I am designing a switching buck converter from 7.5V to 5V using the AP62200TWU-7 converter.
This converter operates at fSW=750kHz Switching Frequency.
The required output current is around 0.5A.
using the formulas provided in the datasheet:
(https://datasheet.lcsc.com/lcsc/2202132100_Diodes-Incorporated-AP62200TWU-7_C2895288.pdf)
I estimated the inductor current ripple βIL to be 0.2A
using the inductor calculation formula:
π=ππππβ(πππβππππ)/(πππβοΏ½οΏ½οΏ½ππβπππ)
I calculated the inductor value to be around 11uH.
In the datasheet they state that it is recommended to select an inductor of approximately 1.2Β΅H to 4.7Β΅H which is below the calculated value.
the question is: should I select a 11uH inductor? or go with the max 4.7uH as per the datasheet?
1 Answer
Generally it is a good idea to use the recommended values, because other aspects such as error loop compensation are internally fixed in regulators like this.
It's not clear that this regulator will have compensation issues. Constant-on-time control is generally stable (as opposed to peak current mode and voltage mode controls, which depend on the filter LC values), but whether they have modified the basic control in a way that it may affect response, is not documented.
The safe plan is to try both. If the transient response is inadequate with the calculated value, use a value closer to recommended instead.
A lossy output capacitor may also be a stability aid: consider using ceramic with external resistor (say 20 to 200mΞ©), or electrolytic or tantalum, for the bulk of the capacitor value. High frequency filtering can be maintained by putting a smaller amount (having low ESR) in parallel with it. The ESR adds a zero to the feedback loop transfer function, increasing phase margin and generally improving stability; the cost is increased output ripple.
