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LM5163-LM5164DESIGN-CALC: Efficiency changes with Full load current Iout

Junjie Mao
Prodigy 50 points
Part Number: LM5163-LM5164DESIGN-CALC
Other Parts Discussed in Thread: LM5164

Hi team,

It's a customer issue that I use the quickstart tool to analysis the efficiency of LM5164 at Vin=12V, Vo=5V, Io=50mA.

Other parameters remain unchanged, change the inductor LF to analyze the efficiency at Io=50mA.

However, I also find that the current range set by Full load current Io will influence the calculated efficiency.

For example, when LF=33uH, and full load current is set to 0.5A, the efficiency at 50mA is  82% as below.

However, when LF=33uH, and full load current is set to 0.1A, the efficiency at 50mA is  77% as below.

I'm confused about why the current range set by Full load current Io will influence the calculated efficiency. 

What factors affected this result？

Which result is the correct one?

Is this a display error of the calculation tool? 

Thanks

  • 0
    TI__Genius 10685 points

    Hello,

    This calculator recommends an inductance value that will result in an inductor current ripple of ~40% of load current. Changing the full load current Io therefore changes the recommended inductor value. 

    For a larger Full load current Io, the maximum allowable inductor current ripple to ensure continuous conduction mode (CCM) will increase. This means that a smaller valued inductor can be used for larger load currents in order to ensure that the device operates in CCM.

    In your conditions, the device will be operating in PFM mode where the frequency is changed to obtain the desired output. Assuming all other conditions remain fixed, the current through a larger valued inductor will take more time to ramp up to the minimum peak current limit threshold. This longer amount of time to ramp results in more energy being transferred to the output and therefore the the device does not need to switch as often (meaning a lower switching frequency) in order to ensure that the output voltage remains in regulation. At light load, efficiency is dominated by switching losses therefore lower switching frequency results in less switching losses and higher efficiency. 

    Regards,

    Harrison Overturf

  • 0
    TI__Prodigy 50 points

    Yes, the recommended inductance will change with the full load current value.

    However, my actual inductance that I selected is fixed. 

    Why the change in the full load current will influence the actual calculated efficiency?

  • 0 in reply to Junjie Mao
    TI__Genius 10685 points

    Hello,

    It looks like this has to do with the inductor core loss calculation.

    The calculator assumes this to be inductor core loss at full load current Io.

    The calculator also assumes full-frequency CCM switching at full load for the core loss calculation. 

    If you adjust the core loss then you will get better matching between the two cases. However, for the 0.1A case the converter is not in CCM with the 33uH inductor so the second assumption is not valid and the core loss calculation will be slightly off.

    Regards,

    Harrison Overturf