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LM5155: How to Design a Boost Converter Using LM5155

Essalhi Abderrahim
Prodigy 30 points
Part Number: LM5155

Hello TI Team,

I'm using the Driver LM5155 for designing a boost converter 12V/24V. For the design calculations, I'm using the application note '' How to Design a Boost Converter Using LM5155", but I have some questions regarding this application note:

 - The output current is 2A which mean the maximum input current should be arround 8.8Amps, while in the appilcation note is considered 2.985Amps. could you please justify ? (refer to app note page 3)

- Why the maximum ripple of the inductor is set to 60% and the equivalent duty cycle is 33%? is there any relationship ?(refer to app note page 3)

- From where it comes the "1.66" for the sense current resistor formula ? (refer to the page 4)

- In The driver datasheet, the soft start formula it seems not correct ? ( datasheet page 16)

Thank you in advance 

Abderrahim,

  • 0
    TI__Guru* 78190 points

    Hi Essalhi,

    Thank you for posting.  The following are answers to your questions, in the same order.

    - The App note is correct.  The 2.985A corresponds to 16.02V input voltage.

    - I will pass this to my colleague to explain.

    - I will pass this to my colleague to explain

    - The formula is correct.

    Please wait a couple of days for the answer to the two remaining questions.

    Thanks,

    Youhao Xi, Applications Engineering.

  • 0 in reply to Youhao Xi
    Prodigy 30 points

    Dear Youhao,

    Thank you for your asnwer

    lets focus on the first question, the input current as you said is calculated at the maximum input voltage which is 16V

    My question is, why for calculation the inductor you considered the maximum input voltage instead of the minimum ? because the maximum input current is obtained at the minimum input voltage

    Best regards,

    Abderrahim

  • 0 in reply to Essalhi Abderrahim
    TI__Mastermind 21720 points

    Abderrahim,

    Please see my responses below.

     - The output current is 2A which mean the maximum input current should be around 8.8Amps, while in the application note is considered 2.985Amps. could you please justify ? (refer to app note page 3)

    The maximum input current will occur at the minimum input voltage. This will be around 8.8A. The maximum ripple ratio of the inductor current to input current happens when the duty cycle is 33%. Calculation the VIN the results in a duty cycle of 33% is about 16.08V. When the input voltage is 16.08V the input current will be around 2.985A. This is the reason that this input current is used.

    - Why the maximum ripple of the inductor is set to 60% and the equivalent duty cycle is 33%? is there any relationship ?(refer to app note page 3).

    33% results in the maximum ripple ratio of the inductor current. 60% is just a good trade off between efficiency, loop response and size as explained in the application note. 60% ripple ratio can be selected on the per the designers preference.

    - From where it comes the "1.66" for the sense current resistor formula ? (refer to the page 4)

    This is related to the falling slope of the sensed inductor current and the internal slope compensation. The 1.66 is there to ensure the ratio between the two is enough to eliminate sub-harmonic oscillation at > 50% duty cycle.

    Let me know if you have any questions.

    Thanks,

    Garrett

  • 0 in reply to Garrett Roecker
    Prodigy 30 points

    Thank you very much

    One more question, Is there any spice model that i can use to simulate the steady state and the transient ?
    Here is my email: Aessalhi@lear.com

    Best regards,

    Abderrahim,