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With all the technology available today with being able to boost voltage efficiently using SMPS, why do we still use 9V batteries? Is there some secret advantage with them that I am unaware of?

If you look at the size as well, the 9V is just big and bulky and I have designed projects where I can use 2xAA batteries and boost the voltage, which will give me longer battery life than a 9V. And it takes up the same amount of space.

A lot of circuits today also need regulating, and the easiest way to do that with a 9V is a linear regulator (usually to about 5V) and I am aware this is not the case for every design, but that right there is energy wasted, and yet again, boosting the voltage from 1 or 2 AA batteries will probably give your product a better shelf life.

I saw a comparison between a 9V battery and some AA batteries, where someone found the energy available, and ended up with this data: enter image description here NOTE: These results were from Energizer Alkaline batteries

So with all this data, why are 9V batteries still used in designs? Are there some applications where it would be advantageous to use them? Or is it usually a better idea just to go for the AA or AAA solution?

There have been times where I have considered using a 9V battery for some of my projects but it always seems after doing my calculations, they just don't hold up as well as others, so am I missing something?

For reference, the datasheets for the compared batteries are here: AA 9V

EDIT: I am not intending this to be an 'opinion-based' question, rather, I was intending to ask from a practical point of view, if there were advantages to choosing a 9V over any other solution (such as boosting AA batteries). Just wanted to make that clear!

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    \$\begingroup\$ Boost circuits can be noisy. Plus, they cost more to manufacture than "Batteries not included" labels. \$\endgroup\$ – Dampmaskin Aug 8 '17 at 13:55
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    \$\begingroup\$ "why are 9V batteries still used in the industry?", what industry would this be? Not my industry at least! \$\endgroup\$ – pipe Aug 8 '17 at 13:58
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    \$\begingroup\$ power electronics wasn't cheap or small or widely available in the past \$\endgroup\$ – user3528438 Aug 8 '17 at 14:01
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    \$\begingroup\$ I would say it is all about manufacturer legacy design inertia and extremely wide consumer availability. \$\endgroup\$ – Michael Karas Aug 8 '17 at 14:07
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    \$\begingroup\$ @Bob There's a lot of room for fudging voltages with anything intended to replace alkaline cells because during discharge each will drop from ~1.5V to ~1V when the cell is dead. a fixed 7.2 or 8.4v NiMH replacement is perfectly in spec. 9.6v is pushing the notional spec; OTOH the intrinsic voltage drop in alkaline cells means that any devices using them either need to be able to handle a relatively broad range of voltages directly or have some sort of voltage regulator. \$\endgroup\$ – Dan Neely Aug 8 '17 at 20:22

10 Answers 10

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First a AA is not a battery.. it's a cell. A battery is a "battery" of cells, i.e. more than one cell. A 9V battery contains six 1.5V cells. Rip one apart and see.

As for why we still use 9V batteries, it really is a matter of the design. The nice thing about 9V batteries is they give you a fairly wide operating voltage range during their life without being overly high a voltage. They also come in a really nice compact package that has a rather usable clip.

I would in no way recommend using one with a linear regulator to generate 5V though unless the current requirement on that 5V is very small. Better to design your circuitry using componentry that will work at the 9V directly.

It also really depends on the nature of your widget. If you have sensors or transducers that require larger voltages and it's battery operated, it's simpler and usually cheaper to go with a 9V battery.

One also has to consider the ramifications of boosting a lower voltage. By doing so you will introduce a lot of new problems not the least of which is the electromagnetic noise you will be adding and needing to deal with. Efficiency is also an issue.

But in the end, there is a lot of to and fro on the decision so this question really comes into the opinion ranks.

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    \$\begingroup\$ I respectfully disagree with your first sentence. A AA cell can be a battery. A lot of devices only take one AA. \$\endgroup\$ – trpt4him Aug 8 '17 at 19:28
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    \$\begingroup\$ @trpt4him Also, they sell "AA batteries", not "battery refill AA cells". Surely when discussing a circuit you can call the part with the AA's in it "the battery", but when discussing the objects you buy at a store and use to power that circuit, I think they are "batteries". \$\endgroup\$ – Darren Ringer Aug 8 '17 at 19:33
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    \$\begingroup\$ In non-EE English, "AA battery" and "AA batteries" are phrases used by people and in advertisements (at least in the USA) every single day. "Two AA batteries not included" appears on so many packages. en.wikipedia.org/wiki/AA_battery \$\endgroup\$ – Todd Wilcox Aug 8 '17 at 19:54
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    \$\begingroup\$ I agree that a battery is an array of cell(s). I see cells as a component of a battery; my car battery has 6. I do not agree that the term "battery" cannot be used for single-cell devices, just as I do not agree an array requires 2 elements to declare. A 1-cell battery is simply a cell which has additional finishing and packaging to render it fit for direct consumer use. \$\endgroup\$ – Harper Aug 8 '17 at 21:21
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    \$\begingroup\$ AA is not a battery or a cell. It's a size. Whether it is comprised of a single cell or multiple cells seems like an implementation detail that people using it don't need to care about as long as other observable characteristics remain the same. \$\endgroup\$ – jamesdlin Aug 9 '17 at 5:57
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A 400mAh 9V battery will last a year with a 40µA current draw.

Now consider a smoke detector. It is low power analog circuitry, most likely drawing less than the 40µA figure above. If you wanted to power it from a boost converter and AAs, then you'd need a converter with very low idle current.

But... when there is fire, now you need quite a bit of power, and enough volts, to drive the piezo loudspeaker. These need voltage. 9V is louder than 3V.

So your very low idle current DC-DC converter also needs to output high current if needed.

You also need to be able to measure state of charge accurately on the AAs.

All this will cost more than the difference between 9V and 2AA. And remember, the customer pays for the replacement batteries, not the manufacturer!

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    \$\begingroup\$ But surely, the really best design would be to have the detection circuit operate at only 1.5 V without any boosting, and only in case of a fire switch on the high-power boost converter to drive the speaker? \$\endgroup\$ – leftaroundabout Aug 8 '17 at 21:46
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    \$\begingroup\$ A few (10? maybe 20?) years ago I'd have said "because they are there". Early transistor radios were made to run on 9 V because of the germanium transistors they used. So "9 volt" batteries were very very common, easy to find in stores. Not so much any more. My local convenience stores stopped carrying them quite a while ago. \$\endgroup\$ – Jamie Hanrahan Aug 8 '17 at 21:56
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    \$\begingroup\$ Not sure it matters but all the smoke detectors I've purchased in the last 10 years run on AA batteries. \$\endgroup\$ – JimmyJames Aug 9 '17 at 19:30
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    \$\begingroup\$ @JimmyJames: exactly the opposite of my experience. All the smoke detectors I bought in the last 10 years run on 9V batteries. Even the backup for the powered ones was 9V. \$\endgroup\$ – Martin Argerami Aug 10 '17 at 13:23
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    \$\begingroup\$ @JimmyJames: People would be less likely to leave detectors without power if there were an easy way to shut off the low-battery chirping for 12 hours or so without disabling the detector. If the detector starts chirping at 2am and the store isn't going to open until 9am, what is someone supposed to do? \$\endgroup\$ – supercat Aug 11 '17 at 14:50
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Boost circuits have a quiescent current; some energy is wasted simply by having the boost converter. They also get very inefficient at low duty cycles.

So if you have a circuit that usually draws a very small current but occasionally needs to draw more, it's difficult to address that with a single boost converter.

The main users of 9V batteries are things like smoke alarms and multimeters that exactly fit this use case: low current some or almost all of the time. If you expect the battery life to be less than, say, 3 years with a 9V battery then it may be a poor choice.

You can see this in that almost anything that has a radio - toys, remote controls, etc - will be using either several AA or rechargeable Li of some sort.

Boost converters aren't free either, they cost parts and space.

(The design that allows cheap wall clocks to run off a single AA battery is quite neat, and I'd like to see a good reverse engineering of it.)

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    \$\begingroup\$ Nothing special about 1.5v electromechanical clocks, really. Here's an ancient datasheet which is pretty detailed: mikrocontroller.net/attachment/197819/ICM7038.pdf \$\endgroup\$ – oakad Aug 10 '17 at 2:28
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    \$\begingroup\$ In addition, boost converters tend to be much more noisy than step-down converters. \$\endgroup\$ – Lundin Aug 10 '17 at 12:01
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My multimeter uses a 9V battery and maybe draws about 10mA to function. It is very accurate.

If it is powered by a 3V battery, the average current will be 40mA with a 75% efficiency. However the surge current have to go up 80mA, thru an inductor. Basically the trace, battery wire and the inductor now act as an antenna giving off energy to my sensitive components inside my multimeter.

Then my readings will jump around and I'll curse and swear for buying a 2 AA battery powered meter.

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    \$\begingroup\$ Very good answer. Like that you used a practical example. Thank you \$\endgroup\$ – MCG Aug 8 '17 at 14:55
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    \$\begingroup\$ My multimeter uses 2 AA batteries and works just fine, yet is even one of the most accurate ones. So, if the circuit is designed to work with low voltages it works - question remains why they used the 9V in the multimeter. Probably because they have done it like that for ages and never change a running system if you don't have to. \$\endgroup\$ – Arsenal Aug 8 '17 at 16:22
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    \$\begingroup\$ The newer accurate models now uses 4 X AA. There are some models that uses just 2 X AA but did not boost to higher voltage. The downside is some function are not included. \$\endgroup\$ – Jason Han Aug 9 '17 at 1:18
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I used to have an old programmable calculator powered by a bunch of AA batteries. The annoying thing about it was that is would easily reboot if you shook it. This never happens to my multimeter running from a 9V battery.

Also, electronics which needs about 5V is usually powered by 3xAA batteries, not by a 9V one. Having a few extra volts does simplify a lot of designs, especially with older opamps which could not provide rail-to-rail output. This is less of an issue nowadays, when most parts have a 3.3V version.

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  • \$\begingroup\$ Don't blame the batteries if your battery compartment is poorly designed. I have exactly the same problem on my VX-8 radio, and it uses an encased rechargeable battery pack. Just touch the radio and it reboots. \$\endgroup\$ – richard1941 Aug 10 '17 at 17:43
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    \$\begingroup\$ @richard1941 Battery design does have an effect on battery compartment design, and AA cells rely on the same structural parts for mechanical support and electrical contact. \$\endgroup\$ – Dmitry Grigoryev Aug 11 '17 at 8:23
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    \$\begingroup\$ Good point about the 9V battery connector. The battery can easily hang loose by the connected cables. \$\endgroup\$ – kaay Aug 11 '17 at 13:46
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PP3 batteries are fine with many meter chips, for example. In many cases the total energy per dollar may not be all that important- if the battery lasts for a year or three it is good enough, and they can save a bit of money with circuitry. Another example would be a backup battery- shelf life is more important than energy density.

You will often see button cells used, which hold even less energy and can be breathtakingly expensive (or quite cheap) in applications where total energy consumed over a reasonable battery life is not that high. Where more energy is required, rechargeable Li-ion cells are more common.

I don't think that many new applications, even in consumer goods, use 9V batteries- but there will be older devices (and older designs) out there for decades to come.

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  • \$\begingroup\$ Thank you for your answer, You are correct in saying that not many new applications use them, although a small few do. This does seem to support that it may be better not to use them do you not think? You are very correct in saying if it lasts long enough, then it's good enough. I suppose it would be cheaper to use one of them with something drawing a very small amount of current, so that is a good point to make \$\endgroup\$ – MCG Aug 8 '17 at 14:11
  • \$\begingroup\$ @MCG Yes, I would suggest it is probably best not to use them in new designs. \$\endgroup\$ – Spehro Pefhany Aug 8 '17 at 15:30
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    \$\begingroup\$ One type of unit that uses them still is a guitar effects pedal. Seems like a good idea when you want a high enough voltage to supply circuitry to deal with line level and have some headroom while not wanting a noisy switcher. \$\endgroup\$ – DiBosco Aug 9 '17 at 11:55
  • \$\begingroup\$ DeBosco--- I have exactly the effects pedal you are talking about (Boss PS2, makes me sound almost as good as Rod Piazza). I always suspected that the designers were in a conspiracy with the battery manufacturers. \$\endgroup\$ – richard1941 Aug 10 '17 at 17:50
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    \$\begingroup\$ I think you mean PP3 batteries - a 9V battery measuring 49x27x18mm. The last time I saw a PP9 (9V, 81x66x52mm) outside of its retail packaging was over 20 years ago. It powered a discrete transistor AM radio. I've never seen anything that large in a meter. The PP9 has largely been replaced by 6xAA batteries (or with modern lower voltage circuits, 4xAA batteries) but the PP3 remains popular for those small applications whose voltage requirements would need more than 2 AAA cells. \$\endgroup\$ – Level River St Aug 10 '17 at 20:17
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When you try to design ultra low noise equipment, a quiet linear regulator on a 9V battery will beat almost any DC-DC converter which invariably will have some switching noise (either in the output, or emitted). That certainly was the rule of thumb in our astrophysics department (admittedly, that's an old argument that may have been overtaken by reality).

Others have pointed out the advantage (in terms of standby current) of using 9V in a system with a low power need but the requirement of long battery life / high instant power (e.g. fire alarms). But you are right - if power density is the main criterion, the old PP9 is not a terrific choice. And that's why you don't see them in many flashlights, for example.

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Considerations of battery selection may include propensity to leak, supportability of a product, maximum current demand, and recycling issues.

  • Leakage of AA cells: In electronic equipment battery leakage can be a big problem unless special design features are incorporated in the battery holder to avoid corrosion. Alas, gold contacts are expen$ive. I do not know if 9 volt batteries are more or less likely to leak, but if you can find out, that might influence your selection decision for equipment that must go unattended for years. I have some gear wrecked by leakage of Duracell AA alkalines.
  • Supportability: Perhaps at some time in the future 9 volt batteries will become as available as Kodachrome is today. In that case, any equipment may require replacement. This may be a marketing, not an engineering decision: your marketing people may like the idea that the product becomes useless in a few years and will need to be replaced.
  • Maximum Current Demand: The maximum current and power are governed by Thevenin and Norton. The battery acts as if it is a perfect voltage source in series with a resistor. The voltage and resistance are variable according to the internal condition of the battery. And the maximum power available is at a load current that drops half the battery voltage across that internal resistance. Some lab work may be necessary to decide the best battery for your application, depending on peak power demand.
  • Recycling Issues: For environmental reason$ only, of course, I like to recycle my AA Duracells by the unorthodox means of recharging them and using them again. I find that the recharged AA Duracell comes back to full voltage, but has a higher internal resistance that prevents my radio from playing at maximum volume. The recharged cells also have a shorter life. But I get many more hours of play at lower volume. With a 9 volt battery, you may have only one dead cell out of six, so recharging to restore the one may degrade the other five by overcharging. Some lab experimentation would be required to determine the best way to go. And whatever the result of your experiments, a change in the battery supplier or the supplier's manufacturing process might invalidate your results.

So, good luck. In engineering their might not be a right or wrong answer. Most decisions are a compromise between competing issues.

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    \$\begingroup\$ I like your first point about the leakage. I have had AA batteries leak in some products too in the past \$\endgroup\$ – MCG Aug 10 '17 at 20:49
  • \$\begingroup\$ 9V batteries are probably less likely to leak since the total volume of electrolyte released from a leaky cell is usually pretty small, and most 9V batteries have enough empty space inside that they could contain the leakage from one or two cells. \$\endgroup\$ – supercat Aug 11 '17 at 14:47
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A couple of points not addressed in other answers:

Your weight comparison is fair, but I think direct volume-volume comparision is unfair, because usually there is some space wasted when packing round objects. The space occupied by the cuboid containing an AA cell is about 10.5cm3. Multiply that by PI/4 to get the cylinder volume and we get close to the 8.1cm3 in your table.

With a PP3 9V battery, you are paying to have 6 tiny 1.5V cells (smaller than AAA) supplied in a convenient package. It is this "unnecessary" packaging that adds the extra weight and bulk. The applications where these batteries are used are small enough that the cost for this packaging often is not perceived to matter.

If an equipment manufacturer (who normally sells his goods "battery not included") is choosing between using 2xAA cells plus voltage converter or 1xPP3 cell, he has to consider not only the additional cost of the converter, but also the battery connector. PP3 batteries have a very neat off the shelf connector that clips on one end, while individual cells need a proper battery holder integrated into the product, to grab the battery at both ends. Given the consumer rarely considers the cost of batteries when purchasing equipment, that gives the manufacturer two reasons to go for the PP3.

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  • \$\begingroup\$ I've disassembled some Radio Shack brand 9V batteries in the 1970s and found that they contained a stack of six almost-rectangular cells, rather than the usual six long skinny ones. I don't know if any modern 9V batteries are made that way, though. \$\endgroup\$ – supercat Aug 11 '17 at 14:46
  • \$\begingroup\$ I may be a rare consumer, then. 9V batteries tend to be a deal-breaker for me when selecting a device: they're still unreasonably expensive compared to AA. For about the same price per volt (CA$20 for 40 AAs or CA$13 for 4 9Vs), I get about 4.5 times the number of Joules with AAs, if the numbers in the table are correct. \$\endgroup\$ – Mathieu K. Aug 29 '18 at 13:38
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Boost regulators (and the engineers to design them) haven't always been cheap or small. Consequently, there is plenty of equipment in the market that uses either an unregulated 9v rail or a linear regulator. Smoke alarms and multimeters are a couple of examples - redesigning would cost money and wouldn't substantially reduce the BOM cost or increase marketable functionality.

(Now, a smoke alarm that recharged a supercapacitor using the alpha emissions from the source and never needed a battery change would have a market advantage - not sure if the laws of physics allow...)

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  • \$\begingroup\$ I think a source that emits that much energy would be illegal to sell to average people. \$\endgroup\$ – Cees Timmerman Aug 10 '17 at 16:31
  • \$\begingroup\$ 12 VDC * 8 mA (piezo speaker) / 4e-6 W (nuclear pacemaker battery) / 60 / 60 = 6.67 hours to beep for 1 second? \$\endgroup\$ – Cees Timmerman Aug 10 '17 at 16:58
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    \$\begingroup\$ I think the amount of material in a nuclear pacemaker wouldn't be licensed for an uncontrolled device, which is one reason they aren't used any more: en.wikipedia.org/wiki/Betacel. The obvious (if boring) energy source for a smoke alarm would be a solar cell or mains pickup coil. \$\endgroup\$ – Rich Aug 11 '17 at 4:57
  • \$\begingroup\$ Or a thermal power generator: installed on the ceiling, it will feel a temperature difference between the inside air and the attic. I recall having a problem where the heat detectors were sounding false alarms because the attic was heating the ceiling! \$\endgroup\$ – JDługosz Aug 12 '17 at 11:21
  • \$\begingroup\$ I like your idea for a power source, but it should only be allowed for above average people like us. \$\endgroup\$ – richard1941 Aug 12 '17 at 14:49

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