When will the AA battery voltage drop?

Question

I have a device which runs on two AA Duracell batteries. The batteries (brand new) together produce 3.2 volts (1.62 volts individually), which is estimated to run for at least 2 years. The device is always running and consumes 300 micro amperes, which isn't a lot.

Could someone tell me how long can the batteries will last on 3.2 volts (1.62 volts individually) before dropping to 3.0 volts (1.5 volts) in years/months/weeks/days/hours/minutes? So basically, how long until the batteries voltage begins to drop?

Answer 1

I don't have good news for you.

Below is the discharge rate for Durcell batteries done by some company. Look at the DC label; this is a Duracell coppertop battery. The complete test can be found here: link.

There was no µA test (it would take too long), but I guess the voltage will drop below 1.5 V after about 1-2% capacity discharge. There will also be battery self-discharge (very slow). My guess estimated time for 0.3 mA (300 µA) before dropping to 1.5 V will be somewhere between 50-150 hours. You can test this; it's not that long.

I would suggest to use different kind of batteries. Alkaline and a 3.0 V requirement is just the wrong battery for this application.

You can also consider different battery chemistry. If you use some lithium AA battery, for example Energizer L91 - they have much more energy "available" before voltage drop below 1.5 V, but be careful - they have also higher initial voltage (about 1.7 V) and they are expensive (in my country they cost 2-4x more than alkaline Energizer or Duracell).

Image from Energizer Ultimate Lithium L91 datasheet: link

However if you want to pull 0.3 mA (300 µA) from battery for two years (over 17000 hours) - you need more than 5 Ah (5000 mAh) before the voltage drops below 1.5 V. This is probably too much for any AA battery available on the market.

There are also nickel-zinc AA batteries, they have nominal voltage 1.65 V, but they have less capacity (50% less than Energizer L91).

Another idea might be three batteries with a low-dropout linear regulator. Three AA batteries in series will provide more than 3.0 V until they are completely empty, but a linear regulator may be necessary for some devices - three new alkalines may have 4.95 V initially.

Answer 2

The batteries (brand new) together is producing 3.2 volts (1.62 volts individually), which is estimated to run for at least 2 years. The device is always running and consumes 300 micro amp which isn't a lot.

Your math is off: A high quality AA cell has about 3000mAh, which gives you about 10,000 hours run time, or a bit more than one year.

But the discharge voltage will drop quickly lower than 3,0V - end voltage is about 0.9V/cell. I would expect the drop from 3.2 to 3.0V to occur within one or at most two weeks.

Answer 3

Using a reulator allow power or current to be stabilised as the battery voltage drops.
Using any regulator you need to define the load that the battery sees unless your load is a resistor. To be able to calculate battery life you must follow the manufacturer's datasheet

Example-only graphs are from here - Duracell AA Alkaline-Manganese Dioxide Battery.

If we use a linear regulator we are following the constant current.
If we use switching regulator we are following constant power.

To be possible to follow these load-lines we must use a test load in the output of the regulator and read the voltage during test .
A single measurement give you different results each time so needed put at least 20 measurements in a buffer and create a moving average (shift so make last out new measurement in and divide by 20) the time between measurement (if your design is for years) can be between 30 mints to one hour. In this way measurement get good stable results

If we are using only DC current in the circuit Average and RMS are the same.
If you are measuring voltage with Microcontroller you must use moving average to get good reading otherwise you have noise.

For the chart here is the polynomial that I have developed of Voltage wrt current and time.

Voltage(t) =- (60.10778E-2 • CRNT• t)⁵ + (84.63365E-2 • CRNT • t)⁴ - (10.91466E-1 • CRNT• t)³ + (12.58045E-1 • CRNT• t)² - (10.32172E-1 • CRNT• t) + 1.600119 • n

CRNT-current in mA
t-time in hours
n-number of batteries in series.

Generally an electrical circuit contains different currents, if the design relies on AVG So can be a big mistake between reality and measurement.