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

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0

$begingroup$

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

power-electronics

share|improve this question

edited Jun 26 at 17:44

Shavan We

asked Jun 26 at 10:04

Shavan WeShavan We

722 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

0

$begingroup$

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

power-electronics

share|improve this question

edited Jun 26 at 17:44

Shavan We

asked Jun 26 at 10:04

Shavan WeShavan We

722 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

$begingroup$

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

power-electronics

share|improve this question

edited Jun 26 at 17:44

Shavan We

asked Jun 26 at 10:04

Shavan WeShavan We

722 bronze badges

$endgroup$

share|improve this question

edited Jun 26 at 17:44

Shavan We

asked Jun 26 at 10:04

Shavan WeShavan We

722 bronze badges

share|improve this question

edited Jun 26 at 17:44

Shavan We

asked Jun 26 at 10:04

Shavan WeShavan We

722 bronze badges

asked Jun 26 at 10:04

Shavan WeShavan We

722 bronze badges

asked Jun 26 at 10:04

Shavan WeShavan We

722 bronze badges

2 Answers
2

active

oldest

votes

11

$begingroup$

Right under the circuit diagram you used in your question it tells you: -

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.

share|improve this answer

edited Jun 26 at 10:50

answered Jun 26 at 10:10

Andy akaAndy aka

249k1111 gold badges193193 silver badges444444 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  

  
  
  
  

 | 
show 1 more comment

2

$begingroup$

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.

share|improve this answer

answered Jun 26 at 10:08

ColinColin

3,25222 gold badges1111 silver badges2525 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

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11

$begingroup$

Right under the circuit diagram you used in your question it tells you: -

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.

share|improve this answer

edited Jun 26 at 10:50

answered Jun 26 at 10:10

Andy akaAndy aka

249k1111 gold badges193193 silver badges444444 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  

  
  
  
  

 | 
show 1 more comment

11

$begingroup$

Right under the circuit diagram you used in your question it tells you: -

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.

share|improve this answer

edited Jun 26 at 10:50

answered Jun 26 at 10:10

Andy akaAndy aka

249k1111 gold badges193193 silver badges444444 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

Right under the circuit diagram you used in your question it tells you: -

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.

share|improve this answer

edited Jun 26 at 10:50

answered Jun 26 at 10:10

Andy akaAndy aka

249k1111 gold badges193193 silver badges444444 bronze badges

$endgroup$

share|improve this answer

edited Jun 26 at 10:50

answered Jun 26 at 10:10

Andy akaAndy aka

249k1111 gold badges193193 silver badges444444 bronze badges

answered Jun 26 at 10:10

Andy akaAndy aka

249k1111 gold badges193193 silver badges444444 bronze badges

answered Jun 26 at 10:10

Andy akaAndy aka

249k1111 gold badges193193 silver badges444444 bronze badges

2

$begingroup$

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.

share|improve this answer

answered Jun 26 at 10:08

ColinColin

3,25222 gold badges1111 silver badges2525 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

2

$begingroup$

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.

share|improve this answer

answered Jun 26 at 10:08

ColinColin

3,25222 gold badges1111 silver badges2525 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

$begingroup$

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.

share|improve this answer

answered Jun 26 at 10:08

ColinColin

3,25222 gold badges1111 silver badges2525 bronze badges

$endgroup$

share|improve this answer

answered Jun 26 at 10:08

ColinColin

3,25222 gold badges1111 silver badges2525 bronze badges

answered Jun 26 at 10:08

ColinColin

3,25222 gold badges1111 silver badges2525 bronze badges

answered Jun 26 at 10:08

ColinColin

3,25222 gold badges1111 silver badges2525 bronze badges