One more question, approx what current will the 10v-15v curcuit draw from Vdd and from battery?
I ask because if its only tiny I'll leave it connected and powered but if it's high I'll dissconect from the batteries and remove supply power to it except when taking a measurment.
About 1.3mA.
Whether or not that makes any difference, depends on how big the battery is. If the battery is small, say 1.3Ah, it will fully discharge in 1000 hours, with this circuit connected, but if the battery is much larger, 26Ah, it will take 20000 hours.
If low power operation is required, the TL431 could be replaced with a different IC, which requires less current and the resistor values all increased, but if that's not good enough, a totally different approach is required.
EDIT:
Where is the 5V supply coming from? If it's off the battery, via a regulator, that might be the main drain on the battery.
If the 5V is off, Q1 will turn off and no current will be taken from the input. To save power, the 5V supply, to the circuit I posted, could be an MCU output pin, which could be set to high, when a measurement is taken and low, when it's not needed.