I am currently planning a project to light up a model of a cruise ship. I've done this before with some success.

To keep things simple I want to wire in Parallel. (Assuming I've understood the difference correctly; Each LED will have a direct connection to the power source)

I will be using a mix of colours; White, Warm White, Blue, Red, Green. Every LED is pre-wired or will be fitted with a resistor (470) to allow them to be run from a 12v source.

Every online simulation I run and my own previous experience says this will work perfectly fine, but every article I read seems to scream "no don't do it" and even then, opinions seem to differ.

enter image description here

Above is an example of my wiring diagram, forgive the crudity. There will be a lot more LEDs used than this, but this gives you an idea of my plans.

Any thoughts?




This is the overall schematic that I have planned:

enter image description here

Sadly I cannot provide the direct link for this circuit as it is too long for browsers apparently.

  • 1
    \$\begingroup\$ What are the objections to doing it? seems fine to me \$\endgroup\$
    – Makoto
    Nov 17, 2017 at 12:27
  • 8
    \$\begingroup\$ If you are saying that every LED has its own resistor then it will be fine. However, with a 12 V supply you could save on wire, electricity, heat, and resistors by wiring groups of two, three, or four LEDs in series with one resistor per group - you would have to calculate the value of the resistor. \$\endgroup\$ Nov 17, 2017 at 12:28
  • 5
    \$\begingroup\$ @AndrewMorton I agree, with the proviso that he only wire up the LEDs of like colour that way. The reds and the warm-whites will have very different current requirements. \$\endgroup\$
    – Trevor_G
    Nov 17, 2017 at 12:39
  • 7
    \$\begingroup\$ The reason everyone screams don't do it is because you write that you are connecting LEDs in parallel - you're not! You clearly show each LED having an individual series resistor. \$\endgroup\$
    – pipe
    Nov 17, 2017 at 13:03
  • \$\begingroup\$ Hello. Yes, 90% of the resistors are "pre-wired with resistor to work at 9v-12v" They are all 3mm. The only ones that come without resistors are 0402 pre-wired. These don't come wired with a resistor so was planning on adding them myself. Again 1 per LED. So wiring them in groups of 2 or 3 per resistor isn't an option for me now. I'll be honest, my knowledge is pretty much limited to + is positive - is negative. Also, will the number of LEDs make any difference this way? There will be at least 126+ Probably closer to 150 time I am done. Thanks \$\endgroup\$ Nov 17, 2017 at 15:57

4 Answers 4


Your schematic is not ok, but your description is.

If you are using one LED and then the proper resistor the connection is fine. For instance (led values taken as example):


simulate this circuit – Schematic created using CircuitLab

Anyway, you will be wasting at least 3/4 of your power on the resistors.

My suggestion, whenever possible, put the maximum amount of leds in series:


simulate this circuit

Just keep in mind that

  1. you should not exceed 9-10V of drop on the leds (so for instance max number of 3V leds is 3)
  2. all the leds in the same line will sense the same current
  3. you will not be able to individually control the leds, just the lines (not a problem if you plan to turn them on
  4. for every branch, you will need to have at least one current-limiting device (a resistor is the cheapest option, but if you want higher quality control you can also use a constant current circuit made by two transistors and two resistors)

Note: as Olin said in the comments, many times you may want different currents for different LEDs. For instance, low brightness red LEDs may need a higher current than the standard blue ones, so it needs some balancing. So putting the leds in series should be the last step; please test the various currents before, in order to choose the appropriate value so that the effect is ok for you, then you can group them by current like shown before.

  • 9
    \$\begingroup\$ This is the right idea, +1. However, I would only put the same kind of LED in each string. That makes it easier to balance the relative brightnesses of the white versus the red, for example. It might be that whites at 8 mA match reds at 12 mA, for example. \$\endgroup\$ Nov 17, 2017 at 12:58
  • \$\begingroup\$ I agree completely with @OlinLathrop, whites can be a factor of 10 or more current than reds. \$\endgroup\$
    – Trevor_G
    Nov 17, 2017 at 13:17
  • \$\begingroup\$ @OlinLathrop I agree, but the premises are that the current inside the LEDs are already chosen at this stage, and that you want the red to have the same current as the 10mA white (that's why I wrote it in the schematic too). In any case this is worth at least a mention, thank you \$\endgroup\$
    – frarugi87
    Nov 17, 2017 at 13:27
  • \$\begingroup\$ Thanks. Could you be more specific re my schematic? Are the connections wrong or simply the symbols I used? (Forgive my crude knowledge in that dept.) \$\endgroup\$ Nov 17, 2017 at 15:58
  • \$\begingroup\$ @DominicJamesSibthorp well, in the schematic you only put some LEDs connected to +12V, while in the text you specified that each LED has its own resistor \$\endgroup\$
    – frarugi87
    Nov 17, 2017 at 21:12

but every article I read seems to scream "no don't do it"

LEDs should not be wired in parallel when driven with a constant current source.

When wired in parallel with a current limiting resistor and powered with a constant voltage source will work okay, but not ideal.

Ideal is where each LED or string of series LEDs are driven with its own constant current source or regulator.

The drawback to resistors is the forward voltage will vary depending on current, and temperature and therefore the current and luminous output will vary. Not too applicable in your project. Resistors should be fine for your application unless it is battery powered.

It appears you are going for 20mA (12V / 470Ω). Keep in mind the forward voltage varies especially between red and the other colors.

12V is kind of high for single LEDs. With 150 LEDs this project will draw about 35 watts where 25 of the watts is due to the resistors. The model will get very warm. If possible 5V would be much better and 3.3V would be ideal.

enter image description here

Keep in mind the luminous output at the same current can vary significantly. You may want to adjust the currents to match luminous output rather than forward current. For example within the same Cree XPE2 product line the luminous flux for Red Blue Green White at 350mA varies from 33 lumens (blue) to 126 lumens (green and white).

To choose the resistance value use a calculator such as: LED Series Resistor Calculator

  • 1
    \$\begingroup\$ It sounds like the "LEDs" the OP has are actually compound devices that contain both the actual junction and a series resistor dimensioned for 12 V in one package with two leads. (Which makes sense, especially when marketed to modeller hobbyists who then don't need to worry about doing their current limiting themselves). With that, putting them in series is not really an option. \$\endgroup\$ Nov 18, 2017 at 8:46
  • \$\begingroup\$ @HenningMakholm at no time did I recommend going with series. My biggest concern is how hot the model will get with 35 watts heating it up. Especially when 25 watts is wasted in the resistors. \$\endgroup\$ Nov 18, 2017 at 16:56
  • \$\begingroup\$ Can you give a bit more information about the statement "LEDs should not be wired in parallel when driven with a constant current source"? Is that because a slight over-voltage to one of the LEDs could cause thermal runaway? Is there something special about a battery power source? \$\endgroup\$ Dec 13, 2017 at 16:30
  • 1
    \$\begingroup\$ @DanEsparza it has nothing to do with battery other than efficiency is more important. Parallel is going to affect efficiency. the question is how much. It's more about the imbalance but in some cases the imbalance will create thermal runaway. At issue is no two LEDs or especially strings of LEDs will have the same forward voltage. That means when connected in parallel they are forced to have the same forward voltage. This stress at lease one LED. When the LEDs are high power and the current imbalance affects the temperature things get crazy. It's the differences in IV curve. \$\endgroup\$ Dec 14, 2017 at 17:08

As long as you have a resistor per LED then this should work fine.

At 12V: say worst case scenario there is 0v drop across the LED, then current is:

$$I =\frac{12\ \mathrm{V}}{470\ \Omega} = 26\ \mathrm{mA}$$

Power through resistor: $$P=I^2R = 0.026^2\times470 = 0.32\ \mathrm{W} $$

The only suggestion I would give is to use 1/2 watt resistors just to be safe

  • \$\begingroup\$ Thanks. For the LEDs that came pre-wired, I have no idea what the wattage of their resistors is. Just that they are done to work at 9v-12v. The LEDs I need to add resistors to will be 470 and I think they are 1/4w \$\endgroup\$ Nov 17, 2017 at 15:59
  • \$\begingroup\$ If the LEDs (with resistors) are rated for 12v that's fine. If you're going to use 1/4 watts, confirm with the specific led forward voltage drop that the resistors won't dissipate more than 1/4 watt \$\endgroup\$
    – Makoto
    Nov 17, 2017 at 16:06
  • \$\begingroup\$ Thanks. Sadly the only information I have regarding the LEDs that require resistors adding is: Forward voltage:2.8V~3.4V, AC or DC Compatible. Forward current (typ/max): 15mA/20mA. But that is a general description as the sale listing was for more than one type. IE 0402/0603 etc \$\endgroup\$ Nov 17, 2017 at 16:09
  • \$\begingroup\$ As others have said, for those LEDs requiring resistor (and given what you just said), you can easily string 2 LEDs in series and have 1 resistor (maybe try 3 and see if brightness changes significantly). Use @Makoto's calculation as example. In your case, say you put 3 in series, assume 9V total drop, that leaves 3V for the resistor. To have about 15 mA, have 3V/15mA=200 Ohm. I'd start with a larger resistor and decrease the value slowly rather than burning an LED. If you put just 1 LED, don't forget, (12V-3V)/15mA=9V/15mA=600 Ohm. So if you use 1 resistor per LED, you'll need around 600 Ohm. \$\endgroup\$
    – nurchi
    Nov 17, 2017 at 21:38

I like your circuit. Keep it simple. If one branch goes out you only loose one LED. Make sure your supply can handle the current. Assuming each branch is designed to 20mA per led. With 20 LEDs in parallel you're up to almost 1/2 amp. Good luck with the display!

  • \$\begingroup\$ Thanks. Glad you brought amps up. That is next on my list to work out. I've looked into trying to calculate that but it either uses equations that are beyond me, or asks for information I don't have/don't know where to find. \$\endgroup\$ Nov 17, 2017 at 16:26
  • \$\begingroup\$ If I have worked this out correctly. It will be using around 1.54 Amps. I assume Amps is cumulative, and not some fancy equation... \$\endgroup\$ Nov 17, 2017 at 16:52
  • 1
    \$\begingroup\$ Yes, assuming you continue adding more braches to your circuit exactly as you've shown then those currents will simply add. \$\endgroup\$
    – wayne2056
    Nov 17, 2017 at 17:04
  • \$\begingroup\$ You mention you will be using up to 150 LEDs. At an average of 20ma, you will need at least a 3 amp supply. I recommend 12V 5A DC power supply. \$\endgroup\$
    – Guill
    Nov 23, 2017 at 22:33

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