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What are benefits and drawbacks of using Buck Regulator with external MOSFET compared to Buck with embedded MOSFET? Is there some general rule?

In my case I need 5V 3A from 9V-30V supply.

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2 Answers 2

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The differences are largely obvious just from the mechanics:

  1. Integrated is smaller since you don't need to add the pass element.

  2. Integrated is a little easier to use since the chip does more for you, particular the driving of the pass element.

  3. External can be better performance. The integrated pass elements don't usually have the best available specs.

  4. External is necessary for high current. When the current is high enough that dissipation of the pass element is a issue, then forcing it to be small and integrated on a chip that also has to do other things is a obvious disadvantage.

In general, get integrated for simplicity and board space, external for high power or high efficiency. It is interesting to note that cost is often not a deciding factor. The chips with integrated pass element are suprprisingly expensive. You can often do better on cost by getting exactly the transistor you need for your particular case, as apposed to the many sizes fits all that had to be integrated onto the chip.

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  • \$\begingroup\$ I read somewhere on Monday that in order to minimize the perturbations in DC DC converters you have to design the inductor mesh as small as possible... I'm working with a converter for a fuel cell (150A max) where the MOSFETs are embedded and I don't have any problems... \$\endgroup\$
    – Yves
    Commented Jan 23, 2013 at 17:37
  • \$\begingroup\$ Basic cheap - fast - good comparison \$\endgroup\$
    – Passerby
    Commented Oct 18, 2015 at 3:36
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If you want maximum efficincy it may influence your external / internal decision.

For maximum efficiency you want synchronous conversion where the freewheel diode is replaced by a MOSFET, thus removing the forward voltage drop of the diode and replacing it with the IR drop of the FET fully on resistive channel. At most current levels even a Schottky diode will drop at least 0.3V and 0.6V is more common that you may expect and rather higher again at substantial scurrents -say 10 A + to ++.

In your case with 5V at 3A out, every 0.1V drop in the freewheel diode represents about 0.1/5 = 2% efficiency loss and 0.3W dissipation. (This depends in part on your duty cycle but at the high end of your Vin range your will be in freewheel mode much of the time.

SO - internal switches "just work" for synchronous switching - whatever drive arrangements, anti shoot through, dead time handling etc that is required is handled by the IC designer. The penalty is that you accept the onboard Rdson of the switches and these are often substantially poorer than you would choose to use if making the decision re external FETs. BUT if you use external synchronous switches the IC designer may make drive reasonably easy, or not. Some ICs benefit from a Schottky diode across the synchronous flywheel replacement FET to catch the signal transitions that the internal drive does not handle well. without this extra diode you can get losses at the switching edges. If the whole lot are internal the designer has to 'do it right' or it will show up in the specs as poorer performance.

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