6
\$\begingroup\$

So it's that time when I have to replace my mouse batteries (2 x AA, 1.5 V each). I used a voltmeter to check how much voltage there is left. Turns out it measures around 1 V each for the 2 batteries.

I assumed that when a battery runs out, there's no voltage left at all. But in this case, there's still 1 V left.

Can anyone explain why this is the case?

enter image description here

\$\endgroup\$
5
  • 1
    \$\begingroup\$ It will go to ~0V if you try to apply a load (DMMs are high impedance). \$\endgroup\$ Sep 27, 2020 at 14:32
  • 7
    \$\begingroup\$ As the battery is depleted, it's voltage will go down. At around 1 to 0.8V most of the capacity is used up. (it's not linear, for the last bit of capacity the voltage will drop like a rock). The electronics in your mouse, as all electronics, need some minimum voltage to operate so it's impossible to keep running all the way down to 0V. \$\endgroup\$ Sep 27, 2020 at 14:38
  • \$\begingroup\$ Your image is 90 degrees rotated. For most devices with two batteries, one of your batteries is inserted the wrong way. You got several answers but didn't comment on, or approved of, none, yet. \$\endgroup\$
    – Roland
    Sep 28, 2020 at 13:47
  • 1
    \$\begingroup\$ Let's rephrase your question. The mouse stopped working. You measured around 1 V on each battery. You did not care to share the actual voltage readings. You did not try new batteries. 1.5 V is 'around' 1V, and perhaps your mouse is defective. Your actual question may be why does this mouse stop working in the middle of this important work. Your problem may be why am I still working with this outdated mouse model. \$\endgroup\$
    – Roland
    Sep 28, 2020 at 14:03
  • \$\begingroup\$ I've seen a battery read -0.6V. It had been installed in series with other batteries of different brand, but luckily had not started leaking yet :) \$\endgroup\$
    – jpa
    Sep 28, 2020 at 16:40

6 Answers 6

8
\$\begingroup\$

That looks like an optical wireless mouse. If so it will have an LED to illuminate the surface you're using it on so that the camera (yes there is one - it's only got, maybe, 64 pixels or so) can see the irregularities in the surface. LEDs have a certain minimum operating voltage and this is about 1.8 V to 2.0 V for red, the most common used in these mice.

enter image description here

Figure 1. LEDs require a certain "forward voltage" before they will pass enough current to turn on. here you can see that to get 5 mA or so through a red LED that you'd need > 1.6 V, not including the voltage drop across whatever switches the LED on when you wake the mouse up. Image source: LEDnique.

In addition the rest of the electronics, including the radio transmitter, will require a couple of volts to work.

Engines using petrol from a tank work at full power until the fuel runs out. Batteries are chemical cells and do not perform in the same way. The voltage remains reasonably constant as the chemistry is used up but at some point the voltage starts to fall off quite rapidly.

enter image description here

Figure 2. Various AA cells output voltage with respect to time for a 500 mA discharge. Your mouse is probably using < 15 mA. Image source: PowerStream.

Note that a 1 V reading on a 1.5 V battery does not mean that 2/3 of its capacity is still remaining. It's nearly flat.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ Is it possible that the electronics don't draw any significant current because the LED threshold is not reached (en.wikipedia.org/wiki/LED_circuit#Power_source_considerations)? In that case even a near-dead battery should show significant voltage. You could make clearer that your (nice) output voltage diagram applies to batteries under load. If you drain them, then detach the load, then measure, the curve would be much flatter (depending on the quality of your voltmeter).. \$\endgroup\$ Sep 28, 2020 at 10:11
3
\$\begingroup\$

Does a person drop dead when they are no longer able to perform useful work? No. A battery is the same way. A battery that has dropped to 0V is like a person that has dropped dead...a recharge of food, rest, or electricity isn't going to get them back up again.

As for why this is the case. It's more of a chemistry or physics question, but basically batteries are not gas tanks. They're operate on a chemical reaction and these chemicals aren't completely depleted by the time the reaction is no longer strong enough to produce useful work.

And for rechargeable batteries, the materials have to be maintained in a particular condition so the reaction can be reversed when you recharge the battery. If the voltage falls to 0V it means you have damaged the chemicals in the battery and the reaction can no longer be reversed.

\$\endgroup\$
3
\$\begingroup\$

Two reasons:

  1. Electronics need some voltage and current to work. The two batteries are wired in series (2*1.5V≈3V when full) and the mouse probably runs on 1.8V internally. It has a voltage regulator to lower the battery voltage to that level. Once the battery voltage drops below 1.8V the circuitry won’t work reliably anymore since the voltage regulator can only lower the voltage, not raise it. The voltage is simply too low to light up LEDs (even red ones) and a 1.8V microcontroller would go into reset due to its brown out detection. However, it probably shuts down before that due to reason #2:
  2. Rechargeable batteries musn’t be deep discharged. The mouse can’t know if you are using rechargeable or non-rechargeable batteries so it stops discharging at a safe cut-off voltage. Keep in mind that batteries have non-linear voltage curves, there is very little energy remaining once you reach <1V.
\$\endgroup\$
3
\$\begingroup\$

The nature of the electrochemical reactions in a typical Alkaline cell is such that when new Voc is about 1.65V. Voltage falls with discharge until almost all available energy is exhausted by Vbat=1.0V. Some devices will operate on 0.9V or even 0.8V and you can obtain some energy at low current at these voltages. Below about 0.8V the current available is in the microamp range and very few devices require that little current.


It has been suggested that Vbattery remains at about 1.5V throughout and that the lower voltages seen are a result of rising battery resistance. This is not the case and the following simple experiment demonstrates that typical Alkaline battery terminal voltage falls with discharge. You could easily try this experiment yourself to satisfy yourself of the validity.

I took an AA Alkaline cell that measured 1.218V on a 10 megohm input resistance DMM.
I measured Voc with the same meter plus 1 megohm and 6.6 megohms in series.
As expected Vmeasured dropped to about 1.102 V with the series 1 megohm and 0.733V with the series 6.6 megohm. These values are consistent with expectations for a 10 meg meter.

THEN I added 1 kohm parallel resistance across the cell.
In each case the Vmeasured fell by a few millivolts due to the load from the 1k load.
ie the meter load of around 1.2/10 meg ~= 0.1 microamps is about 10,000 times smaller than that 1.2 mA of the 1k load and will have utterly negligible effect on the voltage measured. Try it.

\$\endgroup\$
3
  • \$\begingroup\$ If one has two or more alkaline batteries in series, it's possible for one of them to be over-discharged to the point that it can actually produce voltage in the reverse direction at a significant fraction of a milliamp. \$\endgroup\$
    – supercat
    Sep 28, 2020 at 15:28
  • \$\begingroup\$ @supercat Yes. || For interest: I was mainly interested in addressing the claim that battery voltage stays at 1.5V and lower terminal voltage is caused by the meter impedance and battery impedance acting as a divider. That's clearly not true but a small experiment proves it conclusively. \$\endgroup\$
    – Russell McMahon
    Sep 29, 2020 at 12:23
  • \$\begingroup\$ I mentioned the reverse charge because while it may be difficult to show that a discharged battery isn't behaving as an ideal voltage source in series with an impedance so big that an load of even 10^15 ohms would diminish it to almost nothing, a battery that can drive a significant reverse voltage through a 10K resistor would thoroughly contradict that model. \$\endgroup\$
    – supercat
    Sep 29, 2020 at 16:02
2
\$\begingroup\$

No.

The circuitry has detected insufficient voltage and will not work. Usually there will be an early warning - flashing led or similar, good to take notice when it happens.

\$\endgroup\$
-3
\$\begingroup\$

Battery voltage is always 1.5 volt, it's the internal resistance that goes up in time due to corrosion of the electrode, because the internal resistance goes up the battery can source less current. If you load your 1V battery with 1 ohm resistor the voltage will drop to near 0.

\$\endgroup\$
16
  • 2
    \$\begingroup\$ Try monitoring the voltage of an almost flat cell with a DC multimeter with10 M\$\Omega\$ input impedance and see if your theory holds up. (It doesn't.) \$\endgroup\$
    – Transistor
    Sep 27, 2020 at 20:28
  • 3
    \$\begingroup\$ You are mistaken, Moty. You can confirm this by measuring the open circuit voltage and measuring the terminal voltage with a known load and calculating the internal resistance. What makes you think I'm at school? Are you? \$\endgroup\$
    – Transistor
    Sep 27, 2020 at 22:19
  • 2
    \$\begingroup\$ @Moty Your reasoning seems tautological...although I can see how it might be the case since things like galvanic voltages are set by the chemistry and unchanging. If you are going to claim the reaction is always 1.5V internally, then please provide some references such as one involving measuring the dead voltage with a 10M input impedance meter and a 100MOhm input impedance meter. Also, I assume you are claiming that the corrosion mechanism is for all batteries? Or just carbon-zinc? I think you would expect a linear increase with this mechanism but that's not what we see. \$\endgroup\$
    – DKNguyen
    Sep 28, 2020 at 5:20
  • 3
    \$\begingroup\$ @Moty The following demonstrates that typical Alkaline battery terminal voltage falls with discharge state and the effect of battery series resistance is utterly minimal. || . You could easily try this experiment yourself to satisfy yourself of the validlity. || IO took an AA Alkaline cell that measured 1.218V on a 10 megohm input resistance DMM. || I measured Voc with the same meter plus 1megohm and 6.6 megohms in series. As expected Vmeasured dropped to about 1.102 V with the series 1 megohm and 0.733V with the series 6.6 megohm. ... \$\endgroup\$
    – Russell McMahon
    Sep 28, 2020 at 14:09
  • 2
    \$\begingroup\$ ... These values are consistent with expectations for a 10 meg meter and your explanation. THEN I added 1kohm parallel resistance across the cell. In each case the Vmeasured fell by a few millivolts due to the load from the 1k load. ie the meter load of around 1.2/10 meg ~= 0.1 microamps is about 10,000 times smaller than that 1.2 mA of the 1k load and will have utterly negligible effect on the voltage measured. Try it. \$\endgroup\$
    – Russell McMahon
    Sep 28, 2020 at 14:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.