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TPS61087: Design operational range check - schematic review

Part Number: TPS61087

Hello I was wondering if someone could do a sanity check on my design. The Circuit is supposed to pass 5VDC normally to a 1A load, with the TPS61087 operating in discontinuous mode with 4.5Vin and the charger charging (1) 1S 3.2V LiFePO4 cell unless the 5VDC power is lost in which case the booster should provide the 5Vout with 3.2Vin for 30 seconds.  The inductor is 3.3uH  Wurth 3.4 A, fout is 1.2Mhz, Iswpk = 2.12A, Ir = 215-340mA. 

Your input is much appreciated, thanks. - Andrew


  • Hi Andrew,

    1)There is 0.5V voltage drop on D2, so when the 5V is lost the real input voltage for TPS61087 is 2.7V, you need to make sure the voltage will not drop below 2.5V in 30s because the minimum input voltage of TPS16087 is 2.5V;

    2)The output capacitance seems too large;

    3)You can design R4 and C12 according to equation 9 in datasheet, the Cout should be effective capacitance considering DC bias;

    4)With 1A load, TPS16087 works in CCM rather than DCM.

  • Hey thanks for your feedback Yichen, yes I am aware of the voltage drop of the diodes and I have considered changing the Rcomp Value to something more like 56K. Can you be more specific about "too much capacitance" ? Also my concern is that with the input voltage being 4.5V with the output voltage set to 5V that the switcher will not like that small overhead margin, I was assuming or wondering if it would be in DCM or even work at all ?

  • Hi Andrew,

    According to datasheet, 40uF should be sufficient and there is 220uF in your schematic. The device works in CCM for your condition. And it doesn't require much capacitance.

  • Hey thanks for your reply Yichen, BTW I just discovered the new Webench toolset after a long hiatus and I must say what an amazing set of tools those are. I thought Filter-Pro was great but those tools really deliver exactly what I need as a design engineer, thanks TI !!  I was able to run the booster design and find all of the potential pitfalls and I have determined that  1)  a larger inductor is needed and 2)  that there are better TI parts suited for my application.  Thanks for all of your Help  Slight smile