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TPS61178: Design Review

sonali zodage
Intellectual 375 points

Part Number: TPS61178

Hi,

We are using TPS61178RNWR boost converter in one of our design. I have attached an schematic for boost converter which we have design. Please review the schematic and provide your feedback.

  • 0
    TI__Guru 70785 points

    Hello sonali,

    There is public holiday for the persons responsible for this device. Could you please summarize for me what you changed compared to the typical application in the datasheet?

    Best regards,
    Brigitte

  • 0
    TI__Genius 17555 points

    Hi Sonali,

    From your schematic, I find some points might need to be checked.

    1. the input voltage and current range. Note that there is a remark that the current is 0.8A, I guess the BAT need average current limit to be 0.8A ,right?

    the current limit resistor should be checked to meet the lowest input voltage to ensure 0.8A current limit.

    2. the voltage reference is 1.198 typically, so after voltage drop the output shall be nearly 12.5V. Please confirm your demand.

    3. the compensation resistor and capacity is recommended to be lower value, like 15Kohm and 6800pF.

    Best Regards

    Fergus

  • 0 in reply to Fergus He
    Intellectual 375 points

    Hello Fergus,

    Sorry for the late reply.

    1. The 0.8A current is at lowest input voltage i.e. 10.8V. Can you please tell us what should be current limit resistor value for 0.8?

    Also, we have used 86.6k limit value to limit current upto 8A peak current. 

    2. The input voltage (VBAT) is 10.8V to 12.6V and we need boost output voltage upto 12.8V for our application. It is possible to do this?

    3. Okay, we will changed compensation to lower value.

    Thanks,

    Sonali 

  • 0 in reply to sonali zodage
    TI__Genius 17555 points

    Hi Sonali,

    1. it can be calculated that the ripple current is around 1A at 10.8V input, so your peak current is 0.8/2+1=1.4A,

    you can use 3A current limit that the resistor can be 250k.

    2. it is possible to do this. but the chip's current may be too large for the application.

    Thanks,

    Fergus

  • 0 in reply to Fergus He
    Intellectual 375 points

    Hello Fergus,

    Thank you for your feedback.

    1. Where do you find the formula (0.8/2+1=1.4A) which you have used to calculate peak current. 

    Thanks,

    Sonali

  • 0 in reply to sonali zodage
    TI__Genius 17555 points

    Hi Sonali,

    Sorry I wrote the wrong formula.

    The frequency is around 500kHz (the Rfreq=316k) and the duty is 1-Vin/Vo=0.163

    The current ripple Δi=Vin*d*Ts/L=1.07A.

    The right formula is 

    We get the peak current formula Ipk=0.8+1/2=1.3A.

    Best Regards

    Fergus