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Confusion with LM2735 Boost/SEPIC Converter Inductor Selection

Peter Garratt
Prodigy 50 points 
I have some confusion regarding the selection of the inductor for a LM2735 application and the applicable datasheet is adding to the confusion.
I am looking at a boost application with Vin=3.3V, Vout=12V and a load current of about 185mA
Using Example 4 of the LM2735 datasheet, SNVS4853, (which is very close to my needs - the output load is a bit higher than my application) as a vehicle
for discussion, the suggested inductor is 15e-06H.

Now, considering equation (5) from the same data sheet and working backwards to deterrmin delta IL form the selected inductor, this leads to the following:

Ts = 1/520kHz = 1.923e-06 s
Vin = 3.3V
Vout = 12V
D = (Vout - Vin)/Vout = (12-3.3)/12 = 0.725
L = 15e-06H

&

deltaIL = (3.3/(2×5e-6))×.725×1.923e-6 = 153mA

This is in excess of 40% of the 350mA load current and would appear to contradict the guidance given on page 11 of SNVS4853 to use between
10% & 30% of the load current.

Additionally, the data sheet uses the term "ripple of maximum load". Can I take this to mean 92 ×deltaIL)?

Could someone please clarify the above interpretations, so that I can be confident I have selected the right value of inductance?
  • 0
    TI__Genius 15810 points

    Hello Peter,

    The inductor is  usually chosen to have about 30% ripple current. But this does not mean that much percent of the load current. The current through the inductor should have a 30% ripple. For the buck topology, the current through the inductor is the load current. Therefore for a buck, we size the inductor for 30% of the load current. But for a boost topology, the average current through the inductor is given by this formula:

    Iind = Iout/(1-D)

    Therefore, for a 350mA load, you get an average current of 1.27A through the inductor. This means that the entire ripple is given as:

    Iripple = Vin*D*T/L = 300mA. This is about 24% of the average inductor current. 

    The sentence 'ripple of maximum load' should mean that the ripple in the inductor current corresponding to the max load. This means that if your load varies from 80 to 100mA, then the ripple should be set corresponding to the full load of 100mA. I hope this clears up the confusion. Let me know if you have more questions.

    Regards,
    Akshay 

  • 0 in reply to Akshay Mehta
    Prodigy 50 points

    Askay,

    Thank you for the clarification. That makes more sense. In thinking about boost converters in general, one can see that the input current will be higher than the output, by definition from power/energy considerations.

    What was not so clear was whether this was some special constraint for the device.

    If I'm honest, the data sheet could be a bit clearer to lead the thinking and facilitate implementation, particularly for people who are less conversant with the various SMPS topologies.

    However, your explanation resolves my confusion, so thanks once again.

    Regards

    Pete

  • 0 in reply to Peter Garratt
    TI__Genius 15810 points

    Hello Pete,

    Thank you for your honest words. The goal usually is to make the datasheet as clear to understand as possible. We will look at it again and see how we can reword it so it is not confusing.

    Regards,
    Akshay