I have a new piece of electronics (a small amplifier.) It runs off six AA batteries, giving 9V.
What are the practical issues in rewiring it and replacing those with a few 9V batteries in parallel?
Obviously the textbook answer is they are equivalent, but when one sees batteries being used they are almost always in series, not parallel. My gut tells me four 9V cells would last longer and deliver the necessary 9V to the amplifier.
Am I missing something here?
What would prohibit? Nothing, except you likely get less capacity with four 9V batteries in parallel than with six 1.5V batteries in series. And because connecting batteries in parallel is generally not a good idea.
Comparing 9V and 1.5V batteries of the same chemistry from same manufacturer, both drained to 0.8V per cell:
At 25mA load current, a single 9V battery has about 550 mAh and single 1.5V has 3000 mAh capacity.
Assume the six 1.5V batteries just ideally stack up to 9V 3000mAh battery.
Then, at roughly same load, 10mA per 9V battery from three batteries, would equal 1800 mAh. Six 1.5V AA batteries would give capacity of 3000mAh at 25mA, and even 2000mAh at 250mA load.
At 100 mA load current, 1.5V battery has more than 2500mAh, four 9V batteries at 25mA each is less than 2400mAh.
At 250 mA load current, 1.5V battery has 2000 mAh, and with five 9V batteries each delivering 50mA, they would add up to slightly more than 2000 mAh in capacity.
So how many 9V batteries you would need to match running time depends on the current consumed by the device, because 9V batteries have more losses than AA batteries.
All the obvious warnings not to parallel batteries apply, because they never have equal voltates, and thus connecting batteries in parallel will always cause current to flow from largest voltage battery to others, and the battery with least voltage is charged up with the current. And primary batteries must not be charged.
In practice, don't parallel 9V batteries.
The other answers talk about the physics, but there's also the commercial angle to consider, you did ask for practical considerations! AA batteries are made by the gazillion. 9 V batteries have a much smaller market volume, and tend to cost a lot more per watt.hour. If you are going to run your device from disposable batteries, consider the cost.
You need to take into consideration the battery capacity in mAh (milli-amp hours). This is a measure of how much current it can deliver over time. For cells in series, the voltage is additive while the capacity remains the same. For cells in parallel, the capacity is additive while the voltage remains the same. By "capacity" I am referring to the battery being able to deliver current over time.
A typical AA cell will have a capacity of something like 3000 mAh, but this figure varies based on current draw. If you draw only 25 mA, you get something close to this value. If, however, you draw 500 mA, you may get less than 2000 mAh.
Consider that a 9V battery consists of six very small 1.5 V cells. You might correctly conclude that it cannot deliver the same current, and certainly not for the same duration. However, you're on the right track: Connecting multiple 9V batteries in parallel can approach the same capacity. However, there's not much point if you require six 9V batteries to roughly equal six 1.5 V AA cells. Assuming these are all alkaline, they're going to have roughly the same energy density, so you won't achieve a smaller (physical) battery size unless for example you changed to something with different energy density (e.g. lithium-based).
You should measure what current your amplifier uses in order to make a more informed decision. If it uses very little (doubtful), you might be OK using a 9V battery (or a few in parallel), albeit with a shortened run time.
Further thoughts:
The 9V battery datasheet shows a capacity of 600 mAh at a discharge of only 10 mA. This is reduced to 450 mAh at 50 mA discharge. Let's pretend your amp draws 50 mA (likely more, but go with me here). At 50 mA draw, the AA cells would have roughly 2750 mAh capacity. This means you would need six (2750 divided by 450) 9V batteries to deliver the same capacity at a 50 mA draw.
The maximum current that batteries can deliver depends on their internal resistance. The datasheets I've linked are very simple and don't provide these details, but you can infer that the 9V battery is not intended to deliver more than 50 mA for very long, as the capacity chart only goes up to 50 mA.
Because capacity falls off rapidly at higher discharge rates, you might in fact run into problems with multiple 9V batteries in parallel if your amplifier draws high current. What is the run time of your amplifier now? What is its current draw? If your amplifier is drawing 500 mA and your maximum run-time is ≤4 hours, you might discover that with six 9V batteries that is significantly shortened.
If a product has several independent battery receptacles wired in parallel, the user will insert one exhausted battery and one fresh battery, because why not.
When this happens a high current will flow from the fresh battery to the spent battery which has lower voltage. Alkaline batteries are not rechargeable, and rechargeable batteries like a controlled charge current. High current resulting from putting two batteries at a different state of charge in parallel is more likely to damage them than to do something useful.
Many 9v batteries are simply 6 AAAA batteries in series, configured that way to make it compact.
This link shows the makeup of a 9v: http://dacuriousone.blogspot.com/2011/05/batteries-in-most-consumer-electronics.html
So you're essentially putting 1.5v batteries in series and then putting those collections of batteries in parallel. Like others mentioned this is not efficient and the batteries will not discharge at the same rate, and cause issues. You'd get better capacity with the AAs in series.
Of course, you are missing a lot. This is OK as long as you learn.
Here we talk mainly about the primary (non-rechargeable) things.
Why batteries are used traditionally in series: Because of the needed voltages vs what batteries have to offer.
If we skip the "tube" era in electronics, for most purposes one needs few volts to few tens of volts in order to power some transistor-based circuit.
On the other hand, the available electrochemical cells supplied modest voltages, e.g. between 0.5V (Volta pile cell when happy) and 2V (lead-acid battery cell).
Both the cells and the electronic circuits converged somewhat and today a lot of devices operate off a single Li cell (3.0V to 4.2V).
The series connection has its drawbacks - e.g. the unmatched cells (they are always somewhat unmatched) becoming dangerous when overdischarged. For the ordinary zinc-carbon 1.5V cells the danger ends up being electrolyte leaks.
But at least, when left alone, cells connected in series are ok for indefinite period.
What about the parallel connection?
Cells are not made equal even if you want to connect them in parallel. They have subtle (and depending on the quality, not so subtle) voltage differences. If you connect them in parallel, they will try hard to equalize these differences.
The cell with the higher voltage will try to charge the cell with the lower voltage.
But wait, they are not rechargeable! This is why the one with the higher voltage will deplete in these attempts, and the one with the lower voltage will build up gas pressure, heat or just wear off.
This may or may not happen to some visible extent with modern, reputable brand-name, fresh cells. On the other hand, one cannot be sure.
What you will do by paralleling two 9V batteries is actually to combine the drawbacks of both series and parallel connection (they are internally 6 cells of 1.5V).
Even if you manage to get them working right, you won't get much more of energy per volume or mass.
The 9V battery is optimized towards miniaturization. It saves the user from dealing with 6 separate miniature cells and their contacts.
On the other hand, it has additional volume, mass and manufacturing effort spent on the internal connections and the common package. These don't bring you energy, they bring you (supposedly) comfort.
And, the small cells in the 9V battery are worse in energy density in the first place - because of the infamous square/cube effect.
You don't need to believe me. Just get the specifications of these two types and calculate their energy density. For the same chemistry (be it zinc-carbon or alkaline) they differ like 1:2 if you compare 9V to 1.5V AA and maybe 1:3 if you compare to D cells.
There are more parameters to batteries than just voltage. There's also current, energy, and power.
CURRENT: The amp-hours which is how many ampheres the battery can supply for one hour before dying. Voltage is not taken into account so this can only be used to compare batteries of the same voltage.
ENERGY: The watt-hours is the same thing as amp-hours except voltage has been taken into account so this can be used to compare batteries of different voltages.
POWER: Then there is the power the battery can provide. Like above, this can written as the maximum current you can draw from the battery, or you can take voltage into account and write it as the peak watts the battery can supply.
It is possible to get a battery that can provide a lot of energy, but you can only pull that energy from the battery slowly (i.e. low power). Such a battery can be used to power a low power load for an exceedingly long time, such that it might even provide more total energy than a high power load that runs for a shorter amount of time. However, it cannot power this high power load since you cannot pull energy from the battery fast enough to provide the power this load requires.
Not answering the exact question, but if you want to improve on ugly AA cells, I would try to make the device run on a nice cheap and easy rechargeable usb power bank. There is a fair chance that it will operate satisfactory on 5V. If not, there are small boost converters that can turn 5V into 9V (adjustable) with about 85% efficiency and cost only some 1 to 5 euro or dollar, for max 3A.
clarify:
What I really proposed is to wire a USB cable to the device, perhaps with a small boost converter. Then a standard USB power bank, which comes in a lot of sizes, is just one option, another option being running from a USB power adapter. USB is actually the new DC power standard. I am typing this post on a laptop powered by USB-C. USB-C covers several voltages including 9V. The proposed boost converter is an improvisationary fix, the ultimate is a USB-C solution, but, to my limited knowledge, there is not yet a simple device to connect an ordinary 9V apparatus to USB-C. Not yet. Hopefully soon.
In case of powering from a usb power bank, that is better than a rechargeable battery, because:
I find it interesting that noone has mentioned internal resistance.
AA batteries have MUCH lower resistance than 9V batteries do. How will that change your circuit? Only you know.
I actually burned an IC that controlled a solenoid when I connected AA batteries to its 9V battery connector.
Additionally, a rule of thumb: For the same chemistry, energy capacity per volume is about the same. Swapping AA batteries for 9V ones which occupy the same space will do approximately nothing to available energy.
As already noted, the current draw is likely too high for a single PP3, and multiple in parallel is "A Bad Thing"(tm)
So, use a single PP9 or BAT10, both of which are rated for much higher currents.
