Understanding leds in series

I need to convert a strip of 16 leds from 31 V to 12 V. All 16 leds are not in one series. They are divided into two series. Each series has 8 leds and one resistor. Each series gets 31 V. When lit, the voltage drop at one led seems to be 1.75 V. That would be 14 V over all 8 leds. The remaining 17 V drops over the resistor.

But what is actually happening here? When I connect the power, is there a moment when 31 is divided to all 8 leds? That would be some 3.9 V. And that will light up the leds. The current through a led is some 27 mA. And 27 mA to drop 17 V needs a 630 Ohm resistor, which is what I measured. So the leds need the 3.9 V to light up, but when the current starts flowing, each led only "senses" the 1.75 V. Am I on right track here?

I tried to bypass the resistor and connected the 8 leds directly to a 12 V source (led battery). No light. It would have been only 1.5 V over each led. I tried with only 7 leds, which would be 1.71 V. Still no light. Without a resistor, I didn't dare to test only 6 leds, which would have given 2 V.

People are always pointing out the importance of a current restricting resistor, when using leds. But on some page I read that if the voltage source is equal to the forward voltage of the led, no resistor is needed. But could it be that the leds always need the resistor? With the 630 Ohm resistor, my 8 leds light up with 31 V, 3.9 V per led, then keep shining beautifully with 1.75 V.

If I'm right, I might have to divide my 16 leds into four groups of four leds plus one resistor, to be able to use my 12 V source. Four leds need 7 V. The resistor needs to drop 5 V, which an 190 ohm resistor would do with a current of 27 mA.

So there's no Arduino involved - yet. I just want to get this led thing straight.

Your measurements are somewhat fishy!

What colour are these? White LEDs run on something like 3 to 3.5 V.

27 mA is a reasonable value for a luminaire LED - as distinct from the common "indicator" LED rated for 20 mA.

So to run on 12 V, you cannot effectively use four in series, only three.

If you have a DMM it’s surely easy enough to measure the volt drop across the resistor and LEDs in the existing circuit rather than guessing what it might be. If you don’t have a DMM maybe now is the time to get one.

What type of LEDs are we talking about?

Steve

I do have a DMM. And I suspected something fishy, because a few leds showed the 1.75 V drop and one had 3 V drop. The leds are warm white, from a Xmas tree candle set. So maybe the 3 V is the right voltage and 1.75 V was only due to bad connection or whatever. I stick needles through the insulation to get the voltages.

So the forward voltage simply defines at what voltage the led lights up and can stay lit? No need for a resistor, if the Vf equals the power source voltage? Or the sum of Vf of all leds in the series equals the power source?

But on some page I read that if the voltage source is equal to the forward voltage of the led, no resistor is needed.

IF

Look at the graphs Paul__B posted, a tiny change in voltage causes a big change in current. You don't have to get the voltage wrong by very much for the LED not to light or go pop. A resistor limits the current to the desired level and the characteristics of the LED sort the voltage out.

Johan_Ha:
But on some page I read that if the voltage source is equal to the forward voltage of the led, no resistor is needed.

And you also believe what you read on Facebook? :roll_eyes:

I'll leave you with Perry's comments. :grinning:

I'm going with 4 leds in series, without a resistor. It lights up the leds and 18 mA flows through, according to my DMM. The power source is a led battery, a minimal risk of any voltage spikes. I didn't compare the brightness with anything else, but it looks quite good. Sure it could be brighter, but it's for a set of Christmas tree lights to be used in the middle of a forest.

Ready. 51 mA shared on four parallel loops with four candles in series in each loop. No resistor. Cheap Chinese 12 V led battery, 7.2 Ah. I guess it's a moped battery.

OK, well it is the case that most of the battery-operated Christmas light sets do not use resistors.

Part of the problem is that a LED is not linear - i.e. things are a curve so the calculations for me were close but not good enough - after some head scratching I got an assortment of resistors and put a meter in the circuit and changed resistor values until I got the current draw and the LED's were visible outside - so some testing is probably in order.

Paul__B:
OK, well it is the case that most of the battery-operated Christmas light sets do not use resistors.

Might they not be those leds with resistors built in?

fionning_macara:
Might they not be those leds with resistors built in?

Can you actually cite such a thing from the Web? I have not seen such. :astonished:

Usually they are listed as 12V LEDs.
For example, you can see the lump of a resistor on one leg of these LEDs.
Amazon.com: EDGELEC 50pcs 12 Volt 10mm Blue LED Lights Emitting Diodes Pre Wired 7.9 Inch DC 12v LED Light Diffused Colored Lens Small LED Lamps: Home Improvement?

I have also seen LEDs with the current resistor built into the package.

CrossRoads:
For example, you can see the lump of a resistor on one leg of these LEDs.

Link corrected: https://www.amazon.com/EDGELEC-LED-Emitting-Diffused-Colored/dp/B07WRKP3YY/

Indeed, but these are not the Christmas light LEDs. :grinning:

No, the battery-operated LEDs do not use resistors at all. They are powered by two “AA” cells delivering about 3 V - up to 3.2 V with fresh batteries. And if you choose to believe the graph I posted in #1 (should appear on this same page, so I did not repeat it) then you will note this neatly corresponds to about 20 mA per LED.

Similarly, four LEDs in series across a 12 V SLA battery gives the same result.

Here's my original set with a 31 V power source and resistors. The voltage drop over one led was 3.1 V. I think the current was some 20 mA. (In the OP I wrote that the voltage drop was only 1.75 V, but that was wrong.)

And here's my modification. Four sets of four leds, no resistor. The power source is 12 V lead battery.

Total current is 52 mA, that's 13 mA through one led. And the voltage drop over one led is obviously 3 V. Well, my DMM measures 12.64 V on the battery without any load. The brand new battery is obviously sold fully charged.

Anyway, 13 mA, and the leds shine quite nice. Is there any chance the current can increase due to warming or anything? Perhaps I have to be careful after recharging the battery.

And the voltage drop over one led is obviously 3 V.

Not at all obviously. With as an LED is not a liner device the voltage across each LED will not be the same each will be different, and it will change with age and temperature.

Using no resistors with an LED or even many in series is an act of quite utter folly and you should have your hands smacked for even thinking about it.

And how are you going to smack all the manufacturers of the Christmas lights? :grinning:

Well they have a vested interest in selling you a new set every year.