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I have been learning about buck converters and I think I understand the principle of their design. However, I am unsure of how a current rating would be calculated for a particular design.

The schematic below is rated at 3A (Vin=48v; Vout=5v). However, if I want to power a device that requires no more than 500mA, I presume (perhaps naively?) that a 1A rating would be sufficient, being conservative.

I understand that the main contributors to current rating in a buck are the diode, inductor and output capacitor - by reducing the rating of these components I can make a much smaller PCB. So if I use, for instance, a 1A inductor and diode (the two largest components by far), will this affect the function of the circuit, and produce the same 5v output? And would a 1A component confer an overall 1A rating, or should some components be rated more conservatively to handle voltage spikes produced by ripple current?

Many thanks in advance for your wisdom!

3A rated buck converter schematic - 48v > 5v

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  • \$\begingroup\$ If you change anything, you may have to re-calculate the compensation components C6 and R4. \$\endgroup\$ – mkeith Nov 12 '18 at 7:47
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You might be better off choosing an IC that is targeted at providing the level of current you're looking for. I haven't looked at this design in detail but it is normal to select an inductor value that results in a peak to peak ripple current of about one third the output current, so the inducance value will be higher. Inductor needs to be rated for peak current. Smaller diode sounds OK. You can no doubt tolerate less capacitance at the output. All these changes will no doubt impact stability. I suggest you systematically work your way through the datasheet design process.

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  • \$\begingroup\$ Thanks for the explanation, I guess I haven't read enough about the IC itself yet. Can you recommend any resources for the design process? \$\endgroup\$ – Cameron Hyde Nov 12 '18 at 19:43
  • \$\begingroup\$ For this particular device, there is guidance regarding component selection in the datasheet starting at p15. For a good general introduction try ti.com/seclit/ml/slup067/slup067.pdf \$\endgroup\$ – Steve Hubbard Nov 12 '18 at 23:08
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This IC has a fixed value of current limiter, you can not adjust it. It senses the voltage drop accross the integrated MOSFET switch with a preset value.

As for inductor: The output maximal current is not the same as the inductor maximal current. You can google for buck converter calculations. This IC limits the current trough the inductor at 6A, so the inductor has to be rated for peak 6A.

The output current is the mean value of the inductor current, it depends if it works in CCM or DCM mode, see the basics of buck converter.

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