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I am designing a circuit that will be powered by 3 AAA batteries.

I know that the nominal voltage of an AAA battery is 1.5 V, but I also read on the internet that the open-circuit voltage could be higher than 1.5 V (e.g. 1.65 V.)

In my case, 3 batteries are connected directly in series to a microcontroller which has an operating voltage range from 2.0 - 4.5 V. The supply voltage should be 2.7 - 4.5 V (0.9 × 3 - 1.5 × 3)

Question:

  1. Should I worry that the voltage could be potentially higher than 4.5 V and damage the microcontroller?
  2. When the microcontroller sleeps, the circuit will consume 10 μA current from the batteries. Will this drop the voltage to 4.5 V (as it is not open-circuit)?
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    \$\begingroup\$ Which MCU or MCU board it is? Is the 4.5V nominal maximum or absolute maximum? \$\endgroup\$
    – Justme
    Commented Feb 21, 2023 at 6:56
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    \$\begingroup\$ Can you use two AAAs instead of three? \$\endgroup\$ Commented Feb 21, 2023 at 13:12
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    \$\begingroup\$ Putting an additional battery in series isn't going to make your battery life any longer unless your circuit can translate that higher voltage into lower current draw rather than simply dissipating it as heat. You'd likely be far better served by going to 4 batteries in a 2x2 configuration. \$\endgroup\$
    – Gene
    Commented Feb 22, 2023 at 4:36
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    \$\begingroup\$ @Jasen: NiMH batteries are typically only about 1.2V per cell under load for most of their discharge curve, so two of those will only get the supply voltage up to 2.4V for most of the battery life. Fresh out of the charger they're at 1.45V or so, but for example the 1/10th C discharge curve on en.globtek.com/nimh-battery-safety-notes shows it coming down quickly to 1.2V where it stays flat for a good while. That's probably fine if the device can use the full battery voltage so it doesn't get down to the 2.0V minimum until the batteries are seriously flat, but OP is aiming for 2.7V \$\endgroup\$ Commented Feb 22, 2023 at 8:16
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    \$\begingroup\$ The thing to be cautious about here is that AAA is entirely a packaging outline, and there's no reason why some vastly different chemistry could be used internally... or even a stack of smaller cells. From the POV of the question, it would probably be worth explicitly specifying "nominal 1.5V" or similar; from the POV of answers I think it has to be emphasised that a regulator or even buck/boost converter should always be used. \$\endgroup\$ Commented Feb 22, 2023 at 9:26

7 Answers 7

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The 1.8V cell in the paper you linked is a lithium primary cell. Most consumers would consider that a risk-free premium upgrade to an alkaline cell and would insert it without a thought.

Measuring some fresh random Kirkland cells, I'm getting almost 1.7V per cell (1.693 for example).

An SOT23 CMOS linear LDO regulator will likely deal with this problem at reasonable cost (1 cent-ish) and with pretty low Iq considering the AAA battery self-discharge current.

They don't generally offer reverse battery protection, but then neither does a direct connection.

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  • \$\begingroup\$ That's what I am worrying about, with these type batteries, the voltage will absolutely excess the maxmimum value. \$\endgroup\$
    – John
    Commented Feb 21, 2023 at 8:51
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    \$\begingroup\$ A CMOS regulator is possibly a good solution, even if it spends much of its life not regulating at all. \$\endgroup\$ Commented Feb 21, 2023 at 8:54
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A fresh alkaline battery with no load can be about 1.5 to 1.7 V, depending on construction and the purity of the chemicals used, and can have an internal resistance as low as 0.1 Ω.

At that internal resistance you get a voltage drop of 0.1 Ω × 10 μA = 1 μV when the MCU sleeps; not enough to bring the battery's voltage down to the 1.5 V you need (when using 3 batteries in series) to stay within spec of your MCU.

I don't think your MCU will immediately go up in flames, but you will be very close to the abs. max. supply voltage of the MCU, which could well shorten its life span.

Just an observation: a single Li-ion battery's useable voltage range seems to fit your supply voltage specification quite well.

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You may get as high as 5.55-ish V with 3 fresh cells if they are lithium primary.

I've done battery capacity tests on dozens of AAA and AA brands; the results are here. Generally, you'd be safe with alkaline or zinc-carbon cells. You won't be safe with Li-FeS₂ cells, even though those are somewhat rare.

There are tables in that website, and you can sort by U_start, where you'll see the worst it gets is for the Varta Lithium Professional AA. In my tests, they've started at 7.03V when you have 4 AAs in series, and that would translate to 5.27V for 3 cells in series. But in the comments you can see this is after a short 30-s "warmup" discharge, and the voltage before that was even higher.

The tables for AAA don't sport such a high U_start maximum, the most is again a lithium primary at 1.63V/cell, but I haven't gotten hold of Varta Lithium Professional AAAs. If I did, I would imagine that AAAs of the same brand would sport the same chemistry and same potential, so a 5.25V+ would be likely, too.

You need to protect your MCU in some way. I'd use a cheap LDO, as Spehro suggests.

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  • \$\begingroup\$ Thanks, your test data is helpful \$\endgroup\$
    – John
    Commented Feb 24, 2023 at 8:54
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The theoretical maximum, derived from the standard potentials of Zn and MnO2 electrodes, is like 2.0 V.

While the MnO2 electrode is far from ideal, voltages like 1.9 V are possible in practice and 1.7-1.8 V is frequently seen in fresh alkaline or salt cells.

If you want to avoid using a voltage regulator at all costs, you are better off using two AAA cells instead of three and go down to 2 V if the rest of the circuit is OK with 2 V.

1 volt per cell for small loads is as much discharged as practical. Maybe less than 1% of the cell energy is left below 1 V.


p.s. while somewhat offtopic, Li-Ion cells with maximum voltage of 4.2V or 4.35V are also made in AAA size.

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You are correct, a fresh battery could have about 1.65 V. Just take a battery and measure it to confirm yourself.

Using 3 batteries will exceed the nominal maximum voltage range and will be close to the absolute maximum range.

  1. It may not burn immediately, but operation of the chip is not defined or guaranteed outside the nominal limits. It may work but it may degrade faster and have shorter operating life.

  2. 10 μA is very little load. Rough estimation is that it will not be enough to bring a set of fresh batteries down to 4.5 V.

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You should obviously not drive a MCU directly from raw batteries (unless the MCU explicitly has a built-in regulator, some do but it's hardly standard). Most modern MCUs have built-in low-voltage detect and it will trip somewhere halfways through your discharge cycle. Older MCUs will require such a circuit externally - otherwise they will go completely bananas when dropping out of the specified voltage range.

Also, having a variable logic level voltage instead of 3.3V / 5V is just madness.

So you should have a boost regulator in between the batteries and the MCU. Therefore you are looking at the wrong circuit, you need to check what your boost regulator can handle.

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  • \$\begingroup\$ I know what you mean. It's a low price product, so we want the cost to be lower. \$\endgroup\$
    – John
    Commented Feb 21, 2023 at 7:33
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    \$\begingroup\$ @John So use a MCU with built-in boost regulator. I know of several STM32 that has this feature, and very likely many other vendors do as well. \$\endgroup\$
    – Lundin
    Commented Feb 21, 2023 at 7:37
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    \$\begingroup\$ Many products drive a MCU straight from the batteries and work just fine. There is no rule that a regulator must be used. It will work as long as the range of voltages match. \$\endgroup\$
    – Justme
    Commented Feb 21, 2023 at 9:40
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    \$\begingroup\$ @Justme Yeah if you want a MCU which just sits there as a dead piece of silicon, not communicating with any other IC, not reading analog signals, not using serial buses etc etc. All of which require stable voltage levels. \$\endgroup\$
    – Lundin
    Commented Feb 21, 2023 at 9:43
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    \$\begingroup\$ @Lundin If there's no analog involved (other than possibly ratiometric measurements), and other circuitry is powered at the same voltage, it absolutely can work. It's not recommended, but it will do the job. \$\endgroup\$
    – Hearth
    Commented Feb 21, 2023 at 16:43
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Idea: Put a bog-standard silicon diode in line with the batteries. The 0.6V drop across the diode makes your maximum voltage small enough and you also get free reverse-battery protection.

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    \$\begingroup\$ arne - Hi, Re: "The 0.6V drop across the diode" The forward voltage drop across a standard silicon diode isn't constant at 0.6 V but varies with forward current & temperature. Previous Q&A on this include: "What are the disadvantages of using a diode to drop Vcc by a fixed amount?" and "1N4148 not giving correct voltage drop". So depending on the ranges of forward current & temperature, things are more complicated than just adding a diode assuming a (constant) 0.6 V drop. Perhaps it's worth mentioning that phenomenon, if you want to suggest using a diode? Thanks. \$\endgroup\$
    – SamGibson
    Commented Feb 21, 2023 at 17:56
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    \$\begingroup\$ The OP specified a sleep current of 10 micro-amps. The diode forward voltage will be much smaller than 0.6 to 0.7V in that case. \$\endgroup\$ Commented Feb 23, 2023 at 10:42

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