Hello everyone,

I have a constant current switching power supply (LED driver) that takes in 230V AC and provides a constant current output of 700mA. The output voltage is rated at 27V–43V (up to a maximum of 48V). I’d like to reduce the output current to around 400mA. Ideally, I’d love to make it adjustable, but even a fixed lower current would be fine.

Before I dive into any hardware modifications, I was hoping to get some advice from the community. Has anyone here successfully modified a similar driver to reduce its output current? If so, what approach did you take (e.g. swapping a feedback resistor, adjusting a potentiometer, etc.)? Are there any general rules or recommended practices for reducing the output current on these kinds of constant current supplies?

I’d appreciate any tips, warnings, or suggestions to help me do this safely and avoid damaging the driver. Thanks in advance for your expertise!

 

Hi
I have modified a couple of led drivers to reduce the output current , but your driver is a little different. As can be seen from the photographs, this driver has an improved PFC .
The output current depends on the magnitude of the current sensor . It seems that the resistors 1R30 and 1R50 , all in parallel forms the current sensor. Remove one or two of them and you will see how much the output current has decreased

 

Thanks for your thoughts.
You mean the resistors marked RC1 - RC5 on the bottom side of the PCB? Why are these resistors so small? I'd really like to understand how this thing works ...

 

Two resistors 1.5 Ohm and two resistors 1.3 Ohm in parallel gives value of 0.35 Ohm.
Output current is 0.7 Amp, so power dissipated in those resistors is 0.17W only. Therefore, low -power resistors can be used

 

using same rationale, removing two 1.5 Ohm should do the trick and get you to 0.375A
if you need 400mA, put 10 Ohm in place of one of removed resistors.
if you remove two 1.3 Ohm resistors and add one 3.3 Ohm resistor (or two 6.2 or 6.8 Ohm resistors) it would be the same result.

voltage across the shunt resistors is 0.24375V and I=V/R
so each 1.5 Ohm resistor carries 162.5mA
and each 1.3 Ohm resistor carries 187.5mA

 

OK, sounds great. I will do some actual measurements over the weekend. Then probably take off the resistors and solder in a couple of jumpers on the pads. Then test out various combinations. The 400mA target current was an estimation. I want to see how the LEDs perform.

 

The resistors you guys pointed out indeed did the trick. Taking out one 1.3 ohm resistor meant the total resistance of R's in parallel increased a bit and the current dropped. Meaning the power on the LED's now dropped, which is the goal I was after.
Solved! Thanks for your inputs.