I've opened several LED bulbs on the channel so far and I did the same now that another one has failed on me.
However, unlike the previous ones, this one was a bit different. Instead of using a dedicated constant current driver IC, this LED bulb used a capacitive dropper to reduce the current. This is not a preferred way of driving LEDs but it is for sure the cheapest one.
A capacitive dropper circuit works in a way that uses capacitors in series with the LEDs. On each of the mains sinewave half, a charge builds up on the capacitor and with that, it drops the voltage and the current that can pass through the circuit.
As with the previous LED bulbs, I attempted a repair where I wanted to reduce the current that goes through the LEDs so the resulting light module can then be used in a project. Based on my understanding of the individual LED chip properties, I did some math to calculate a value for a series resistor, and that kind of worked until the resistor started burning.
With a few trials and errors, I managed to reduce the output of the bulb by 4 Watts by removing one of the capacitors in the capacitive dropper.
You can check the entire process on the video below:
Tools and materials used in the video:
- 10W LED Bulb - https://s.click.aliexpress.com/e/_A7Xqi7
- JCD 8898 Soldering station - https://www.banggood.com/custlink/K3vE4gUOEJ
- Riden RD6012 Power Supply - https://www.banggood.com/custlink/3G3y0ezOyl
- Multipurpose screwdriver pen - https://s.click.aliexpress.com/e/_A6Ycd1