Electronic – Reverse current protection for a battery-operated circuit

Here is a low power LDO voltage regulator (TPS79733) that consumes less than 2uA quiescent current whilst delivering up to 10mA to keep your MCU ticking over when it's asleep.

It has a fixed output of 3.3V and I think the clincher would be to arrange for your TPS63061 switcher to generate maybe 50mV more output (3.35V) so that when it gets activated, it sources all the higher current when you are taking 400mA.

In effect the LDO would not source any current because it would see an output voltage across its terminals that forces it to "open circuit" its pass-transistor.

Thus you don't need to enable/disable the LDO regulator.

My question is how to handle this switch-over phase. The LDOs which I looked at have reverse current protection and may be put into standby/shutdown mode. Is this sufficient to not destroy the LDO when the switcher starts working?

Handling the switch-over phase - enable the switcher when the MCU powers up Standby/shutdown for LDO protecting the LDO - the LDO will always be enabled.

The circuit as shown will work. Remember though that it's a linear regulator, which means that the voltage is dropped by turning excess energy into heat. At 4.5V the drop is small, but if you're dropping to 1.5V (3.5V drop) at 500mA (you shouldn't expect to draw more from USB) then you'll have to deal with 1.75W of heat. The amount of heat that the IC you linked can dissipate depends on the PCB design (did you mean to link a surface mount component?), but in any case 1.75W would be the upper end of what you could expect a TO-220 package component to dissipate. I'd probably use a heat sink or ensure that my load wasn't drawing as much current.

Resistor selection basically doesn't matter. The two aspects that you need to normally consider when selecting resistors are the power rating (1/4W, 1/2W etc) and the tolerance (1%, 5%, 10%). The power rating isn't important in this case (see below) and because you've got a manually adjustable potentiometer the tolerance really isn't important either. Almost any resistor of approximately the right value would do for \$R_2\$.

Both resistors can be low wattage ones. As you've said, the \$I_{ADJ}\$ current is negligible and can be completely ignored. There will also be a current flow through the two resistors from the output voltage to ground, which you can calculate with Ohm's law (\$V=IR\$). In all cases this will be about 7mA.

The capacitors are there for filtering/ripple reduction purposes and their characteristics aren't too important. If they're approximately the same as those suggested in the datasheet (and the voltage rating is above what they'll see) then there shouldn't be any issues.

Your calculations seem to correct. This is just a basic voltage divider calculation. The regulator adjusts the output voltage until it sees 1.24V on ADJ. You can confirm that your calculations are correct with the equation given on wikipedia:

$$V_{div} = V_{in} \times \frac{R_2}{R_1+R_2}$$

Where \$V_{div}\$ is 1.24V, \$V_{in}\$ is your target voltage and you want to solve for \$R_1\$.

$$R1 = \frac{V_{in} \times R_2 - V_{div} \times R_2}{V_{div}}$$

for the 4.5V case therefore:

$$R1 = \frac{4.5 \times 180 - 1.24 \times 180}{1.24} = 473 \Omega.$$

There's inherent variability in electronics which mean that whatever values you calculate won't be exactly right when you wire it up. Wire it up, connect the output to a volt meter and adjust the potentiometer until you have the right voltage.