LED driver identification

[Diver300]

This is on a PIR LED lamp.
I've had a couple of LEDs fail, and I've put other ones in their place, but I would like to cut down the current a bit.

There are two strings of LEDs, each of 14 LED assemblies, and each assembly is 5 or 6 LEDs in series. Each string runs from 230 V ac mains via a rectifier, smoothing capacitor and the IC in the picture. There is a smoothing capacitors and one of the ICs for each string.

I was hoping that there would be a current control resistor that I could increase to cut the current down, but I don't understand why there are two resistors for each IC.

I tried and failed to find the data sheet for the IC, and I didn't recognise the manufacturer's logo.

Has anyone got any idea on how to find the datasheet?

 

[Tony Stewart]

Current sensing by measuring the V across the 3R3 and adjust as required.

Your LED specs also affect results, which you can edit.
https://tinyurl.com/27qdz5f4 (Vf @ If)

 

[Diver300]

Current sensing by measuring the V across the 3R3 and adjust as required.

View attachment 146374
Your LED specs also affect results, which you can edit.
https://tinyurl.com/27qdz5f4 (Vf @ If)

Unfortunately, it's not quite that circuit.

The two resistors are connected together at one end, and that point is the -ve supply. Also the voltage that has to be dropped is quite large. It's somewhere around 60 - 80 V.

There must be the datasheet for the IC somewhere.

 

[Tony Stewart]

You can figure this out by reading the current sensing resistor.

 

[Diver300]

The LED current is 22 mA, and there is 220 mV across the 10 Ohm resistor, so that implies that I could increase the 10 Ohm resistor to reduce the current.

There is 71 V across the IC.

There is 55 mV dc across the 3.3 Ohm resistor and around 100 mV ac. That implies and average of 17 mA, but far from steady. I am confused by that. I have seen current limiting ICs that turn on different banks of LED at different parts of the mains cycle to improve the power factor, but this lamp is all run from quite large smoothing capacitors. The power factor is quite good at 0.82, so there may be some clever design feature that gets the power factor that good.

 

[Diver300]

I've changed the 10 Ohm resistors to 15 Ohm, and which has reduced the current in the LEDs to about 15 mA, Changing the resistor is a bit of a pain on an aluminum PCB.

The power is down from 14.6 W to 10 W, and the power factor is slightly worse at 0.78.

The lamp is a couple of years old, but it would have only run for a few minutes a day as is controlled by the built-in PIR sensor and it doesn't get many people come past. It's a bit of a disappointment that it needed fixing and I hope that it will last better now that the current is less.

I still don't know what the 3.3 Ohm resistors do.