LED resistor values.

[tangodownNZ]

Hi, I am just wondering if you can please help me.

I have four LEDs connected to a 9 volt battery. They have little circuit boards in them. They will run between 9 volts and 15 volts.
I have them wired up in parallel to a 9 volt battery. They work fine. They are drawing 30mA at 9 volts. So I think that is 0.27 watts. I can’t get the resistor calculators to work because they ask for supply voltage and forward voltage and the forward voltage must be lower than the supply voltage. But I am supplying 9 volts and the led is using 9 volts, but when I type in Vs = 9 and Vf = 9 I get a null answer,

My question is I want to make them dimmer using a resistor but I don't know what resistor to put on them. I was thinking just one resistor soldered onto the positive wire just before the battery.
This is where it gets confusing because I read somewhere that if I decrease the current I will decrease the voltage, but I think these need a minimum of 9 volts to run. Can somebody please tell me if I can use a resistor to do this, and what size resistor I need please.
I was looking at .25 watt resistors but I don't know how many ohms, I would just be guessing. They are currently drawing 30mA so I would like to get that down to around 15 - 20mA

Here they are Little Dot SMD LED Accent Light - 30 Lumens | Super Bright LEDs - Little Dot SMD LED Accent Light - 30 Lumens | Super Bright LEDs - https://www.superbrightleds.com/moreinfo/led-wired-bolts/little-dot-smd-led-accent-light/639/1928/

 

[Lucien Nunes]

The first question is whether there is an IC / active control circuit on the board to stabilise the current / brightness against changes in voltage? If there is, you might not be able to dim them with a resistor because the electronics will fight back.

Assuming not, the current is set by internal resistors on the board. A nearly fixed (forward) voltage is dropped across the LED chip(s) themselves and the remainder, making up the 9V or more that is specified, is dropped across the resistor(s). What you want to do is add resistance so that the total of the internal and your external resistance is suitable to set the current that you require at the voltage you are using.

You can't use a simple online calculator because that doesn't know about the internal resistor in the module, it assumes a bare LED chip. In fact you can't calculate it either unless you know how many chips of what colour are in series, because otherwise you don't know what fraction of the 9V is across the resistor(s) and what fraction across the LED(s).

You could try to trace it out with a voltmeter across the pins of the LED package, or post pics of the tracks and resistors of the module here. Or we could make an assumption about how it is wired and suck it and see. The example in your link shows what seems to be 3 green chips each of which will have about Vf=2V, for a total of 6V leaving 9-6=3V dropped in the resistor. If it takes 30mA the resistor must be 3/0.03=100 ohms. You want to drop 3V at 20mA so the total resistance must be 3/0.02=150 ohms, therefore you need to add an external resistor of 150-100=50 ohms. So I would start with 47 ohms and see what visible difference that makes. You might find very little difference and need to increase that to 100, even 220 ohms.

If you want one resistor to drop the feed to 4 modules, then it needs 1/4 of the resistance. This will only work because the modules have internal resistors. You can't connect bare LED chips in parallel to share a resistor as the current can't be relied upon to divide equally between the chips due to temperature differences and production tolerances. But the internal resistors in your modules will ensure the current shares correctly.