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BQ25570: BQ25570 Efficiency

kaleb chen
Prodigy 65 points
Part Number: BQ25570
Other Parts Discussed in Thread: TIDA-00242

Hi there,

I took a look at the TIDA-00242(by using BQ25570) test report link: https://www.ti.com/lit/ug/tidu382/tidu382.pdf?ts=1655081493736.

On the First page, the power level is 1.8mW and the capacitance of the supercap is 47mF. Based on the graph on the right, the charging time to charge this 47mF supercap from 0Vto 4.2V took around 4041s. By using E=1/2*c*(V2^2-V1^2), the average output power calculated based on this graph is around 0.103mW. by using n=Poutput/Pinput, 0.1025mW/1.8mW*100%= 5.5%? 

May I know if this calculation is correct? so that the bq25570 charger efficiency is only 5.5%? or have I made a mistake?  if so, how to improve the efficiency of BQ25570 in charging a supcap?

Massive Thanks.

  • 0
    TI__Guru*** 143880 points

    HI Kaleb,

    When charging the super cap from 0 to 1.5V, the charger is in cold start mode.  Cold start is very inefficient.  From 1.5V to 4.2, the boost charger is much more efficient.   Using the BQ25570 to charge a super cap, it is faster and more efficient to connect the super cap to the output of the buck regulator.  The time in cold start is reduced and the buck regulator prevents the boost converter from dropping back into cold start.

    Regards,

    Jeff

  • 0 in reply to Jeff F
    Prodigy 65 points

    if in this case(connecting the supercap to the output of the buck), shall I connect my load to Vstore or to the output of the buck as well(in parallel to the supercap)?     And Another question is, When charging the super cap from 0 to 1.5V, the charger is in cold start mode. is it 1.5 or 1.8? Thanks 

  • 0 in reply to kaleb chen
    TI__Guru*** 143880 points

    Hi Kaleb,

    Yes. The load would have to be at the output of the buck as well.

    The spec to exit cold start is:

    Not sure why Gordon reported 1.5V.  

    Regards,

    Jeff

  • 0 in reply to Jeff F
    Prodigy 65 points

    Hi Jeff,

    In this case, (I only use supercap and no battery) and connect supercap and the load together at Vout(output of the buck), will the Vbat still have any functionality? Can I leave it as a open circuit when I design the PCB? Thanks

  • 0 in reply to Jeff F
    Prodigy 65 points

    Hi Jeff,

    If I connect the supercap and load both at Vbuck, is that mean the voltage on supercap can only be charged up to 1.8V? (if the load is 1.8V).  and then The supercap will directly provide power to the load? the voltage on the supercap needs to maintain at 1.8V? Thanks

  • 0 in reply to kaleb chen
    TI__Guru*** 143880 points

    Hi Kaleb,

    You can change the buck output voltage by changing the VOUT_SET pin resistors.

    Regards,

    Jeff

  • 0 in reply to Jeff F
    Prodigy 65 points

    Thanks, Jeff, But how can I provide the output to the load if the load need a constent stable 1.8V? when both supercap and the load connect parallel to the buck output?

  • 0 in reply to Jeff F
    Prodigy 65 points

    Hi Jeff,

    Would love to make my question clear, so please just reply to this one. If, in order to charge the supercap much faster, you recommend me to connect the supercap to the output of the buck. But I also have a 1.8 load which requires a stable 1.8 v. How and where shall I connect the load to? I can't connect the supercap directly to the load right? since the supercap has an almost linear drop of voltage with current draw. If I connect the load to Vstore, the Vstore doesn't have a step-down circuit to step the voltage down to 1.8V? Thanks for your reply

  • 0 in reply to kaleb chen
    TI__Guru*** 143880 points

    Hi Kaleb,

    You fastest charging you can connect the supercap to VOUT but you are correct in that the load must also be connected to VOUT.  If you connect the supercap to VBAT, then the supercap will take a long time to charge up to at least 2.1V to exit cold start and give enough headroom to the buck converter to provide 1.8V.

    Regards,

    Jeff