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LMR16030: LMR16030 - output inductor selection

Part Number: LMR16030
Other Parts Discussed in Thread: TPS54060, LMR16006, LMR14006,

Hi team,

I'm working on an LMR16030P-based step-down circuit with the following specifications:

Input voltage: 9-40V
Output voltage: 8V
Output current: max. 2A (typ.: 0.5-1.5A)

Webench recommends a 10µH inductor for this design. Changing the output current to 1A or 0.5A does not affect the inductance and only changes the recommended inductor's current rating.


The datasheet gives the following equation (p.20, Eq.10) to calculate the minimum output inductance:

L_min = [ (V_in_max - V_out) / (I_out * K_ind) ] * [ V_out / (V_in_max * f_sw) ]

Using the same input/output voltages as above, a switching frequency of 541kHz (as calculated by Webench) and K_ind = 0.4, this equation yields the following results:

For I_out = 2A: 14.79µH

For I_out = 1A: 29.57µH

For I_out = 0.5A: 59.15µH


The example design in the datasheet is apparently based on this equation with I_out = I_out_max. Therefore I would assume that it's save to set L_min = 14.8 uH and choose a 15uH inductor.
Either way, the calculated L_min does not match the Webench recommendation.

Why is that? Is it safe to choose components based on the Webench BOM or is the datasheet more accurate in this case?

Are the calculations based on I_out_max correct or is it necessary to choose an even higher inductance based on calculations for typical/minimal loads?


Thank you!
Best regards,
  • Hi Zhonghui,

    Webench may design it with small solution size option, you can try it with Balanced option.

    For wide input condition, more practical way is to design inductor with 30% current ripple at nominal Vin, then check the current ripple at Max Vin if the peak below current limit peak and output ripple meet requirement, normally 50%~60% current ripple is acceptable at Max Vin.

    In you case, 15uH is better choice if no size limitation

    B R

    Andy 

  • Hi Chen,

    Thanks for the reply.
    Size is not a concern in this case.

    I have a follow-up question regarding this: "Webench may design it with small solution size option, you can try it with Balanced option."


    The values in my first message were actually already based on the Balanced optimization and Webench does not choose an inductance larger than those values with any other option. Here's an example:

    For 9-40V input and 8V 2A output, it chooses the following inductors:

    Small footprint: 3.6uH, 4.9A, 22mOhm (f = 1.589MHz)

    Balanced: 10uH, 3.1A, 50mOhm (f = 541 kHz)

    High Efficiency: 10uH, 5.3A, 5.7mOhm (f = 554kHz)


    For the same input/ouput voltages but only 0.5A specified output current, it chooses:

    Small footprint: 2.5uH, 1.3A, 143mOhm (f = 2.129MHz)

    Balanced: 10uH, 0.88A, 276mOhm (f = 541kHz)

    High Efficiency: 8.2uH, 3.8A, 15mOhm (f = 696kHz)

    The latter makes no sense because it directly contradicts the equation given in the datasheet.
    When calculated with 55% current ripple, the formula results in the following values for Lmin:

    2A, balanced option (f = 541kHz): Lmin = 10.75uH [Webench = 10uH]

    2A, small footprint option (f = 1.589MHz): Lmin = 3.66uH [Webench = 3.6uH}

    For max Iout = 2A it makes sense if a higher acceptable current ripple is chosen, like you suggested.
    For 0.5A, however, there are still huge differences despite the higher ripple:

    0.5A, balanced option (f = 541kHz): Lmin = 43.02uH [Webench = 10uH]

    0.5A, small footprint option (f = 2.129Mhz): Lmin = 10.93uH [Webench = 2.5uH]


    According to the datasheet, Lmin needs to be significantly higher if the circuit is designed for small loads (Iout = 0.5A).
    In order to get results that are close to Webench's design, Kind would have to be specified as >200%.

    This can't be right?

    Best regards,

  • Hi Zhonghui,

    For light load 0.5A, small footprint option (f = 2.129Mhz): Lmin = 10.93uH [Webench = 2.5uH.

    For reliable design, you may have to consider the over load, output short, load transient condition where the switching current peak may trip the max current peak 4.75A,  that means you still have to use a big10uH rated 5A inductor for a 0.5A load,  that's why Webench recommend 2.5uH for 0.5A load if require small solution size. 

    B R

    Andy

  • Hi Chen,

    Thanks, but unfortunately this doesn't really answer my question.

    For the balanced and high effiency options (which are not focused on having the smallest possible size) Webench chooses a 10uH / 8.2uH inductor and switching frequencies of 541kHz / 696kHz for a 0.5A output current.

    According to the datasheet, the minimum(!) inductance for this current and frequencies - even with a current ripple of 60% at max Vin (as you suggested) would have to be 39uH / 31uH.

    That's a fundamental difference. According to the equation in the datasheet, the inductor size recommended by Webench would mean a current ripple of ~190% at nominal Vin and ~230% at max Vin.

    Why is that?

    Best regards,

  • Hi Zhonghui,

    You can review data sheet of some low current regulators like TPS54060, LMR14006 or LMR16006, you will find your calculation is correct and also meet Webench result.

    LMR16030 is a 3A converter and normally not recommended for 500mA application, for inductor 39uH and 500mA load, the ripple current is 500mA*60%=300mA which is just 10% for 3A load, which means the 300mA ripple current is almost the minimum value required by LMR16030, The peak current control mode require adequate ripple current information for stable loop control. 

    Of course you can use 39uH in LMR16030 for 500mA load, but need to fine tune the feedforward cap for loop stability.

    B R

    Andy