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LM2576T-5.0 with non-zero Feedback

Craig Renaud
Prodigy 40 points
Other Parts Discussed in Thread: LM2576

Hi All;


I'm working on a repair.  The board has an LM2576T-5.0 simple switcher but instead of having the Feedback pin (4) connected directly to the Output (2), the board designer has a 5K6 resistor in between the pins.


I'm trying to find out what he is doing.  I am assuming that he is using the feedback signal to provide somewhat more than the 5.0 volts output (the designer could have used the -ADJ version of the LM2576; I have no idea why it was decided not to) in order to trim the output voltage using some form of remote sensing, even though the 5K6 is soldered to the PCB.

My look thru datasheets, etc has not (yet) provided me with an answer, which is where I hope someone out in the Community can help.


What would the output be with 5K6 between the Feedback and Output pins?


I can't measure it, because the part is defective, with a near short (<100 ohms) between the in and out pins, and my customer is desperate to get the board going.  My plan is, once I find out what the output would be, to use a -12V version or -ADJ version to get the board back to life, as it is 2 days delivery for the replacement -5 version.

Thanks!

Craig P. Renaud

  • 0
    TI__Guru 72695 points

    I think you are correct; the designer was trying to get more than 5V out from the 5V device.  In the LM2576-5V the bottom FB resistor is 1k; the top is 3.1k.

    This gives 1.23V*((3.1k/1k)+1)= about 5V.  With the external 5.6k we have: 1.23*(((5.6k+3.1k)/1k)+1)= about 12V.  So it seems like he was trying for 12Vout. 

    It is not a good idea to increase the output voltage with an external resistor in this way.  The external resistor will not track the internal divider, causing an error in the output and drift.  Also, the internal resistors match each other well, however their absolute value tolerance is quite poor. 

    I would use either a 12V device or the ADJ, as you mentioned.

     

  • 0 in reply to Frank De Stasi
    Prodigy 40 points

    Hi Frank;

    Thanks for the reply.


    I have a 12V device that I think I will put in, but first I have to make sure there is no reason not to.  Obviously I will have to remove the 5K6 and put in a zero-ohm link.

    Designer has the output from this device going to a standard 7805 that is powering other logic, etc (another reason I couldn't figure out why he had a -5 device: why would he put 5V from the simple switcher into a 7805?  Makes no sense.)

    The output is also going to a charge pump converter to generate a negative rail.

    At any rate, I will do the substitution and let you know how it works out.

    Best Regards,

    Craig

  • 0 in reply to Frank De Stasi
    Prodigy 40 points
    Hi Frank;

    One thing that I have to check is the output inductor to make sure it will be OK to use it with the -12 rather than the -5.

    Craig
  • 0 in reply to Craig Renaud
    TI__Guru 72695 points

    Yes check the inductor

  • 0 in reply to Frank De Stasi
    Prodigy 40 points

    Hi Frank;


    In the end I had to go back to the original -5V part...time was against me and the new part (5V) arrived early.

    Looking at the data sheets I think I would have needed a different inductor as well.

    Thanks for your Help!

    Craig