This 20 A buck regulator module from TI is 6.5x7.5x4 mm and has an internal 330 nH inductor.

This 6.5x6.5x3.1 mm 330 nH inductor from Coilcraft has an RMS current rating of 18.8 A and is the smallest I could find on their website at that inductance.

I have a couple of questions:

  1. How does TI manage to fit the inductor inside the package?
  2. Why are standalone inductors so big?
  3. What is the point of using a buck regulator without an integrated inductor?
  4. How does TI get the module to work with what seems like a really low inductance?
  • \$\begingroup\$ It's worth noting that the required decoupling caps for the regulator module are quite a bit larger than the module itself. Main use case is probably using several in parallel for multi-phase converter, to reduce the decoupling needs. \$\endgroup\$
    – jpa
    Commented Feb 16 at 8:54

2 Answers 2

  1. They just used a small one. The Coilcraft XGL4025-331MEC is a 330nH inductor with a 4mm x 4mm footprint and 2.5mm height, for example. It can handle 20A with a 40°C temperature rise. Its inductance will of course drop slightly at such high currents, but that's okay for a buck converter. By using such a small inductor, TI of course has to accept that higher temperature rise.

  2. Standalone inductors aren't necessarily big, see above.

  3. You can get lower losses with an external (larger) inductor that has less DC resistance. Additionally, buck converters without integrated inductors are a lot cheaper, even when you factor in the external inductor. A design based on a TPS53353DQP ($3.60) and an external inductor ($0.40) will cost approximately $4 in 1k quantity, whereas the TPSM843B22E costs $14 each according to TI.

  4. 330nH isn't unusual at all at 10A and above. For example, the ratio of ripple current to output current of a 20A converter with a 330nH inductor is identical to that of a 1A converter with 6.6µH (assuming both run at the same frequency). The higher the output current, the lower the inductance values you'll typically find in such designs.

  1. There are some hints in TI patents such as this one. Looks like most of the volume of the converter is the inductor.

  2. They need a certain volume of ferrite and more space for the copper winding to meet specifications (saturation current, inductance and DCR, for example).

  3. Usually the cost is much, much lower if you can afford the space. If cost is of no concern you can use whatever you like.

  4. High frequencies (500kHz to 2MHz) mean the inductors can be lower value, all other things being equal.


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