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Buck-Boost Circuit - via WebBench

Brian Cavanagh
Intellectual 285 points
Other Parts Discussed in Thread: LM25118

Hello, this is the first time I've used WebBench for circuit design recommendations. 

I set my criteria as follows:

Vin: 16 - 30 VDC

Vout: 24VDC regulated

Iout: no more than 5Amps.

With the input above, the recommended part was the LM25118.  It's rated to source 20Amps, as a result, the components are very, very large. 

Can someone guide me to a more "size appropriate" design? 

Thanks,  Brian

  • 0
    TI__Mastermind 25570 points
    The current that the PSU can supply is set by external parts rather than by the LM25118. You need to scale the current sensing resistor so that the current limit comes in at the level you want. The external components, inductor, MOSFETs, diodes etc are then sized to handle the 5A output current you need.
    Regards
    Colin
  • 0 in reply to Colin Gillmor
    Intellectual 285 points
    Hi Colin, thank you for the guidance. I downloaded the datasheet for the LM25118 (SNVS726D...revised Aug 2014) and I'm following Section 9 Application and Implementation (page 23). On page 26 (14) and (15) there's Iripple(buck) and (buck-boost) values, 2.86 Amps and 1.17 Amps, respectively. I'm not clear how those values were determined. Finally, in the next paragraph, there's a comment "Because the inductor selected is lower than ... for CCM in buck mode is 1.42 Amps at maximum VIN". I'm not sure where the 1.42 amps came from. I calculated my buck mode inductor L1 = 80uH and buck-boost mode inductor L1 = 160uH. I chose the 160uH to favor the buck-boost only because the example did. Can you advise on this?

    Here's my parameters : Vin(min) = 16VDC, Vin(max) = 30VDC, Vout = 24VDC, Iout = 5 Amps maximum. Thanks, Brian