This is a 9V power supply which will work even on power failure. It uses a rechargeable battery and regulators. A transformer with 15-0-15 AC volts output is required.
In the first regulator U1 the output is lifted up by 1.4V and in the second regulator U2 by a resistor divider. In the second regulator the voltage across resistor R3 is 5V, so the current is 5V / 1K = 5mA this adds to the quiescent current of 5mA from the regulators ground terminal and flows into the resistors R1 and R2 in parallel which form 404 ohms, 10mA thru 404 ohms is 4V. So the output will be 5 + 4 = 9V. Note that the charge and discharge paths of the battery are separated with diodes.
Battery is constatly charged and will cause it’s death after some time.
You must be a more complicated. Battery should ‘float’ after it’s fully charged, and should be back again in case of bad input voltage or charge time.
It means – supervisor circuit and some battery control is missing.
Cheerz.