You've not explained or shown how you have your batteries arranged, nor provided any specs on them. But it seems likely they are in series - thereby providing ~ 9V output.
Batteries in series can provide no more current that a single battery, and even a good quality AA cell isn't capable of much more than 1 amp. And so your battery pack will be unable to source more that 1 amp to the converter - that's not likely to be sufficient as the calculations below will show.
It's not possible to do a calculation with much accuracy because there are several variables missing - the battery specs and the regulator efficiency being two important ones. However, we can make assumptions as we step through the calculations:
1. All of the Power comes from your batteries: P = V x I
For a series arrangement of 6 AA cells, we get ~ 9 VDC
The current capacity in a series arrangement of batteries however is no more than that of a single battery. Also, placing batteries in series means that the effective internal resistance of the series arrangement is 6X that for a single battery. Some estimates state the internal resistance of a AA cell will be at best 0.15𝛀, so about 0.9𝛀 total for 6 in a series arrangement. Also - this resistance increases as the battery discharges - perhaps to 0.75𝛀 at 90% discharge.
I found this graphic depicting how much current a good AA battery can source:
As can be seen from inspection, voltage goes downhill rapidly at any discharge rate needed to sustain an RPi. And while Duracell's graphic shows that 1 Amp may be sourced, this will be be for only a very short period of time, and with a fresh, new battery. Let's choose 500 mA as a practical limit for the maximum current the batteries may source, and use that figure to calculate the power that the batteries can source:
Pbatt = V x I = 9V x 0.5A = 4.5 Watts
2. Losses are incurred before the power reaches the RPi:
The calculations show the batteries are optimistically capable of supplying 4.5 Watts. However, some of this power will be lost due to the internal resistance of the battery, and the inefficiency of the the buck regulator (any regulator actually). These losses must be subtracted from the power delivered by the batteries. We'll estimate these losses as follows:
Rinternal = 0.3𝛀 (for one battery; 1.8𝛀 for 6 in series)
Efficiency of buck regulator = 75%
Power lost in internal resistance: PR = I2 x Rinternal
PR = (0.5A)2 x (6 x 0.3𝛀) = 0.45 Watts
Converter efficiency loss = Pin x ( 1 - Efficiency )
Conversion loss = (4.5 W - 0.45 W) x ( 1 - 0.75) = 1.0 Watts
3. Power delivered to RPi: Pbatt - PR - Conversion loss = 3.0 Watts
Again - the battery discharge curves above inform us this will decrease rapidly over time; the situation is exacerbated by the fact that during startup/boot the RPi will draw more power than it will in a quiescent, idle state. Measurements on my RPi 4B show that it will draw approx 2.5W in an idle state, but the "official documentation" places that figure at 3.0W (600mA). Given that the low voltage threshold is 4.63V, you won't get much run time out of 6 new AA cells.
The "best way" to solve this is to forget about trying to power an RPi from a bunch of AA batteries. Instead, get a rechargeable Li-Ion battery pack. Some are called "portable chargers" or power banks", and supply 5V at their USB output. This eliminates the need for a separate regulator, and the internal batteries can be recharged many times. The other option is a DIY approach - as you've planned, except using a Li-Ion pack instead of AA cells.
However, if you have reasons for pursuing this particular approach with the AA batteries that you don't wish to disclose, and you don't mind laying out more cash for a good quality regulator you could consider a boost or a buck-boost converter. These converters have the advantage of allowing you to put your batteries in a parallel arrangement. Some of them are also more efficient than the buck converter. This table provides a fairly comprehensive overview of the characteristics of the various topologies. You want a topology that has a
Yes in the right-most column ( Vout > Vin ). But you should keep your expectations very low - the cost will certainly be higher, but the performance advantage over your current setup will be slight.