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I am designing a buck converter using LMR36015 (Texas Instrument). The specifications are:

  • Input voltage: 24V (+/-10%) = 21.6-26.4V
  • Output voltage: 3.3V
  • Output current: 0.4A
  • Fsw = 1MHz

This leads to a on-time of 125ns. If I calculate the inductance value using the rule of thumb of 40% ripple current ratio, I find a value of 18µH. I decided to check the result on the TI software Webench and I was surprised to find a much lower inductance value of 2.2µH with an inductor ripple current percentage of 257% (although there is an option to choose the inductor ripple up to 200%)

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I dug into the datasheet and found the note that the ripple current percentage must be higher for output current lower than the maximum current capacity of the IC.

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On Webench, when increasing the output current closer to the IC current capability, the ripple current ratio approaches 40%. When decreasing the output current, it increases a lot (>250%).

My questions are:

  • Why when operating at low output current (in comparison to the IC current limit), the ripple current ratio increases? Is it specific to this TI controller?
  • How can I choose a ripple current ratio when operating in DCM? I believe the software purposely reduces the inductance value to operate in DCM.
  • Is it correct to state that a 200% ripple current ratio is the transition point from CCM to DCM (BCM)? Does it mean that a ripple current ratio of >200% means going further in DCM?

Thank you.

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I decided to check the result on the TI software Webench and I was surprised to find a much lower inductance value of 2.2µH

That value seems in the right ball-park for the boundary condition: -

  • Your power output requirement is 3.3 volts at 0.4 amps (\$I_{AV}\$); that's 1.32 watts
  • Your switching frequency of 1 MHz implies an energy transfer of 1.32 μJ per cycle
  • Energy \$=\frac{1}{2}\cdot LI_{PK}^2\$ and, at the boundary we know that \$I_{PK}\$ will be twice \$I_{AV}\$
  • So, L will be 4.125 μH

On Webench, when increasing the output current closer to the IC current capability, the ripple current ratio approaches 40%. When decreasing the output current, it increases a lot (>250%).

and

Why when operating at low output current (in comparison to the IC current limit), the ripple current ratio increases? Is it specific to this TI controller?

On high load currents, the converter will be running in CCM and this means that the ratio of ripple (pk-to-pk) to average current decreases. At the boundary condition the ripple current has to be twice the average current hence the ratio is 200%. As the converter drops into DCM ripple increases more. It naturally happens this way.

How can I choose a ripple current ratio when operating in DCM?

You are letting the tail wag the dog it seems; do the math on energy transfer at the boundary condition and you end up with a more suitable candidate inductor.

Is it correct to state that a 200% ripple current ratio is the transition point from CCM to DCM (BCM)? Does it mean that a ripple current ratio of >200% means going further in DCM?

Yes and yes.

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    \$\begingroup\$ Thank you for the explanation. \$\endgroup\$
    – Ultra67
    Sep 14, 2023 at 8:47

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