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TPS61088: TPS61088RHLR

Alan Varghese
Prodigy 150 points
Part Number: TPS61088


Hi ,

Good day,

We are using TPS61088RHLR for converting a battery output (3.6V) to 5V.

Attaching the Screenshot of the implemented circuit below

In this, the Vin is taken from Battery. During the normal operating condition, (ie battery voltage= 3.6V), the output is stable and constant. But if the battery become full charge ( ie around 4.1-4.2V) the output of this boost is Changing. Is there any possibilities of occuring situation like this,

May i know the reason for that.

Kindly help me to resolve this issue,

  • 0
    TI__Genius 16885 points

    Hi Alan,

    There are three reasons that can lead to output changing.

    1. In SCH ,I see the MODE is floating, it means the chip will goes to PFM in light load.

    please refer to datasheet Page13

    you can connect the MODE pin to GND to prevent voltage changing on the cost of reducing light load efficiency.

    2. the loop at 4.1V input voltage is unstable. You may find the output ripple in low frequency.

    Please check which one accords with your situation.

    3. the current limit is trigged. you can have a look at the inductor current ,to check if the current reaches the preset limit.

    Best Regards

    Fergus

  • 0 in reply to Fergus He
    Prodigy 150 points

    Hello Fergus He,
    Thank you for your reply.
    The issue seems to be because of the PFM pin being floating.

    Another doubt.
    Is it possible to get a 5V output from any boost converter when our input voltage is 4.9V?.
    Do we need any headroom between the input and output voltages?.
    Simillarly, is t possible to get a 4.9V output voltage from a buck converter for a 5V input?.
    What is the parameter to be considered here if the input and output voltages are within the specified voltage ranges?.


  • 0 in reply to Alan Varghese
    TI__Genius 16885 points

    Hi Alan,

    The parameter to be considered shall be the chip's min turn-on time limit .

    when considering the time to be 180ns and the frequency to be  500k, we can calculate a minimum duty,0.09.

    It means input voltage above (1-0.09)*5=4.55V is not able to regulate to 5V.

    You have to leave some margin for all conditions.

    Best Regards

    Fergus

  • 0 in reply to Fergus He
    Prodigy 150 points

    Hi Fergus,

    Thanks for the response. I understood the case. Is the same equation and criteria applicable for buck also?

  • 0 in reply to Alan Varghese
    TI__Genius 16885 points

    Hi Alan,

    For buck, the minimum turn on time corresponds to the minimum output voltage(or for fixed output voltage the maximum input voltage).

    The maximum turn on time corresponds to the maximum output voltage(or for fixed output voltage the minimum input voltage).

    Best Regards

    Fergus