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LM25085 Minimum Iout

dflohr
Expert 1140 points
Other Parts Discussed in Thread: LM25085, LM25085A

Implemented  LM25085 in a new design.    Vin  27-34V    Vout  3.9V

Output current will occasionally spike to 1.5 amps   but usually operates continuous  .25- .5amp.

Is the continuous low output current OK?

Thanks

  • 0
    TI__Expert 7395 points
    Daniel,
    Thanks for your interest in TI.
    I'll get the appropriate engineer to comment.
    Regards
    John
  • 0
    TI__Mastermind 40040 points

    Hello,

    The LM25085 is typically configured as a Voltage regulator.  This is configured as shown in the datasheet, using Feedback resistors to set Vout.  If configured this way, the LM25085 inly regulates the voltage.  It has no control on the current.  The current is determined by your load.  If you are asking if a minimum Iout of 0.25A is OK, the answer is typically speaking, yes.  Running out Iouts even lower than this is perfectly fine.

    I hope this helps?

  • 0 in reply to David Baba
    Expert 1140 points

    David:

    Thank you.

    When we first ran Webench on LM25085, we set the current at 2 amps, thinking a good factor of safety.  When we then looked at the efficiency in very low current draws, it was extremely low.

    We then reset Webench to something more representative of the actual current draw. When we did, , a number of parts, including the inductor changed (Rt, Radj, Cff).      Efficiency in the new configuration GREATLY improved.

    Dan

  • 0 in reply to dflohr
    TI__Mastermind 40040 points
    Hi Dan,

    The efficiency profile will change significantly based on your target max load current. You will need to design for the Iout max. I suggest you look at various designs and not necessarily the "most efficient" design in webench. Because the most efficient design may give you the worse lighter load efficiencies.

    Hope this helps.

    David.
  • 0 in reply to David Baba
    Expert 1140 points

    Hi David:

    Our first boards came in and we are trying to qualify all parts of the design.

    The LM25085A switcher is running but interestingly, our 3.9 V  target  using 10K and 4.7K feedback resistors only produced 2.8V  (no output load in case that matters)

    Used Webench and the datasheet  RFB2/RFB1 = (VOUT/1.25V) - 1   to calculate and confirm  feedback R values.

    Interestingly, did another Webench with different parameters Mostly higher Vin  and lower Iout.  It gave me 10K and 3K to get 3.9V.

    Closest value I had around was 2.7K so installed that with the 10K  and Vout went to 4.2V.  

    So my question is:  Is something not correct,   does equation above not always set the feedback R values correctly, or something else?

    Thanks

    Dan

       

  • 0 in reply to dflohr
    TI__Mastermind 40040 points

    The 4.7k Resistor  is correct for 3.9V out.  It should not change with load.  Not sure why webench is giving you different values.  Can you use the calculated  values using the formula you described and measure the voltage at the feedback node.  Thanks.

  • 0 in reply to dflohr
    TI__Mastermind 40040 points

    The 4.7k Resistor  is correct for 3.9V out.  It should not change with load.  Not sure why webench is giving you different values.  Can you use the calculated  values using the formula you described and measure the  voltage at the feedback node.  Thanks.

  • 0 in reply to David Baba
    Expert 1140 points

    David,   when we used 4.7k, Vout was only 2.8V.  

    We changed  to the 3k resistor as webench suggested when we put in Iout < 1 amp.

    Got the 3.9V.  Something else going on.

    Dan

  • 0 in reply to dflohr
    TI__Mastermind 40040 points
    Please can you use the 4.7k bottom feedback resistor and measure the voltage at the feedback pin and ground. You should read 1.25v and it should not drop with iout. Can you please measure and confirm. Thanks.
  • 0 in reply to David Baba
    Expert 1140 points

    OK will change and check again.

    Right now with 10K &  3.2K as FB resistors    voltage at FB  is .91  and Vout = 3.8

  • 0 in reply to dflohr
    Expert 1140 points
    David while I am doing that, when you get time, would you please run webench on LM25085a? Vin 30-38V Vout 3.9V Iout 1A
  • 0 in reply to dflohr
    Expert 1140 points
    David:

    Just read the LM25085A data sheet. That part is what we are using and it does indicate FB V=.9 Made a call to the guy who purchased.

    Board layout for WSON package. Digikey no stock. Bought from Mouser. Not sure why purchased LM25085A and not LM25085.

    As its just in prototype now, we can use either. Any suggestion? Will never operate near the 10A rating. Not sure soft start speed much matters in the application.

    Thanks

    Dan
  • 0 in reply to dflohr
    TI__Mastermind 40040 points

    The Feedback  voltage for the LM25085 is 1.25V, the FB voltage for the LM25085A is indeed 0.9V.  The LM25085 uses a COT architecture and as such requires a ripple voltage at the FB node.  0.91V DC sounds about right when you consider you have ripple voltage superimposed on the DC feedback voltage.  There will be a slight error in vout comparing the loaded condition to very light load.  

    Regarding the part selection of LM25085/A, you can use either part, I would go with the IC that the webench design was completed with, as the component selected were based on either the A or non A version.

  • 0 in reply to dflohr
    TI__Intellectual 2320 points

    Hello Daniel,

    If the DC voltage at FB pin measures less than 1.25V , then the part is definitely not in regulation. When the part is in regulation, the FB voltage pin should measure at 1.25V (Vref).

    For the same values of resistors that you have mentioned here (10k, 3k) you should have measured 5.41 volts  (1.25V/3k x (10k +3k) ) at the output, if the part was in regulation.

    1. 10k and 3k will never give you a regulated output at 3.9V, as suggested by the Webench. Please verify that design again in Webench again when you get the chance. As suggested by David, 10k and 4.7k should be the appropriate FB resistor values (for 3.9V output).Also please let us know what is the FB voltage you measure once you change the RFB(bottom) back to 4.7k?

    2. Do you see your output voltage change (reduce) as you increase your load current in your present board? If so, I believe the part is then in an overload condition. Can you please let us know where have you set the current limit threshold value in your design? Since your application can see 1.5A load spikes, I would suggest you maintain your current limit threshold value above 2A (at least), in order to maintain the regulation over the entire load range.

    3. Can you please share your schematic/layout files as well when you get the chance. It would help us verify the other components in your circuit, notably the RADJ used, the inductor value (and its sat ratings) and the output capacitance as well.

    Regards,

    Sourav

  • 0 in reply to Sourav.Sen
    Expert 1140 points

    Sourav:

    Thanks for note.  David and I reviewed over the weekend.    Part  actually used is LM25085A .  That part indeed does have FB   .9 V

    The purchasing guy  bought that part because of availability in WSON packaging thinking is was direct substitute.  Is pin compatible but slightly different.

    Dan

  • 0 in reply to dflohr
    TI__Intellectual 2320 points
    Hello Daniel,

    Thank you for pointing that out. The pitfalls of not reading the entire exchange! I just read your initial post and the header topic and thought it was a LM25085 issue.
    I guess then with the 4.7k you should indeed have the 2.8V at the output, with the LM25085A. So I guess we do not have an issue here then. :)
    However, please ensure your RADJ is selected appropriately as well so that you can ensure full load regulation over the entire load range.

    Regards,
    Sourav
  • 0 in reply to Sourav.Sen
    Expert 1140 points

    David and Sourav:

    Thanks for all of your assistance.

    Hopefully more question on the input power supply.

    Our main power is from a 24VAC supply.  We have built the design with both half wave rectification as well as using a bridge. Filter cap now is 270uF.

    Max power 3.9V we are making with LM25085A looks now to be under 700mA.

    For a number of reasons, we would prefer to use half wave rectification.  The ripple on Vin would obviously be greater.

    Is this be an issue with LM25085A?

    Thanks

    Dan

  • 0 in reply to dflohr
    TI__Mastermind 40040 points

    As long as the trough formed by the ripple voltage is above Vout, you should be fine.  If the Vripple is so large that the instantaneous Vin in lower than Vout, the LM25085 will operate in Drop out mode.  the output will drop to the input voltage, less the voltage drops across the Top side MOSFET and DCR of the inductor.

  • 0 in reply to David Baba
    Expert 1140 points
    Great, thank you again.