter I was linked to and realized something:
Instead of taking Voltage A in and spitting Voltage B out, it completes a circuit with Voltage A and creates a new circuit with Voltage B.

But I had a thought:
SATA provides three voltages: 12V, 5V, and 3.3V.

The RGB LED I found has a common anode, with 2.2V red and 3.3V blue and green.

So, I should be able to connect the 3.3V from SATA to my RGB LED anode, and then the R, G, and B cathodes to the R, G, and B pins on the header.

I can also then take the 12V and ground lines on my fan and connect it to that, instead of using a "3-pin" fan connector that I have to reattach every time I take the side panel off.

Does this work or no?
 

[cachethrash]

I would use the 5V line instead of 3.3, if you're going to just use a line direct from the power supply, for some overhead. And you would still need resistors for each LED color, like in the figure above. And if you use 5V (or 3.3) instead of 4, you might need to recalculate the resistor values. And to be safe, I would then connect them to the ground wire in the sata power connector, not through the motherboard.

Also, to be pedantic, Ohm's Law is (mostly?) for linear circuits and devices, and LEDs are technically non-linear devices, so it's easy to get confused trying to apply Ohm's Law to LEDs and often things don't work out the way you think they will.

[WhitetailAni]

18 minutes ago, cachethrash said:

I would use the 5V line instead of 3.3, if you're going to just use a line direct from the power supply, for some overhead. And you would still need resistors for each LED color, like in the figure above. And if you use 5V (or 3.3) instead of 4, you might need to recalculate the resistor values. And to be safe, I would then connect them to the ground wire in the sata power connector, not through the motherboard.

Also, to be pedantic, Ohm's Law is (mostly?) for linear circuits and devices, and LEDs are technically non-linear devices, so it's easy to get confused trying to apply Ohm's Law to LEDs and often things don't work out the way you think they will.

I'm trying to mod my fan into an RGB fan that I can connect to a 12v RGB header.

That's why.

[mariushm]

Those 4 pin LEDs are not optimal for what you're trying to do, . 

 

Best option would be to get 6-8 pin LEDs where each color element has its own anode and cathode. 

 

For example, something like this: https://www.digikey.com/en/products/detail/stanley-electric-co/ARGB1313HS-TR/6798001

 

It had diffuse top, so the three leds blend together nicely resulting in a single color,  and you can easily solder it to a prototyping board if that's all you have... you can cut thin copper wires from an ethernet cable for example and bend them down and solder them to the pads of the led. 

The corners are at approximately 2.54mm apart (0.1") so you could have wires coming out of prototyping board and bend to touch the corners. Then have wires to the middle of the led ... something like this but even more space between the through holes, led is tinier : 

 

 

 

As independent leds, this will allow you to connect 3 of them in series ... each color its own chain of 3 leds. 

So then you'll have 3 times the forward voltage of each led  .. 2.2v for red,  3.2v for green and blue ... so  6.6v  and 9.6v

Now you can use the formula V = I X R  to calculate the resistor, for let's say 10mA because it would be enough ... so

 

Voltage - 3 leds of same color x forward voltage for that color  = Current x Resistor 

 

R = (12v - 6.6v) / 0.01A = 540 ohm , so you'd use either 470 ohm for a bit more current, or 560 ohm for a bit less than 10mA

R = (12v  - 9.6v) / 0.01A = 240 ohm , so you'd use either 210 ohm (or 2x100 in series) for a bit more current, or 270 ohm for a bit less than 10 mA

 

The resistors can have a low rating, the power dissipated is P = IxIxR  ..so for red for example if we go with 470 ohm, there's gonna be 0.01x0.01x470 = 0.047 watts dissipated as heat, so this means a standard surface mount 0.1w or a tiny leaded 0.125w resistor would be plenty adequate. 

 

 

So you'll have groups of 3 leds and the resistor for each color, and then after the resistor that wire goes to R, G or B in the connector. 

 

 

[WhitetailAni]

On 2/2/2021 at 2:24 AM, mariushm said:

Those 4 pin LEDs are not optimal for what you're trying to do, . 

 

Best option would be to get 6-8 pin LEDs where each color element has its own anode and cathode. 

 

For example, something like this: https://www.digikey.com/en/products/detail/stanley-electric-co/ARGB1313HS-TR/6798001

 

It had diffuse top, so the three leds blend together nicely resulting in a single color,  and you can easily solder it to a prototyping board if that's all you have... you can cut thin copper wires from an ethernet cable for example and bend them down and solder them to the pads of the led. 

The corners are at approximately 2.54mm apart (0.1") so you could have wires coming out of prototyping board and bend to touch the corners. Then have wires to the middle of the led ... something like this but even more space between the through holes, led is tinier : 

 

 

 

As independent leds, this will allow you to connect 3 of them in series ... each color its own chain of 3 leds. 

So then you'll have 3 times the forward voltage of each led  .. 2.2v for red,  3.2v for green and blue ... so  6.6v  and 9.6v

Now you can use the formula V = I X R  to calculate the resistor, for let's say 10mA because it would be enough ... so

 

Voltage - 3 leds of same color x forward voltage for that color  = Current x Resistor 

 

R = (12v - 6.6v) / 0.01A = 540 ohm , so you'd use either 470 ohm for a bit more current, or 560 ohm for a bit less than 10mA

R = (12v  - 9.6v) / 0.01A = 240 ohm , so you'd use either 210 ohm (or 2x100 in series) for a bit more current, or 270 ohm for a bit less than 10 mA

 

The resistors can have a low rating, the power dissipated is P = IxIxR  ..so for red for example if we go with 470 ohm, there's gonna be 0.01x0.01x470 = 0.047 watts dissipated as heat, so this means a standard surface mount 0.1w or a tiny leaded 0.125w resistor would be plenty adequate. 

 

 

So you'll have groups of 3 leds and the resistor for each color, and then after the resistor that wire goes to R, G or B in the connector. 

 

 

I don't have the skills for this kind of thing.

That's why I was thinking the 4-pin ones, because I can use my breadboard wires, cut the ends off, and splice them together with some solder and electrical tape.

Anyway, my refund money came in, so I'd like to buy the parts.

I'm using 4-pin common-anode LEDs with 2.2v on red and 3.3v on blue and green, with a 3.3v or 5v power source.

What resistors should I buy?

[mariushm]

7 hours ago, ragnarok0273 said:

I don't have the skills for this kind of thing.

That's why I was thinking the 4-pin ones, because I can use my breadboard wires, cut the ends off, and splice them together with some solder and electrical tape.

Anyway, my refund money came in, so I'd like to buy the parts.

I'm using 4-pin common-anode LEDs with 2.2v on red and 3.3v on blue and green, with a 3.3v or 5v power source.

What resistors should I buy?

 

You'd need resistors for each individual led.  edit : for each color in the led.. a resistor between each cathode and ground. A CA led will have a common anode pin (the input voltage) and the 3 separate cathodes (the negatives).. you put resistors between negatives and ground or minus of a battery (or COM for common) 

 

 

The formula is simple :  Input voltage -  number of leds x forward voltage of led  (or individual color in led)   =  Current  (in A)  x Resistance 

 

From here, you can flip the formula  and say  Resistance x Current = Input voltage - number of leds x forward voltage

And you can divide both sides with current and you get 

 

Resistance =  [ Input Voltage - (number of leds x forward voltage ) ]  / Current 

 

You can use same formula to figure out current if you have a known resistance value : 

 

Current = [ Input Voltage - (number of leds x forward voltage ) ]  / Resistance

 

You pick the maximum current you want ... for a typical led with pins that would be 20mA  or 0.02A  but it may be too bright at that amount of current, and you'll also get more life better blending of colors at lower current. I'd suggest 5-10mA, you don't want your fan to be a flashlight. 

Also, keep in mind that human eyes are more sensitive to red (hence why the stop lights are red), so you may not want as much current on red as you have on the green and blue leds.  For example, you may aim for 10mA on red and 15mA on green and blue. 

 

Once you get a value, you round it up and down to a E series value (E12 or E24 are most common series, easiest to find)    : https://en.wikipedia.org/wiki/E_series_of_preferred_numbers#Lists

 

So let's say 10mA for red and 15mA for green and blue, and let's say 2v for red and 3v for green and blue.   

 

red : Resistance =  (5v  - 1 led x 2v ) / 0.01 = 3 / 0.01 = 300 ohm   (This is E24 series value, 3.0 x 100, easy to find, alternatives would be 270 Ohm or 330 ohm ) 

green, blue  : Resistance  = (5v - 1 led x 3v) / 0.015 = 2/0.015 = 133.33 ohm ( this is not E12 or E24 series, so we pick closest ... 120, 130 or 150 ohm)  

 

if you go with 120 ohm, you'll get a bit higher current, but still below 20mA  :  Current = (5v - 3v ) / 120 = 2/120 = 0.016A or 16mA

A 150 ohm resistor will get you a bit lower current ... 2v / 150 = 0.013A or 13mA 

 

edit:  you also need to know how much power is dissipated in the resistors ... formula is derived from ohm's law  V=IxR  ... 

so power is = IxV  = IxIxR =I²xR  

 

So if you pick 300 ohm and 0.01A then Power = 0.01x0.01x300 = 0.03 watts dissipated in resistor, which means you can use small ones rated for 0.1w or 0.125w  - if the value is around 75% or more of the resistor rating, pick resistor with higher rating... for example, if you dissipate 0.1w that's pretty close to the standard 0.125w rating of a resistor, so pick a 0.25w rated resistor for safety.

 

 

You don't have to be anal about resistor values, because resistors are a crude way of limiting current ... go up and down and pick E series values. if you want really accurate current, you use led drivers, not resistors, and a mA variation won't really be noticeable with these cheap leds. 

 

Basically , the forward voltage of a led varies with temperature.  When the led is room temperature, let's say 20c , the forward voltage may be 3v (for the blue color)  but after 10-20 minutes of operation when it gets warm at let's say 50 degrees celsius, the forward voltage may lower to around 2.8v 

Unless the resistor has a very poor temperature coefficient, the resistance remains the same, so as the led gets warm, it will consume slightly more current. Your blue led may consume only 15mA when you turn on the PC, but may consume 15.5mA after half an hour. 

 

This is one of the other reasons why it's always a good idea to design something to use slightly less current that maximum values. 

  

    

 

[WhitetailAni]

6 hours ago, mariushm said:

-snip-

If I'm running them in parallel, do I need 1 resistor for each cathode (for a total of 9) or can I just get 3, putting it between the respective ground pin and where the cathodes merge?

[mariushm]

You can, it's not ideal, but it's acceptable. You would have to wire like this ... 

 

As there's the current of 3 leds going through each resistor, you simply have to reuse the formula but multiply the current. 

 

So for red with 2v forward voltage and 5v input and 10mA  you now have 30mA or 0.03A of current because there's 3 red diodes in parallel : 

 

input voltage - (1 led / 3 in parallel) x 2v  = 3 x 0.01 A x Resistance  so Resistance = (5v - 2v)/0.03 = 100 ohm 

but don't forget about the power dissipated in resistor ... now you have triple current so P = IxIxR = 0.03 x 0.03 x 100  = 0.09w  ... 0.125w resistors would still be acceptable

 

for green and blue, assuming 3v and 15mA  ...  (5v - 3v ) / 0.045A   = 2 / 0.045 = 44.45 ohm , so I'd probably use an E24 series 47 ohm resistor, resulting in around 14.5mA per led

The resistor would have to be rated for P = IxIxR = 0.045x0.045x47 = 0.095w .. so 0.125w rated resistor still works.

 

 

 

 

 

 

The resistors connect to ground (like in the picture above) or to the RGB pins in a header, if your motherboard has rgb controller built in. In that case, the resistors simply configure the maximum current and then the motherboard uses PWM  (connects or disconnects the wires' connection to ground thousands of times a second) and so the amount of color brightness changes.

 

The reason this is rarely suggested is for two reasons : 

 

1. There's no two identical leds, due to manufacturing differences, some diodes will have slightly different forward voltage: the red diode in first LED could have 1.8v forward voltage, the one in the second LED could have 2.1v forward voltage.  This means the first led will consume slightly more current than the other LED so it may appear brighter and may also be damaged in time  - not the case with these small leds, but it's a serious risk with leds designed for 100mA or even 500mA  ... 

With these generic cheap rgb leds, the differences are usually fairly small, so it's not a big concern. 

 

2. If someone one of the led diode dies or the wire disconnects, the other two LEDs have to carry the full current "configured" by the resistor.  So in the example above, if one led dies, the other two "share" that 30 mA of current, so each led diode will use 15mA instead of 10mA and will be much brighter.   

If you had 3 x 100mA leds and one died,  each remaining LED would have to carry 150mA and if they're only rated for 120mA or so, they could break down and die. 

 

In your case, there's minimal risk, especially if you set the current conservatively, lower than the maximum recommended value. 

 

[WhitetailAni]

2 minutes ago, mariushm said:

You can, it's not ideal, but it's acceptable. You would have to wire like this ... 

 

As there's the current of 3 leds going through each resistor, you simply have to reuse the formula but multiply the current. 

 

So for red with 2v forward voltage and 5v input and 10mA  you now have 30mA or 0.03A of current because there's 3 red diodes in parallel : 

 

input voltage - (1 led / 3 in parallel) x 2v  = 3 x 0.01 A x Resistance  so Resistance = (5v - 2v)/0.03 = 100 ohm 

but don't forget about the power dissipated in resistor ... now you have triple current so P = IxIxR = 0.03 x 0.03 x 100  = 0.09w  ... 0.125w resistors would still be acceptable

 

for green and blue, assuming 3v and 15mA  ...  (5v - 3v ) / 0.045A   = 2 / 0.045 = 44.45 ohm , so I'd probably use an E24 series 47 ohm resistor, resulting in around 14.5mA per led

The resistor would have to be rated for P = IxIxR = 0.045x0.045x47 = 0.095w .. so 0.125w rated resistor still works.

 

 

 

 

 

 

The resistors connect to ground (like in the picture above) or to the RGB pins in a header, if your motherboard has rgb controller built in. In that case, the resistors simply configure the maximum current and then the motherboard uses PWM  (connects or disconnects the wires' connection to ground thousands of times a second) and so the amount of color brightness changes.

 

The reason this is rarely suggested is for two reasons : 

 

1. There's no two identical leds, due to manufacturing differences, some diodes will have slightly different forward voltage: the red diode in first LED could have 1.8v forward voltage, the one in the second LED could have 2.1v forward voltage.  This means the first led will consume slightly more current than the other LED so it may appear brighter and may also be damaged in time  - not the case with these small leds, but it's a serious risk with leds designed for 100mA or even 500mA  ... 

With these generic cheap rgb leds, the differences are usually fairly small, so it's not a big concern. 

 

2. If someone one of the led diode dies or the wire disconnects, the other two LEDs have to carry the full current "configured" by the resistor.  So in the example above, if one led dies, the other two "share" that 30 mA of current, so each led diode will use 15mA instead of 10mA and will be much brighter.   

If you had 3 x 100mA leds and one died,  each remaining LED would have to carry 150mA and if they're only rated for 120mA or so, they could break down and die. 

 

In your case, there's minimal risk, especially if you set the current conservatively, lower than the maximum recommended value. 

 

I don't think I can control the current - I'm just taking the 5v or 3.3v line from a SATA plug.

But I see your point.

I'll get 10, then, since getting 10 costs less than nine.

Just to clarify - I'd need a 120 ohm or 150 ohm resistor for the red line, and then 270 or 330 ohm for blue and green?

[mariushm]

Buy the RGB leds.

Resistors should be very cheap, so buy a bunch of 47 ohm, 68 ohm , 100 ohm, 120 ohm , 150 ohm  resistors, maybe get 10-25 packs of each.

Then see how bright each element is with various resistor values and settle on the values you like most for maximum brightness.

 

I gave you the formulas, put the voltage and current you want in the formula as I showed you.

 

You can get different values by using 2 resistors in series or in parallel. 

Two resistors in series results in sum of the resistances .. so 100 ohm + 47 ohm = 147 ohm

Two resistors in parallel results in lower value .. 1/R = 1/R1 + 1/R2  ... so 1/R = (R1+R2)/R1xR2  so R = R1xR2 / (R1+R2) ...

so for example 100 and 47 ohm in parallel  => R = 47x100 / (100+47) = 4700/147 = 32 ohm 

 

Buy some lower value resistors, like 10 ohm , 47 ohm, 100 ohm.  You can put two or three resistors in series or parallel to get a different value resistor.

 

... They're 2 cents a piece if you order packs of 50, so 1$ for each value .. packs of 25 are around 2.5 cents so around 60 cents

Here's link with most common resistor values, cheap (for digikey): https://www.digikey.com/short/4tddrn

 

If you put 10 leds in parallel and use only 3 resistors, make sure to measure the dissipated power and get properly rated resistor (use the formula i showed you)

[WhitetailAni]

1 hour ago, mariushm said:

Buy the RGB leds.

Resistors should be very cheap, so buy a bunch of 47 ohm, 68 ohm , 100 ohm, 120 ohm , 150 ohm  resistors, maybe get 10-25 packs of each.

Then see how bright each element is with various resistor values and settle on the values you like most for maximum brightness.

 

I gave you the formulas, put the voltage and current you want in the formula as I showed you.

 

You can get different values by using 2 resistors in series or in parallel. 

Two resistors in series results in sum of the resistances .. so 100 ohm + 47 ohm = 147 ohm

Two resistors in parallel results in lower value .. 1/R = 1/R1 + 1/R2  ... so 1/R = (R1+R2)/R1xR2  so R = R1xR2 / (R1+R2) ...

so for example 100 and 47 ohm in parallel  => R = 47x100 / (100+47) = 4700/147 = 32 ohm 

 

Buy some lower value resistors, like 10 ohm , 47 ohm, 100 ohm.  You can put two or three resistors in series or parallel to get a different value resistor.

 

... They're 2 cents a piece if you order packs of 50, so 1$ for each value .. packs of 25 are around 2.5 cents so around 60 cents

Here's link with most common resistor values, cheap (for digikey): https://www.digikey.com/short/4tddrn

 

If you put 10 leds in parallel and use only 3 resistors, make sure to measure the dissipated power and get properly rated resistor (use the formula i showed you)

Thanks!

[WhitetailAni]

I'm looking at solder to get, and I'm not sure what I'm looking for.

I've got an old soldering pen that starts up as soon as you plug it on, a really old soldering gun that's super heavy, and experience that amounts to moving solder around to get capacitors out.

If I'm wanting to solder power wires carrying up to 12v at 1.5a, what type and size of solder should I be looking for?

[mariushm]

Those soldering pens are shit. Takes a lot time to heat up and they cool down as soon as you put the tip on something.  Should be good enough to solder wires on leds though, it's not a lot of metal.

The soldering guns are better, but they're designed for use a few seconds at a time. The transformer inside and the copper wire tip are not designed for continuous use.

Hold switch down, solder tip heats up in a few seconds and you solder then release switch and tip cools down.  With these, you kinda need to add flux to anything before attempting to solder because the tip temperature is too high and burns the flux inside the solder wire before it gets a chance to do its job (it attacks the surfaces like an acid, removing oxides from surfaces and makes it possible for the metals to chemically bind).

Using flux with the soldering pen would also be a good idea, it really makes soldering much easier, but it's not a must, the tip is usually not so hot that it would burn the flux before it works. 

So yeah, if you can afford to buy some liquid flux, buy it, it really helps.. apply a drop on the surfaces before you apply solder.

 

What solder, 63/37 or 60/40 if you can still buy leaded solder. It's better than lead free.

As for thickness... I'd say anything between 0.5 .. 1mm thick .... though 1mm would be a bit too thick for soldering thin wires to led leads... no idea what that is in imperial units... let me see digikey for some suggestions

 

IF you want something cheap and enough to solder a few leds and make some repairs : https://www.digikey.com/en/products/detail/chip-quik-inc/NCSW-020-0-3OZ/13531439

It's 0.5mm thick ,63/37, no-clean flux, it's good but relatively small quantity at 8 grams of solder. But it's only 3$.

 

You can get 55 grams of 0.5mm solder at 7$ : https://www.digikey.com/en/products/detail/chip-quik-inc/RASW-020-2OZ/9681985

 

You get more value out of 250g or 500g rolls, at around 20-40$, and those would last you for years, but it's too much for just a few leds.  

 

You don't need thick wires for a few leds, especially if they only consume 10-20mA each... with 10 leds you'd be looking at less than 2-300 mA  ... wires as thick as the ones used on fans would be plenty thick

 

 

[WhitetailAni]

6 hours ago, mariushm said:

Those soldering pens are shit. Takes a lot time to heat up and they cool down as soon as you put the tip on something.  Should be good enough to solder wires on leds though, it's not a lot of metal.

This one doesn't.

I actually have to leave it out for 5 minutes to let it cool down.

It's maybe 20 years old.

6 hours ago, mariushm said:

The soldering guns are better, but they're designed for use a few seconds at a time. The transformer inside and the copper wire tip are not designed for continuous use.

Hold switch down, solder tip heats up in a few seconds and you solder then release switch and tip cools down.  With these, you kinda need to add flux to anything before attempting to solder because the tip temperature is too high and burns the flux inside the solder wire before it gets a chance to do its job (it attacks the surfaces like an acid, removing oxides from surfaces and makes it possible for the metals to chemically bind).

Way too heavy to use accurately. This one's from my great-great-grandpa (who I met when I was 2 months old). He died one day before his 105th birthday.

6 hours ago, mariushm said:

Using flux with the soldering pen would also be a good idea, it really makes soldering much easier, but it's not a must, the tip is usually not so hot that it would burn the flux before it works. 

So yeah, if you can afford to buy some liquid flux, buy it, it really helps.. apply a drop on the surfaces before you apply solder.

If I can afford it, I will get some.

6 hours ago, mariushm said:

What solder, 63/37 or 60/40 if you can still buy leaded solder. It's better than lead free.

As for thickness... I'd say anything between 0.5 .. 1mm thick .... though 1mm would be a bit too thick for soldering thin wires to led leads... no idea what that is in imperial units... let me see digikey for some suggestions

 

IF you want something cheap and enough to solder a few leds and make some repairs : https://www.digikey.com/en/products/detail/chip-quik-inc/NCSW-020-0-3OZ/13531439

It's 0.5mm thick ,63/37, no-clean flux, it's good but relatively small quantity at 8 grams of solder. But it's only 3$.

 

You can get 55 grams of 0.5mm solder at 7$ : https://www.digikey.com/en/products/detail/chip-quik-inc/RASW-020-2OZ/9681985

 

You get more value out of 250g or 500g rolls, at around 20-40$, and those would last you for years, but it's too much for just a few leds.  

 

You don't need thick wires for a few leds, especially if they only consume 10-20mA each... with 10 leds you'd be looking at less than 2-300 mA  ... wires as thick as the ones used on fans would be plenty thick

Thanks for the help!