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12v LED wiring info needed

Hi guys,

Apologies in advance if I'm asking a dumb question or don't have a crucial piece of info here. I'm working on an antique radio cabinet PC build, and will be wiring four to six orange LEDs on one circuit. I've already got the LEDs on hand. They're 12v, 18cm pre-wired. I have near-zero experience with electronics. I can solder wires, but don't know the theory at all. For testing purposes, I've got two old power supplies, both are 12v output. One is 2.5a and one is 2a. Eventually, the lights will be wired into a PC power supply 12v rail. I have no idea how many amps that puts out. Is 2a or 2.5a too much for that many LEDs? Do I need some resistors for long-term installation? Should I do these in series or parallel? Will that still be correct once I wire them into my PC power supply? Thanks!

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In general, components don't draw more current than they use. LEDs draw next to nothing. You will want to wire them in parallel though, as the voltage drop in series will be pretty significant.

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Amps on a PSU like this are only given as its needed. It won't force the amps in and blow the LEDs. If the LEDs are designed to run on 12V no resistor should be required. How much they draw is miniscule. We're talking mA. You could wire about as many as you like (within reason). And if you want all of them to light up at once parallel would be better.

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So basically, I can just connect up the wires, and if that's done properly I'll be good. I'll give that a try this weekend. Thanks guys, that was super helpful!

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55 minutes ago, Epitome_Inc said:

So basically, I can just connect up the wires, and if that's done properly I'll be good. I'll give that a try this weekend. Thanks guys, that was super helpful!

If you want to air on the side of cation you can try connecting just one LED and let it run for a while. If it's fine connect the others.

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Just now, Windows7ge said:

If you want to air on the side of cation you can try connecting just one LED and let it run for a while. If it's fine connect the others.

Meh, I've got like... 50 of them and only need 6. I'll just go for it. Thanks again!

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Just a word of warning LEDs are not current limiting. They will eat as much as they can, until the die. If you are really sure that they are rated for 12v and have current limiting resistors inside of them, then you are fine to go. The analog RGB LEDs inside of PCs that use 12v are wired with 3 LEDs in series, each color also has a resistor to limit the current. Not having a picture of link makes it hard to help.

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We need to know the kind of 12v led you're going to use. 

 

By default, a basic LED does not have any current limiting mechanism, so if you just connect a led like you would connect an incandescent light bulb, the led will suck so much current it will damage itself. 

Some LED bulbs with sockets have current limiting mechanism built in, same for the led bulbs for cars which have sockets. 12v led strips will almost always arrange leds in groups of 3-4 leds and each of those groups will have current limiting mechanism in the form of a resistor. 

 

The easiest way to limit current is by adding a resistor in series with the LED or with a series of leds. 

Your led will have a default forward voltage, which you should assume to be around 2.8v...3.2v depending on the chemicals used to make the led. 

Some led packages have multiple led dies inside connected in series, very common packages will have a voltage of 6v (2 led dies in series, super common in monitor and tv backlights), or 9v (3 led dies in series).

 

If your LEDs have a forward voltage of 3v, the easiest for you considering you plan to use a 12v power supply - is to make groups of 3 leds in series and use a resistor for each group, to limit the current going through the leds. 

 

You can calculate the resistor value using this formula  Input voltage - (number of leds x forward voltage of led )  = Current x Resistance 

 

So if your power supply is 12v, you have 3 leds each with 3v forward voltage, and you want 20mA (0.02A) going through the leds, then : 

 

Resistance = [ 12v - ( 3 x 3v ) ]  / 0.02A  = 150 ohm

 

The amount of energy being dissipated inside the resistor can be calculated with formula  P = Ix R  = 0.02 x 0.02 x 150  = 0.06 watts , which tells you it's fine to use a standard 0.125w rated resistor to limit the current, without the resistor overheating itself. 

 

If you have more powerful leds, like let's say 100mA each, then you'd use (12-3x3/0.1 = 30 ohm (you'd round to 27 or 33 ohm, standard values, or put 2x15 or 3x10 in series, or maybe two 68 ohm resistors in parallel, giving you 34 ohm)   and the power dissipated would be 0.1 x 0.1 x 30 = 0.3 watts, which means you should use a resistor rated for 0.5w, or if you parallel two resistors to get the value, 0.25w resistors would work (the current is split across the two resistors in parallel, so each resistor will heat less.

 

 

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Oh wow. This is why I asked. I got these LEDs in orange (which doesn't show on the site anymore)
https://www.amazon.ca/gp/product/B01J74ZEAK/ref=ppx_yo_dt_b_asin_title_o00_s00?ie=UTF8&th=1

"TrendBox 1 Pack (100 Bulbs) 5mm Orange LED Flash Light (Steady) 18cm 12V DC Pre-Wired Round Top Bulb Lamp for DIY Car Boat Toys Flashing Parties" 

No other specs are given on the page. I know different colors have different specs, but that's about it.

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You need current limiting on LEDs because they will burn them selves out otherwise. This can be either a current limiting power supply or a resistor. Little known fact is LEDs have a dynamic resistance which is a tangent line between a graph of forward voltage vs forward current. So as the voltage increases across the LED the current exponentially increases because the resistance decreases causing it to burn out. This is because Ohms law is I = V / R or Current = Voltage over Resistance. So by decrease the resistance while keeping the voltage the same you increase your current. 

 

This is pictured here. The graph on the left is forward voltage vs forward current while the graph on the right shows the resistance of the LED.

image.thumb.png.24fc1ac12fd73e4b435b5337b2a33e48.png

 

So in essence the resistor will drop off any voltage thats left over after the forward voltage with the desired forward current. The formula to calculate the resistor value is derived from Ohms law. Its R = (V- Vf) / If

So its Resistance  = voltage supply - voltage forward over forward current. To find the forward voltage and acceptable forward current of your LED you will need to look at the datasheet for the LED. Also a resistor will have to be placed in series with each LED because diodes don't always behave identically to each other. So if you had one resistor to limit the current of all the LED's one of the would have the potential of taking all the current while the others don't even light up.

Your schematic would look like this. You have a common 12V rail and then a common ground rail but each LED has its own resistor.

image.thumb.png.afa73109b4d2674a6b7b00c9f404c267.png

As an example:

So if you're LED has a forward voltage of say 3.2V and a rated current of 20mA with a supply voltage of 12V. 

R = (V- Vf) / If  = (12V - 3.2V) / 0.02A = 8.8V / 0.02A = 440Ω resistor

From here you will need to calculate the amount of power flowing through each resistor so that you can choose the correct wattage rating.

You can do that with the formula:

 

P = I2R or Power = Current2 x Resistance

P = 20mA2 x 440Ω = 0.176W or 176mW

Resistors commonly come in 1/8W 1/4W 1/2W 3/4W or 1W sizes. In this example we would use a 1/4W resistor because that gives us breathing room in the design.

 

 

You will need to do these 2 calculations for the leds you have. If you want to some help deciphering the datasheet let me know and I can walk you through it. Or if theres anything you would like to know more about!

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I don't have a datasheet for them unfortunately. Is there a way to determine the necessary info using a multimeter?

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14 minutes ago, Epitome_Inc said:

I don't have a datasheet for them unfortunately. Is there a way to determine the necessary info using a multimeter?

If you have the part number I may be able to find it. If not you can make a somewhat educated guess with a multimeter. If you hookup your LED to power with a larger resistor around a 1K (so you don't burn anything out at all) and the put the test leads like I've shown in the schematic then you can measure with your multimeter in voltage mode the voltage drop across the LED. its not perfect because nothing about the circuit will be ideal but it can give you a good guess. You will not know how much current the LED is rated for so I would stick to 10-15mA with the absolute max being 20mA in your final design because you simply don't know what they're good for.

 

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2 hours ago, Epitome_Inc said:

Oh wow. This is why I asked. I got these LEDs in orange (which doesn't show on the site anymore)
https://www.amazon.ca/gp/product/B01J74ZEAK/ref=ppx_yo_dt_b_asin_title_o00_s00?ie=UTF8&th=1

"TrendBox 1 Pack (100 Bulbs) 5mm Orange LED Flash Light (Steady) 18cm 12V DC Pre-Wired Round Top Bulb Lamp for DIY Car Boat Toys Flashing Parties" 

No other specs are given on the page. I know different colors have different specs, but that's about it.

The leds themselves should have a forward voltage of 2.8v .. 3.2v and a current rating of let's say 20mA ... but i would use them at 10-15mA to give them a longer life.

If you look at them, you can see each led has a resistor hidden inside the positive wire, limiting the current and therefore protecting the led, so you can parallel as many of these orange leds as you want, each led will be limited to some amount of current based on the value of the resistor inside:

image.png.d2b29f072cf0a14b8a52488cff406397.png

 

So if your leds are like this ... I'd suggest taking a very sharp blade or a knife and cutting the heatshrink (the black sleeve) over the red wire, to get access to the hidden resistor.

You can now either look at the colors on the resistor to determine the resistor value or simply use a digital multimeter in resistance mode (put the probes on the ends of resistor). Once you know the resistor value, you can figure out the voltage and the current you need.

 

For example, let's say that's a 470 ohm resistor.   You have Ohm's law : Voltage  = Current x Resistance

 

So:  Voltage of power supply  - number of leds x Forward voltage of led  = Current x Resistance 

 

Let's assume they go with 20mA or 0.02A ...   Voltage - 3v = 0.02 x 470  = > Voltage = 0.02x470 + 3 = 9.4 + 3  = 12.4v  which would be reasonable for their title where they say 12v input voltage

 

You can go the other way where you calculate how much current you'd have with various power supplies :

 

with 5v psu :  5v - 1 x 3v = current x 470 = > current = (5-3)/470 = 2/470 = 0.004 or 4mA which is quite low and the leds will barely light up.

with 9v psu :  9-3 = current / 470 => current=  (9-3)/470 = 0.012  or 12mA which is reasonable.

 

 

You can also connect a few of them in series, if you want to use them with a higher voltage power supply, and the formula remains the same but you multiply the resistance value. 

For example, if you want to make groups of 3 leds in series and each led has a 470 ohm resistor on its positive wire :

 

psu voltage - 3 leds x forward voltage  =  current x  (resistance x 3 leds)

 

psu voltage -  3x3 = current x 3 x 470  => psu voltage = 1410 x current + 9 

So for 10 mA , you'll want psu voltage  = 1410 x 0.01 + 9 = 23.1v ... so a 24v power supply would be great, each led would get around 11 mA

 

 

So yeah ... if your individual leds each has one of those resistors on the wire, then you can use them like incandescent light bulbs, each led is current limited by that resistor and won't blow up.

You can find out the actual resistance value and with the formulas above you can determine how bright each led will be depending on the voltage of the power supply.

 

 

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Yes, I think you're right with the resistor value. I found that it looks like yellow-purple-black-black-red. An online calculator tells me that's 470 ohms. The power supplies I got from the scrap pile at work show as 12v 2.5A and 12v 2A. I'm still utterly lost on the calculations from everyone above, but it looks to me like 2A would be overkill for these. Correct me if I'm wrong (which is likely).

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Well, yeah, in that case each led will consume around 20mA ... so 10 leds will be 200 mA, and 100 leds will consume around 2000 mA (2A) 

 

Won't be quite accurate because the wires themselves will have some resistance, which will be added to the resistance of those resistors, and the resistors are not exactly 470 ohm (they probably use cheap +/- 5% tolerance resistors)  and the further you are from the actual power supply (long wires) the more losses are in the wires - for example, I suspect if you put 50 of those leds in a strip (all in parallel) with about 5-10 cm (2-4 inches) between them, I suspect the last leds in the chain will start to be a bit less bright due to the length of the wire adding resistance on top of the individual led resistors.

 

You can use those wallwarts even if they're overkill, they'll only produce as much as the leds need, so it's not like you're wasting electricity by using a higher wattage adapter.

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