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TPS54A20: higher inductance related to better peak efficiency?

Part Number: TPS54A20

Hi , 

I was looking at the training videos: 

and wondering why higher inductance related to better peak efficiency. 

Can you help to explain to me for the reason. 

Great thanks for your help. 

  • Hi User,
    This is pretty straight forward. Consider a series of inductors with the same physical package size. A smaller value inductor will require fewer turns of wire than a larger value inductor. So for the same package size you can have fewer turns of heavier gauge wire which will result in a lower DC resistance. This lower DC resistance is why the lower value inductor tends to have higher full load efficiency as at higher loads the conduction losses dominate. The higher value inductor has lower delta IL (or ac peak to peak inductor current, ILp-p). The RMS inductor current is given by:

    IL(RMS) = SQRT(((Io^2)+((1/12)*(ILp-p)^2)))

    The Io^2 term obviously increases with higher Iout. The ILp-p^2 term is constant with respect to load current. So at higher currents the Io^2 term dominates while at lower currents the ILp-p^2 term dominates. That is why the larger inductor value with lower IL-pp can have better peak efficiency at lighter loads.

    I hope this helps you better understand this topic.