Not exactly. The reason for the higher voltage than expected on the +3.3 volt supply is because of current from your pull-up resistor. The capacitors absorb the current, which charges the capacitors up a voltage limited by the pull-up resistor and the load, which in this case are the feedback network and apparently the microcontroller.

I cannot make out the last word in the label for the signal going into the left side of the microcontroller (please see attachment). I can read "3.3V GPIO..." but can't make out the last word. If this is the power supply to the microcontroller things make even more sense, since you wrote "+4.2 @ A sleep" and "+3.3 @ A normal".

If that is the positive power supply input of the controller, then it all makes good sense. When the controller is sleeping it draws only a tiny amount of current and the pull-up resistor pulls the +3.3V power supply up to 4.2 volts through the electrostatic discharge protection diodes on the pulled-up input, limited by the high resistance voltage divider in the regulator and to a greater extent, by the power supply input current requirements of the sleeping controller. The capacitors just keep the voltage from changing quickly, which, as bypass capacitors is what they are there for in the first place.

When the controller wakes up it draws more current and the pull-up resistor cannot supply enough current for the controller when the power supply is above 3.3V so the power supply voltage drops down to +3.3V and the power supply adds any necessary current to power the controller. (I added this paragraph to make the description complete).

So...to summarize, the source of the extra 0.9 volts is current from the pull-up resistor. The capacitors just keep the voltage from changing quickly, and the inductor plays no significant role in this phenomenon. Try it -remove the capacitors and you should still see the 4.2 volts remain while the controller sleeps, provided it can get into the sleep state without bypass capacitors.

One thing you might want to check just to be sure is whether the small current on the I/O pin is close to the maximum specified input current for that pin when it is driven above the positive power supply voltage, and whether or not everything else that see this higher voltage, such as the capacitors, is safe at that voltage.

If I were doing this, and it turned out that the small current, which I see as about 40 uA, turned out to be a danger I would probably use a larger pull-up resistor if conditions permit or put a large resistor right between the I/O pin and the rest of the circuitry, or better yet, use a +3.3V pull-up and side-step the problem completely. The reason these would be the preferred solutions is that it might be difficult to find a Zener diode with the right current-voltage curve to fit your application because the leakage is very small.

Thank you very much for the detailed answer. Cleared my doubts

 

Not exactly. The reason for the higher voltage than expected on the +3.3 volt supply is because of current from your pull-up resistor. The capacitors absorb the current, which charges the capacitors up a voltage limited by the pull-up resistor and the load, which in this case are the feedback network and apparently the microcontroller.

I cannot make out the last word in the label for the signal going into the left side of the microcontroller (please see attachment). I can read "3.3V GPIO..." but can't make out the last word. If this is the power supply to the microcontroller things make even more sense, since you wrote "+4.2 @ A sleep" and "+3.3 @ A normal".

If that is the positive power supply input of the controller, then it all makes good sense. When the controller is sleeping it draws only a tiny amount of current and the pull-up resistor pulls the +3.3V power supply up to 4.2 volts through the electrostatic discharge protection diodes on the pulled-up input, limited by the high resistance voltage divider in the regulator and to a greater extent, by the power supply input current requirements of the sleeping controller. The capacitors just keep the voltage from changing quickly, which, as bypass capacitors is what they are there for in the first place.

When the controller wakes up it draws more current and the pull-up resistor cannot supply enough current for the controller when the power supply is above 3.3V so the power supply voltage drops down to +3.3V and the power supply adds any necessary current to power the controller. (I added this paragraph to make the description complete).

So...to summarize, the source of the extra 0.9 volts is current from the pull-up resistor. The capacitors just keep the voltage from changing quickly, and the inductor plays no significant role in this phenomenon. Try it -remove the capacitors and you should still see the 4.2 volts remain while the controller sleeps, provided it can get into the sleep state without bypass capacitors.

One thing you might want to check just to be sure is whether the small current on the I/O pin is close to the maximum specified input current for that pin when it is driven above the positive power supply voltage, and whether or not everything else that see this higher voltage, such as the capacitors, is safe at that voltage.

If I were doing this, and it turned out that the small current, which I see as about 40 uA, turned out to be a danger I would probably use a larger pull-up resistor if conditions permit or put a large resistor right between the I/O pin and the rest of the circuitry, or better yet, use a +3.3V pull-up and side-step the problem completely. The reason these would be the preferred solutions is that it might be difficult to find a Zener diode with the right current-voltage curve to fit your application because the leakage is very small.

the word on the left side of the box is 3.3V GPIO Domain