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enter image description here

Using LinkSwitch AC-DC converter as shown.

Output voltage is 12V, 120mA.

Converting 12V to 5V using a regulator will dissipate a lot of power as heat.

Is there any way to adjust the component values to get 5V output?

Datasheet of Linkswitch

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  • \$\begingroup\$ what dos datasheet say? there is a ratio in the feedback. Does adjusting those 2 resistors usefully change Vout? \$\endgroup\$ – analogsystemsrf Jun 26 at 10:11
  • \$\begingroup\$ For any other readers: No, that is not a mistake in the schematic - the LNK304 really has no ground pin. \$\endgroup\$ – immibis Jun 26 at 22:08
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Right under the circuit diagram you used in your question it tells you: -

enter image description here

So with the values shown in your circuit, 1.65 volts is produced at FB with respect to pin S. For a 5 volt output you'll need to lower the 13 kohm resistor - you do the math. Also check this resource (Application Note AN-37 LinkSwitch-TN Family) to establish what inductance value changes are needed when operating at an output voltage of 5 volts.

Be also aware that this design is dangerous to the uninitiated - it doesn't provide any galvanic isolation from live AC voltages and can easily kill someone.

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    \$\begingroup\$ Good point about not providing any isolation. \$\endgroup\$ – Colin Jun 26 at 10:13
  • \$\begingroup\$ Can you simply adjust the resistors without recalculating the inductance etc? \$\endgroup\$ – Huisman Jun 26 at 10:14
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    \$\begingroup\$ @Huisman page 7 of the data sheet tells you how to select the inductor value (same page as the stuff above): The typical inductance value and RMS current rating can be obtained from the LinkSwitch-TN design spreadsheet available within the PI Expert design suite from Power Integrations. \$\endgroup\$ – Andy aka Jun 26 at 10:18
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    \$\begingroup\$ @Andyaka If a newbie reads this, he/she will think you can simply adjust the the resistor divider to get a different output voltage. Maybe you should give the full answer to "is there any way to adjust the component value to get 5V output?" \$\endgroup\$ – Huisman Jun 26 at 10:21
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    \$\begingroup\$ @Huisman there will be some loss of efficiency but unless you go measuring the consumed current accurately you may have a hard time detecting this. You can expect is to work over some range without any other changes and you could do some careful tests (as mentioned this is a dangerous type of power supply unless the device it is in is very-well/double insulated and there is no way to touch any exposed connections even on the low voltage side). \$\endgroup\$ – KalleMP Jun 26 at 10:22
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As the datasheet you linked says:

The values of R1 and R3 are selected such that, at the desired output voltage, the voltage at the FEEDBACK pin is 1.65 V.

R1, and R3 form a voltage divider, you need to work out values for them such that they generate 1.65 V at the feedback pin.

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  • \$\begingroup\$ If a newbie reads this, he/she will think you can simply adjust the the resistor divider to get a different output voltage. Maybe you should give the full answer to "is there any way to adjust the component value to get 5V output?" \$\endgroup\$ – Huisman Jun 26 at 10:21

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