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I have a AC/DC adapter P-1650-22 (used to power a notebook) laying around. I would like to use it for other purposes, i.e. as a constant voltage source for LEDs.

The problem is, that the output voltage is set to 19 V, though I would like it to operate at 11 V - 12 V. The power rating would not exceed the actual rating of 65 W. It would be even less (45 W).

Currently I have no schematic, but I opened up the device already to take some measurements.

Simultaneously I searched the web and have already seen similar products and also some schematics and block diagrams. Please excuse the bad quality of the schematics, but I couldn't find a better one (relevant information is visible though).

From the information I have gathered so far I assume this is a flyback converter.

Would you say that it is worth to tweak some components in order to reduce the output voltage of this device?

For sure I would need to:

  1. Adjust feedback resistive divider for the shunt voltage reference

    What else to think of?

  2. What about the need of adjusting the compensation loop (if there is one)?

  3. Circuitry around the LNK6777E (the IC connected to the second winding)

Concerning point 3, I am not really sure what the LNK6777E exactly does and where to tweak (please see the block diagram):

Block diagram for the LNK6777E

What is the purpose of the "FB" pin? I am aware, that it serves for the regulation of the second winding. So the LNK regulates its own supply voltage. But how is the feedback coming via the optocoupler incorporated via pin "CP"?

So my question is:

What should I consider which I have not already written about?

1st edit: @Bimpelrekkie, why dou you not recommend changing anything? There are safe ways of working on circuitry connected to mains, i.e. isolation transformer, etc. BTW, reading your answer again I think you missunderstood me. My question about the "FB" pin relates to LNK6777. I am well aware this is the feedback niput but I wanted to know, relating to the block schematic I have posted above, how exactly the regulation works, etc.

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    \$\begingroup\$ You could change the voltage feedback network, to lower the output voltage. But as you have to ask what the "FB" pin is for and this is partly a mains connected circuit I would strongly advise against changing anything. Instead, simply get a DCDC converter module like: ebay.com/itm/… these are cheap and easy to use and will just work. There can be unexpected issues if you do change the power brick's feedback network. \$\endgroup\$ – Bimpelrekkie Jan 16 at 9:35
  • \$\begingroup\$ Bimpelrekkie, I am aware of the feedback network built around the shunt voltage reference. I already adjusted the resistive divider. I am asking if there is anything else to consider. Especially around the loop compenstaion and the regulating IC (LNK6777). \$\endgroup\$ – stowoda Jan 16 at 9:51
  • \$\begingroup\$ You can see the output current as fixed, so 3.4 A at whatever voltage you set it at. Depending on how the aux winding is made, flyback or forward, and how much margin was designed into the thing from the start, you may cut out some of the lower end of the operating range. Is it marked 100-240 Vac? \$\endgroup\$ – winny Jan 16 at 9:54
  • \$\begingroup\$ Winny, yes it is marked 100 - 240 Vac. But I do not understand what you are stating. Hmm are you saying that we do not really know if it is a flyback or forward topology? \$\endgroup\$ – stowoda Jan 16 at 10:04
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    \$\begingroup\$ How sure are you it’s an straight implementation of the design note from Power Integrations? I have designed flyback converters with both flyback and forward aux windings. It’s as always a trade off between different things. To you, this may not matter. What may matter is if the lower portion of your input range gets affected by lowering the output voltage or not. Test it! Without knowing the transformer specification, it’s pretty much the only way to tell. \$\endgroup\$ – winny Jan 17 at 7:37

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