You want to get constant current at 12V to run the LED's from 1.5V batteries.
You will need a boost converter, as others have suggested, to raise the 1.5V to 12V. Power conversions are not ideal, and all voltage-changing devices consume/waste some of the input voltage doing their job. The constant-current chart in the data-sheet is meaningless since a boost converter draws a variable amount of current at a variable frequency and/or duty-cycle.
At the LED's, what you are asking for is:
Constant voltage(E) * constant current(I) = constant power (P)
Regardless of the conversion method, you will require the same input power from the battery as you want the LED's to output. Boost converters are very efficient, but do have associated losses.
Pbattery = PboostLoss + Pled
and
P = I * E
So, for a boost converter to draw a constant P from the battery, as E decreases, I must increase, discharging the battery even faster, causing its voltage to drop faster, etc., etc. That is the reason there is not an easy or definitive answer as to how long they will last for your application. LTSpice is helpful for time-integrating things like this but I'm not sure if you can model a AA battery's discharge properly.
There are only two "absolutes" I can think of for this question:
In identical circuits, twice as many batteries in parallel will run for twice as long. Batteries discharge constant power, together.
Twice as many batteries in series will require less intense and/or frequent boosting to reach 12V, reducing switching losses. Batteries discharge constant power, together.
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A boost converter: https://www.pololu.com/product/799
Comparison: https://www.pololu.com/search/compare/132
When you go looking for similar trinkets, check SparkFun, Adafruit, and Pololu first!
How can I find power consumption of TPS7A7001 regulator?
Assuming that Vin = 2.5 volts and Vout = 1.8 volts with a 1A load (just me putting some numbers together), the first thing to note is that the voltage dropped across the regulator is 2.5v - 1.8v = 0.7v and this voltage multiplied by the 1 A current will produce heat directly from the device at a power of 0.7 watts.
The feedback resistors R1 and R2 (in series across the output) are, according to table 1 in the data sheet something like 100kohm and this will consume 32 micro watts (barely nothing compared to the main power dissipation above).
There is some ground pin current and this is only loosely specified as 3mA - this is drawn from the incoming power rail hence it will dissipate 2.5v x 3mA = 7.5mW (again this is trivial compared to the 700mW main dissipation.
Best Answer
This gives you a guideline: -
Every time the OR-gate output rises to "1" you can regard the capacitor (in the table above) being charged-up to the supply voltage. This takes energy (\$CV^2/2\$) and that energy is unrecoverable because when the OR output falls to "0" the capacitor is discharged. A small amount of heat is produced.
Along comes another "1" and you put energy into the capacitor only to lose that energy when the output goes low. So, if your Vcc is 3.3 volts then the energy injected into the capacitor is 114 pico joules.
However, you do that 24,576,000 times a second hence the power is 2.81 mW or, the average current taken from the 3.3 volt DC supply is 0.85 mA.
It's a rough guide.