LED Bulb Conversion: Efficiently Connect 12V to 230V Electrical System for LED Chip Functionality

Increasingly, commonly used LED bulbs refuse to function - they simply stop shining. It turns out that the electrical system that converts 230V alternating current into = 12V fails. On the other hand, the lighting system itself, consisting of many LEDs, is efficient and could be successfully used. The problem is to know where the voltage = 12 V should be connected.

Gwardian wrote:

The problem is to know where the voltage = 12 V should be connected.

If the LEDs are connected in parallel, connect to any LED.

In the photo there is a very simple type of LED lamp and the power supply is based on a serial capacitor and it usually fails, however many of this type of lamps contain a complex system, often with an inverter, and 12V is not enough to illuminate it. There are series circuits consisting of a large number of LEDs, which need about 140V to shine. That's when things get more complicated. One more point is important about the subject of the photo; if 12V is enough to light up, you need to insert a series resistance not to exceed the limit current of the LED. In a fully functional circuit, this function is performed by this red capacitor connected in series with the resistor.

sq9fmc wrote:

A series resistance must be inserted in order not to exceed the limit current of the LED.

A current source can be used, this is a better solution.

As far as I can see correctly, we are writing about power supply from 12V.

The problem is that the network of connections is masked and it is difficult - even if not - for a layman to distinguish whether the diode connections are in series or in parallel. It would be best if an expert on the subject indicated specifically where to apply (power) such a system as in the photo.
From the front, the team looks like in the photo.

sq9fmc wrote:

There are series circuits consisting of a large number of LEDs, which need about 140V to shine.

With such a number of LEDs, it is not 12V for sure. Unfortunately, it is impossible to repair even such simple devices without a minimum of knowledge and a meter. How, for example, do you know that the power supply is damaged. The diodes are connected in series and if one fails, nothing shines. To find out, you need to measure the voltage on the electrolytic capacitor because it takes the voltage to power the diodes.

szon wrote:

With such a number of LEDs, it is not 12V for sure.

After all, you can see on the handle (first post) that this "bulb" is powered by the mains voltage.
The cap houses a typical transformerless power supply system.

Gwardian wrote:

. The problem is to know where the voltage = 12 V should be connected.

Nowhere ... Bring the power supply system to working order (simple as the construction of a flail) and measure the voltage that goes to the diodes.
There are 80 diodes. Counting 3V per diode (The voltages may be different - depending on the type / design of the diode) we have 240V - it can be assumed that they are all connected in series.
And one more thing - the power supply used in this "light bulb" does not require a current limiting resistor (this is to some extent stabilized at the appropriate level by the circuit of the power supply itself), in case you want to connect the power supply - you MUST make sure that the current flowing from the power supply does not exceeded (depending on the diode) 20 mA per structure.

I sacrificed a similar "light bulb" (12 diodes) - it is working - 230V - there are three resistors in the system (as in photo No. 1) - two small ones under the red capacitor and one (larger) R4.
On the other hand, in the circuit of the lamp inoperative (photo No. 2) there are also four - with only one defective resistor - R1.
Hence the conclusion that it causes the lack of light. The problem is to determine its value to solder the right one and the whole thing can be used in some new housing.
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I can send such an arrangement to the interested person if he or she wishes to experiment.
Address on priva.

You need to learn to read the wiring diagram. the diagram of this power supply is in the topic https://www.elektroda.pl/rtvforum/topic2707264.html. You write that there are four resistors in your bulb, so the one that is blown is probably in the place of the fuse. If this resistor burned out, it was either incorrectly selected, which happens, I had such a case in my light bulbs, or the rectifier or the capacitor was damaged. If it is the resistor I am writing about, use it with a value of a few ohms.