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LM61480: Question on Inductor Selection for LM61480

erik neyland
Prodigy 40 points
Part Number: LM61480

Hello,

 

I have a question on calculating the Inductor value for the LM61480.

 

Equations 5 and 6 in the datasheet don’t seem to correlate to each other.  Can you clarify what equation 6 is for?

 

My variables are Vin = 12V; Vout = 9.5V; Fsw = 400k; K = 0.3; Iout(MAX) = 8A (actually 6A, but says to use max of part).  This gives me an L = 2.06uH, which makes sense.

 

Equation 6 says L > 5uH.  What is Irated? 8A?.   This is why eq 6 doesn’t make sense to me.

 

Erik

  • 0
    TI__Mastermind 47060 points

    Hi Erik,

    Equation 5 is a good starting point to for most typical applications. However for peak current mode control devices where applications operates above 50% duty cycle, there is a potential for subharmonic oscillation. As such, it is suggested to use Equation 6 instead of equation 5 here. 

    For your application, you should use Equation 6. Irated = Imax = maximum rated current for the device chosen.

    Regards,

    Jimmy