Modifying a buck converter for lower current rating

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?

• If you change anything, you may have to re-calculate the compensation components C6 and R4. – mkeith Nov 12 '18 at 7:47