LED lighting power consumption - Page 1
Forum section: electronics / beginners

author: symbianas
author_rank: Regular Contributor
timestamp: September 05, 2021, 11:54:29 am
content: Hi, I have bought way too bright car trunk LED lighting module from "ebay". I have tested it initial consumption to be 150mA. I have decided to remove some LED's, to make it less bright and was expecting lower power consumption. But power consumption stayed the same 150mA! Why is that?LED module back (I have marked connection paths):LED module front (with 4 rows of LED's removed):Should I reduce brightness and power consumption some how differently?

author: Manul
author_rank: Super Contributor
timestamp: September 05, 2021, 12:15:29 pm
content: Hold your horses! There is a constant current driver on the board, so if you remove LEDs, current is still same. The danger is that now this current is shared by smaller amount of LEDs, which means that each of them are getting more (around double more). That could damage LEDs, so be careful. You can adjust current by changing some resistor(s). I suggest to draw schematic diagram first.Edit: R3 and R2 in series must be the current sense resistor. So replace with higher value resistor proportionally how much less current you want. For example if R3+R2 is 5 Ohms and you want to reduce current by half, solder 10 Ohms.

author: mariush
author_rank: Super Contributor
timestamp: September 05, 2021, 12:42:47 pm
content: Like Manul says, you have a simple led driver which limits the current going into the LEDs. It seems like the leds are arranged as 8 strings of 3 leds, all 8 strings connected in parallel.So, if the led driver limits current to 80mA, then each string of 3 leds gets 80mA / 8 = 10mA.By removing half the leds, each string gets now double the amount of current, so they leds will get hotter, and may overheat and eventually get damaged/burn out.The amount of current is controlled through those resistors, my guess is R2 and R3. Increasing the value of one of the resistors should lower the brightness, as the current should be lower. For example, lift one end of the R2 resistor (seems to be a 2.4 ohm resistor) and solder a 1 ohm resistor in series with that resistor, to increase the resistance value, or replace it with a 3.3 ohm resistor if you have one handy.

author: symbianas
author_rank: Regular Contributor
timestamp: September 05, 2021, 01:33:15 pm
content: Quote from: Manul on September 05, 2021, 12:15:29 pmThanks! R2 and R3 seems to be connected in series 1R6 and 2R40 which makes 1.6+2.4=4ohm resistance. So I need 8ohm then. I found only 3R9 3.9ohm in the shop, I hope 0.1ohm difference is not that important. Will buy it and include it in series.

author: Manul
author_rank: Super Contributor
timestamp: September 05, 2021, 02:22:37 pm
content: Quote from: symbianas on September 05, 2021, 01:33:15 pmIt should be at least double because you removed half of LEDs, but it can be more. You need to look how bright it is and how bright you want it. It might still be too bright. So two 4R7 or two 6R8 might be optimal, try and decide yourself.

author: Zero999
author_rank: Super Contributor
timestamp: September 05, 2021, 02:47:46 pm
content: Quote from: symbianas on September 05, 2021, 01:33:15 pmReplacing 1R6, 5R6 would be better, because you only need to change one resistor.I recommend soldering back the LEDs you've removed, because running the LEDs at a lower current will increase the reliability.