I'm developing a backup power supply using a supercapacitor and a TPS61030 boost converter based on this Texas Instruments circuit design. This is the circuit shown in the reference guide: and it is explained that the VOUT is set through the feedback resistors R6, R9 and R10. When the Vsys is present, the voltage at the LBI pin is above the threshold and the pin LBO is at high impedance so the resistor R9 is not affecting the others feedback resistors, setting the (so called in the guide) Vout_re to 4.5V, so lower than the Vsys after the D1 diode. In the case Vsys disappears, the pin LBO is connected to GND, modifying the feedback network in such a way to obtain 5V at the output. You can find the calculation at page 5 of the reference guide. This is functional block diagram from the datasheet: I don't understand why they are differentiating VOUT and Vout_re, because since the feedback resistors are used to set the VOUT voltage, those 4.5V should be also present at the output, lowering the value of the Vout when the Vsys is present. What I would expect, since this is a backup PS, is that as long as the Vsys is present, the boost converter goes in some kind of high impedance state, and not output 4.5V. I think I'm missing something, probably the IC is not outputting this voltage, but I cannot find a proof of that neither in the datasheet nor in the reference guide. Thank you in advance for your help.
If you naturally set the output to regulate at 4.5 volts and a slightly higher voltage is present (due to Vsys and D1), the regulator will immediately shut-down it's output stage in order to try to regulate to 4.5 volts. Clearly Vsys holds the output higher at 4.5 volts hence the output stage of the regulator shuts down.
those 4.5V should be also present at the output, lowering the value of the Vout when the Vsys is present
You have to remember that the output from this voltage regulator is not some kind of push-pull output that can force a lower voltage; it is a single series transistor between Vbat and Vout and will just go open-circuit when overdriven at its output.