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TPS61089: Output current calculation

Ashwini M S
Prodigy 90 points
Part Number: TPS61089

Dear Team,

I am using TPS61089RNR Boost regulator in my project. I am providing Vin= 5V and adjusted resistor divider to get Vout=12V.The inductor value used is 1.8uH. Switching frequency, fsw= 500KHz.

I am planning to give output voltage of this regulator to drive the LCD backlight, which requires maximum current of 1.5Amps.

Please let me know what is the maximum output current I can get from above design. Please suggest whether any design changes required to get required output current. Also please can you suggest how do I understand what is the maximum output current that I can design for the above part.

Regards,

Ashwini M S

  • 0
    TI__Mastermind 34201 points

    Hi Ashwini,

    Thank you for reaching out.

    TPS61089 peak switching current limit is set by ILIM pin resistance. With 100kohm Rlimit, the peak switching current limit MIN spec is 9A. Considering the inductor current ripple is 30%, average inductor current is 9*(1-30%/2)=7.6A. Assume efficiency is 93%, the maximum output current is 5V*7.6A*0.93/12=2.9A. This is a quick and rough calculation.

    For your application, 

    1. The calculated inductor current ripple is 3.3A so the inductor AC core loss would be high. Suggest use 3.3uH inductance.

    2. With 3.3uH inductor, calculated inductor peak current is around 4.8A. Leave 1A margin, set the peak switching current limit = 6A, ILIM pin resistor = 172kohm.

    3. Change C88 to 4.7nF to have higher phase margin.

    For details, please refer to TPS61089 datasheet.

  • 0 in reply to Zack Liu
    Prodigy 90 points

    Dear Zack,

    Thank you for the explanation.

    1." The calculated inductor current ripple is 3.3A so the inductor AC core loss would be high. Suggest use 3.3uH inductance." 

    Can you provide explanation for how you have calculated inductor current ripple here.

    2. "With 3.3uH inductor, calculated inductor peak current is around 4.8A. Leave 1A margin, set the peak switching current limit = 6A, ILIM pin resistor = 172kohm "

    With 3.3uH inductor calculated inductor peak current you have got is 4.8A. Please provide explanation / Calculation for this.

    Regards,
    Ashwini M S

  • 0 in reply to Ashwini M S
    TI__Mastermind 34201 points

    Hi Ashwini,

    You could find the inductor AC ripple current and Peak current calculation in TPS61089 datasheet page 15.