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TPS61288: What happens if I calculate Rc, Cc and Cp incorrectly?

Štepán Šípal
Prodigy 10 points
Part Number: TPS61288

Tool/software:

Hi,

I am designing a BOOST application circuit around aTPS61288, and I have problems determine the compensation loop components (Rc, Cc, Cp). Can something go wrong if I potentialy calcualte them incorrectly?

-----------------------------

My specific design paramters are:

Vin: 2.7 - 4.2 V(nominal: 3.6);

Vout: 5 V;

Iout: up to 8 A (nominaly much less);

Inductor: 4.7 uH;

-----------------------------

I have been following the datashet and tried to substitue the values to get Rc, Cc, Cp, but I have a problem determining Co, since the 5 V output bus is charging a 10mF electrolytic capacitor. Shall I take that in to account or just use Co as the output capacitor of the device? And is there something, that could go wrong, if I calculate the loop compensation incorrectly?

* I also looked into the excel calculation sheet that is available on the TI / TPS61288 website, but there the fields for Cc, Rc, Cp, which I had been looking for, were just marked as "user input", so I unfortunately did not take much from that.

Thank you very much for your time, and have a nice day.

yours faithfully,

Štěpán Šípal

  • 0
    TI__Mastermind 32350 points

    Hi 

    Since our expert of this device is out of office this week, please expect a late reply here. 

    Regards

    Tao

  • +1
    TI__Genius 16285 points

    Hi Štěpán,

    Sorry for reply late.

    When there is a 10mF capacitor at output side, the capacitor must be taken into account when calculate the loop.

    That is, use Co as the output capacitor of the device.

    Actually, the Rc,Cc and Cp will be somewhat arbitrary as there will be a quite low cross frequency.

    You can just use EVM parameter to have a try.

    Please take care below two points:

    1. the device is not true load disconnected, and there is a body diode of the upper mos.

    This means the device cannot control the inrush current when Vin is higher than Vout, like first connection.

    2. the device will limit its current at when starting.

    the output capacitor will get a relative long rise time as the charging current will be limited.

    Best Regards.

    Fergus

  • 0 in reply to Fergus He
    Prodigy 10 points

    Hi Fergus,

    thank you very much for your time and help. I have just used the typical application values from the datasheet, and everything seems to work just fine.

    Best Regards,

    Štěpán

  • 0 in reply to Štepán Šípal
    TI__Genius 16285 points

    Hi Štěpán，

    Sounds good and you are welcome.

    Best Regards.

    Fergus