Help for an idiot with regard to power in LED circuit

Thread Starter

Father Cool

Joined Dec 5, 2018
12
So I have a project that I started 10 years ago involving a circuit of LEDs sealed inside a model. My electronics knowledge was poor back then and when I wired them up I connected the 18 white LEDs in 6 series banks of 3 LEDs which was to run on 12V. These banks were then connected in parallel. Horrible I know

These banks were then protected with a bank of 3 resistors soldered in in parallel to each other to give the right resistance and that soldered in series with the banks of LEDs. These are not high power resistors.

The problem is now that I know more about electronics (I've dug out the project to finish it) I am very very concerned that the resistors will not be able to take the power required for the circuit. I estimate the circuit will draw probably about 1.8W maybe?

My issue is I have 2 wires going into the model to power the circuit and was wondering if anyone could suggest a method of keeping the power down in the circuit be external means to stop the resistors inside melting themselves.

I cannot get inside the model to change the circuit and the circuit will be controlled by an Arduino board via Mosfets or similar.

Help!!
 

Thread Starter

Father Cool

Joined Dec 5, 2018
12
This is the circuit give or take. Excuse the fag packet sketch on the phone. I have a further resistor on each set of 3 LEDs I just descovered. I cant remember the values of any of them though. They all look like 0.25w resistors though. All I know is it was designed to run 12v.

The LEDs are white and I'm sure they run on 3.6v with a max 30mA but I'm happy to run them at 10 to 15mA.

20181205_134117.jpg
 

AnalogKid

Joined Aug 1, 2013
8,927
I don't think there will be any resistor melting. Here is how to design the circuit.

3.6 x 3 = 10.8 V across three LEDs in series
12 - 10.8 = 1.2 V across the resistors
1.2 V / 15 mA = 80 ohms.
80 x 3 = 240 ohms
Three 240 ohm resistors in parallel will limit the LED string current to 15 mA.
(.015 ^ 2) x 80 = .018 W

Total power dissipation in the resistor group is 18 mW, or 6 mW per resistor. 0.1 W resistors will be barely above room temperature. If your resistors are running hot to the touch, something is wrong somewhere.

ak
 

Thread Starter

Father Cool

Joined Dec 5, 2018
12
I dont know if they are hot. I cant touch them. Its just that I dont want them to blow because I will have no way to repair them if they do.

Thanks for the calcs. That makes me feel much better. Not sure if they are 240ohm resistors though!

Also, your calc didn't allow for 6 banks of 3 LEDs. Still the power dissipation in the resistors would still only be 6x 18mW which is 108mW. Or 36mW per resistor.
 
Last edited by a moderator:

radiohead

Joined May 28, 2009
508
You can parallel series strings of LEDs as long as each series string has its own resistor. AnalogKid's calculations are correct.
 

bertus

Joined Apr 5, 2008
21,137
Hello,

It is not wise to parallel leds directly.
The leds will not have exactly the same forward voltages.
The string with the lowest forward voltage will draw most of the current and will fail as first.
Then the other strings will get more current and again the string with the lowest forward voltage will draw most of the current and then that string will fail.
etc. until all strings are dead.

Bertus
 

bertus

Joined Apr 5, 2008
21,137
Hello,

The only way I see to protect the leds from failing, is to limit the TOTAL current to about 30 mA.

Bertus
 

Thread Starter

Father Cool

Joined Dec 5, 2018
12
Hmm ok. Well on the really poor photo I have I do have a resistor on each of the banks as well as the main bunch. I assume I did the resistor for each and then decided that it wasnt enough so added the 3 one the main feed. I think I'll stick my multimeter across the circuit and try to gauge whats going on.
 

iONic

Joined Nov 16, 2007
1,650
Essentially what you have is a "Black Box," well, maybe gray. You have a light is series with an unknown resistor.

Lets assume the following:
1) you used the "max" If current of 30mA
This is generally a rating for PWM regulation or for very short durations and not continuous current

2)you used the "max" Vf(voltage drop across the LED)

Given these conditions you would have come up with a series resistance of ~47 Ohms

Lets now assume we need the following conditions:
If = 20mA
Vf=3.2V

Given these conditions you would have a series resistance of 120 Ohms
Adding an additional series resistance of ~70 Ohms would make for a safer system

Lets now assume a very SAFE conditions:
If = 15mA
Vf = 3.0V

The light from these LED will really not be obviously noticeable to us under most conditions.

The resultant resistance would now be ~220 Ohms
adding an additional series resistor of ~170 Ohms would get you to a safe zone

----

Another option would be to place a variable resistor(pot) between the source and the circuit (200 or higher, but not too high that the resolution would not be fine enough to calibrate.) Adjust to ~200 Ohms and lower if the light is still acceptable.Measure the pot resistance and replace it with the appropriate resistor.

I usually take the "typical" If(forward Current) and calculate a series resistor for a few mA lower to account for individual variations of LED I/V.

Also, noticed you said you added a resistor in each branch of 3
LEDs, you must be close. You could just use the variable pot method and find an acceptable light level.

P.S. Do you have any of those LED's left??
 
Last edited:

ebeowulf17

Joined Aug 12, 2014
3,275
Here's an example of how a series-parallel LED circuit should be structured:
upload_2018-12-5_11-38-35.gif
The key point is that there's a separate resistor is series with each series string of LEDs. You've given conflicting accounts within this thread, but your most recent comments seem to indicate that you do in fact have a resistor in series with each series group of 3 LEDs. As long as that's true, the rest is easy. The fact that you have a resistor (or three paralleled resistors) in series with the power source is totally fine and won't hurt anything.

So, if you expect each string to draw 30mA, and you've got 6 strings in parallel, you'd expect to draw 180mA for the whole system. Since you don't actually know what resistors you've got in there, I'd suggest you set up a quick test measuring current through the whole lighting system (if you're not sure how to put your meter in series for a current measurement, look it up or ask here - don't fry your meter making the usual mistakes!) Once you know how much current your lights are drawing, you'll know how much (if any) adjustment you need.

I would plan to do a very short test at first - even if your original circuit was over driving the LEDs, they'll be fine for a few seconds at least. Assuming the system is more or less functional, adjust the values of the shared resistors you have access to and retest until you get the total current down to maybe 90mA or so (15mA per string.)
 

Thread Starter

Father Cool

Joined Dec 5, 2018
12
Ok I'll sort out a test rig.

I found an old picture of the inside before it was sealed. It looks like I have a resistor for each bank as you have drawn plus also the 3 resistors in parallel inserted in series with the circuit as I drew in the beginning.

I just dont know what values the resistors are in there. I do know it ran 10 years ago on 12 v and worked. I am just nervous that if I power the lights and leave them on a while something will die and ruin the model.

Thanks for the input I will report back.
 

AnalogKid

Joined Aug 1, 2013
8,927
If you have a DMM with an Amps or milliamps mode, use it to measure the total current between the external DC source and the entire LED group. This can be used to estimate what is going on inside the device.

ak
 

Thread Starter

Father Cool

Joined Dec 5, 2018
12
Guys I know what the circuit should be. I'm not a novice any more. I was back then.

Thanks for the diagrams but thats not what the circuit is and I cant change it now because its sealed in a model. I just need to power whats there without the fear of killing the lights.
 
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