Electronic – powering +5v device with usb and/or battery

A 4.2V buck converter isn't good for charging a lithium ion battery unless it can't source too much current. There's three stages that a li-ion battery needs to be charged correctly. Messing any of those up can result in exploding batteries. See charging lithium ion batteries for more info on that.

Your question about the charge entering and leaving a battery isn't really an issue. I think in that regard, your question is a dupe of another on here a month or so ago that asked what happens when you charge and discharge a battery at the same time (I can't find it now...). At any rate, the point of the answer was that any extra current you need from the 4.2V buck will be sourced from it, and any current from the battery that it can source will be sourced.

If the battery is discharged lower than 4.2V, then the buck converter will supply both the battery some current and it will source the boost some current. This assumes that you have connected the battery in parallel to the boost converter.

Yes but, but, but ... . The basic method is correct as far as it goes but there are more factors to consider than are covered by that formula.

The figure is essentially correct

• for an ideal 5V supply

• and a battery whose Ah capacity is numerically much larger than Imax_A,

• if you wish to charge at true constant current in the CC phase,

• as long as your assumed resistances are correct.

If you can tolerate a decreasing current during CC then Icharge can be greater when Vbattery is < 4.1V, provided that you do not exceed the maximum available current from the "USB" source. eg a 1000 mAh battery may allow C/1 max charge rate = 1A. Your formula may give 600 mA in a specific case, but when Vbat is below 4.1V Imax will increase. Your formula gives Reffective = V/I = 0.6 / 600 mA = 1 Ohm. At Vbattery = 3.7V the formual yields 1A and for Vbattery < 3.7V you will need to current limit in CC mode t present the C/1 rate being exceeded.

One assumption that you made but did not explicitly state is that V_USB === 5V.
Actual is ~= 5V. You may look at the spec to see what the range of "~+" is in theory. In practice it can be somewhat higher than 5V and may be less.

Note that while that gives the maximum theoretical current for a 5V USB supply, the available current may be restricted to less than this theoretically and/or actually by the supply itself.

Also, maximum battery current must not be more than the manufacturer allows - usually C/1 but eg C/2 in some cases and I've seen 2C allowed for some "special cells.