20
\$\begingroup\$

When changing the batteries of kids toys, remote controls, etc., I test the batteries and have noticed that they drain differently. One will be dead and the other is still in the green.

Why does this happen? Also, at this extreme, why not just have a single battery?

They are typically AA and can be the ones with the product or rechargeables.

Also, why does the device stop and not start using the good battery?

\$\endgroup\$
  • 15
    \$\begingroup\$ Manufacturing tolerances \$\endgroup\$ – Chu May 19 '17 at 12:37
  • \$\begingroup\$ Varying conditions of heat, etc during shipping and the lifetime of the battery. \$\endgroup\$ – user56384 May 19 '17 at 14:44
  • 6
    \$\begingroup\$ Note: A "battery" is a collection of multiple "cells". AA "batteries" are really just one cell, and multiple AA cells together form a battery. 9V batteries really are a battery ... if you open one up, you'll see 6 individual 1.5V cells in series. \$\endgroup\$ – Steve May 19 '17 at 17:59
  • \$\begingroup\$ Cost. Adding a battery management system is expensive. \$\endgroup\$ – Mast May 20 '17 at 10:37
23
\$\begingroup\$

There will always be some variation in capacity between batteries. In toy use the current is typically drawn at a high level compared to many other applications, so the internal resistance of the batteries is very important.

In that service the voltage under load drops quite rapidly as the end-of-life approaches, so with minor differences in capacity the cells can have quite different voltages.

However, the ones that appear to be good are likely very close to their end of life as well, unless the cells were not all replaced at the same time with equally fresh cells. The mismatch in quantity between packaging and use (cells packaged in 4's and used in 3's for example) may contribute to a mismatch in capacity.

enter image description here

In the above graphs for a Panasonic AA cell you can see that the voltage drops over a period of a few hours from 1V to 0.8V. A radio lasts for 130 hours or so and the decline is much gentler. Once the voltage drops below the 800mV end of life, it will rapidly drop to a much lower voltage and may even reverse (under load, or even after the load is removed) because of the large reverse current supplied by the other cells in toy-type service.

Battery testers typically work by applying a load to the battery and measuring the voltage. The load applied is necessarily a compromise between the requirements of disparate applications.

If your battery tester draws a relatively small current it may still think the cells that are almost depleted are fine (and they are fine, in a radio!) but it can't mistake the cells that have been reverse charged to death. This is a reason why some battery testers have a choice of currents for testing- at a higher current the apparently 'green' cells would likely be well into the amber range.

In a typical toy the batteries are simply connected in series and the toy requires a certain total voltage at a relatively high current to operate. The series chain of batteries is only as strong as its weakest link, so if one battery is very dead the motors etc. won't get enough voltage to operate.

\$\endgroup\$
  • \$\begingroup\$ I have frequently observed that switching the batteries in, for example, a remote control will extend the operation of the batteries. It seems to me that the explanation for this observation and the explanation for the OP's observation should be similar. \$\endgroup\$ – Jack Aidley May 21 '17 at 14:01
20
\$\begingroup\$

Toys are often supplied with the cheapest battery cells which may be lower capacity, but also have the widest tolerance. This tolerance and the "weakest cell's fails 1st" is the general rule.

To understand this, model each battery as a charged capacitor, say 1000 farads +/- 20 %. Then put them in series with the same initial voltage and load them with, say 100 ohms, and simulate or calculate the end of charge using worst case tolerances, say three cells +20, 0, and -20 %.

Then you will understand. Lithium polymer and car batteries on the other hand start at 0.1% then age towards 10% before single cell death at more than -90% tolerance of the initial charge.

ESR rises and C drops in values towards end of charge rapidly.

If you analyze this, then you will understand.

I pulled the C value out of thin air for simple understanding, but it depends on the mAh hour rating. Ic = C * dV/dt where dV is usually 1.6 to 1 V = 0.6 V depending on chemistry, but the mAh rating is I* dt = C * dV, so 50 mAh = 50 mA * 3.6 ks when rated for some period like 1 C = 20 h, but every and all batteries have some ESR value too, where a wide tolerance applies and ESR generally drops with RISING mAh rating which changes with both low SoC and long age.

The rule of thumb is that ESR is sometimes related to Ah, BUT not a ways.

In the end the ANSWER to your question is the battery cell with the lowest C [F] value DECAYS FASTEST. When in series, if sustained it will also become reverse charged. next question please...

OK here is a comparison of AA battery capacities.

http://www.batteryshowdown.com/static/images/mah_large_200mA.png

Note above the "DURACELL standard" is 2000 mAh or 2Ah or 7200A-seconds

  • EE common wisdom:

    • \$f_{-3dB}=\dfrac{0.35}{t_R}\$ rise time 90% to 10% or 10 to 90%
    • RC=T or L/R=T is defined as asymptotic time,
      • for T = RC decay from 1 to 1/e * 100% ~37% or 0 to ~63%
  • but battery marketing info sometimes uses Ah capacity using 100% SoC to 0%, so there are some differences in measurement time standards.

However in both cases the energy stored is E= ½CV² but in the case of a battery I recommend E=½C(Vi²-Vf²) as Vbat from initial, Vi at 100% Soc to final Vf but rather 10% SoC to avoid deep discharge rapid aging.

Each voltage is very dependent on cell temperature; but as I recall;

if you put a slight load to remove short term memory, the measure Vbat (100%)

  • Lipo ΔV=0.7 from 3.6 to 3.0 V
  • Lead acid = 12.5 to 11.5 V
  • Alkaline = 1.5V to 1V

enter image description here

\$\endgroup\$
  • 1
    \$\begingroup\$ credits to Peter for fine edits, \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 19 '17 at 19:57
  • \$\begingroup\$ You can't model a battery as a capacitor. They have wildly different discharge characteristics. Try making an LC tank circuit but use a battery instead of a capacitor ... \$\endgroup\$ – Jamie Hanrahan May 21 '17 at 18:42
  • \$\begingroup\$ sorry @JamieHanrahan but the correct model for a battery is just like a supercap or "double electric-layer" cap with two RC values, Rp and dynamic ESR*C which rises in ESR and drops in C with age. If you look at the supercap graphs from Maxwell you will see the same memory effects although much less than a lead acid battery which drops from 14.2 to ~12.8V this time constant is much shorter than the main storage but in LiPo and supercaps that ratio of T's is much closer. admittedly there are more complexities with battery losses at high discharge currents, but not so in 1C range \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 21 '17 at 20:52
  • \$\begingroup\$ Yeah? Try making an LC resonant circuit with one and get back to us. \$\endgroup\$ – Jamie Hanrahan May 21 '17 at 21:55
  • \$\begingroup\$ @JamieHanrahan Q must be >>1 to resonate with L have you tried series or parallel? for 1kF Supercap, if so then a small LiPo cell might not work at 100kF because of the ESR * C product ( related to 1/Q) is too high. It is not my assumption that is wrong but yours. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 22 '17 at 0:06
15
\$\begingroup\$

In addition to the notes on battery manufacturing tolerances, I have seen (at least one) device that would split off battery electrical connections to different parts of the circuit.

In my case, it was a cheap alarm clock with 3 AAA batteries, 2 were in series to run the electronics, and then a third was in series with the first two to run the LED/Alarm.

schematic

simulate this circuit – Schematic created using CircuitLab

I don't think this is common, but it is another possible explanation.

\$\endgroup\$
  • 3
    \$\begingroup\$ I've seen this in TV remotes. The higher voltage is used to drive the LEDs while the lower voltage powers the electronics as well. \$\endgroup\$ – David Schwartz May 19 '17 at 20:12
  • \$\begingroup\$ IR LEDs only need 1.2V but can be pulsed to 1.8V typically \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 21 '17 at 21:01
3
\$\begingroup\$

re: Why do batteries in consumer electronics get used unevenly?

  1. Poor cell quality &/or cells came from different batches. This can result in different dis/charge times. Devices (toys) usually require a certain minimum V & A to operate properly. Once the weaker cell/s fail to provide their share of the necessary V & A to operate the device then the device will begin to under perform or stop all together. Depending on the smarts of your charger & tester, as the cells age, that difference can become magnified to the point that when you measure the cells, you find one cell looks 'green' on your tester while the other cell looks dead or near dead.

  2. Users often mix & match cells in their chargers & in their devices -- often times mixing older rechargeable cells with newer cells -- and sometimes even mixing rechargeable cells with primary (disposable) cells. As rechargeable cells age, they tend to lose charge capacity & they tend to have different dis/charge profiles (compared to new cells). Depending on the charger used, that could mean a user uses multiple cells with different charges in her/his device -- which of course will result in the device only operating as long as the weakest cell is able to provide sufficient power (which is usually a much shorter time than the newer, higher capacity cells would provide energy to a device).

re: ...why not just have a single battery?"

A battery is a collection of cells. A larger cell could of course be manufactured for your specific application; however, that cell would likely be less useful for building batteries for other applications when compared to current cell designs.

re: "why does the device stop and not start using the good battery?"

Sometimes (depending the cell connectivity) energy from a good cell can keep a device operating as the other weaker cell/s Peter's out. Other times that is not possible because the power (VA) output capacity of the battery has decreased below the minimum threshold that is required to operate the device.

\$\endgroup\$
  • \$\begingroup\$ I assume this is part of why manufacturers recommend not mixing batteries with different compositions as well? \$\endgroup\$ – JAB May 20 '17 at 19:34
  • \$\begingroup\$ That and mixing cells with greater capacity & lower capacity can create a reverse charged condition in the lower capacity cells after the lower capacity cells are depleted. \$\endgroup\$ – DIYser May 20 '17 at 23:18
2
\$\begingroup\$

It was suggested a battery cannot possibly be modeled as a capacitor because you cannot resonate with a choke (aka reactor).

This is false and only proves the Q is too low to resonate and is not a requirement to be a capacitor (with a chemical offset junction voltage like a diode that must be correctly polarized)

We know for a series tank circuit, the ratio of reactance to resistance = Q must be >1 to resonate and >>1 for a long time, where Q=0.7 is defined as critical damping.

I can safely say, batteries are over-damped ultra (poor) low Q capacitors.

proof

Using a single C model for a battery ( 1st order approximation)

Q=|Xc(ω)|/ESR = \$\dfrac{1}{ESR*2\pi fC}~~~Q=\dfrac{0.16T_r}{T} \text{ for T=ESR*C} ~~ \$

\$~ T_o=\frac{1}{f_o} \text{ res. freq. e.g. ~LiPo: T=100kF*5mΩ=500s {min.theoretical drain time}}\$

  • but NEVER short circuit an un-fused LiPo, as the heat rise will cause it to blow up.

For Q>>1 or 0.16Tr/500 >1 so cycle period Tr>>3125 s or >> 52 minutes

Then choose choke similar or better Q>>1 L/R>>52 minutes e.g. e.g.>>100H/30mΩ

That's why its not possible to make an LC oscillator from a battery.

Some transformers like below have a Q near but < 1 with long time constants. enter image description here

BTW for fun I applied an 18650 LiPo cell across the primary windings of a transformer like above in a factory and as expected V=LdI/dt was an OVERdamped slow ramp until saturation. What a Tank!

The reality is Batteries are huge SuperCaps with tens of thousands of Farads but very long time constants and thus low Q.

But to answer your question.

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
0
\$\begingroup\$

To answer your last question:

If a device runs on 2 AA batteries, these are usually connected in series, meaning the voltages add up to become the operating voltage. In this case 2 * 1.5V = 3V. The device can operate only until a certain threshold, which differs from device to device. Let's say the device is designed for 3V and can operate with voltages greater then 2V. In this case one of the batteries can still have nearly full charge (say 1.4V), but the other is completely drained at 0.2V. This equals a combined series voltage of only 1.6V which will not power the device any more.

\$\endgroup\$
  • 2
    \$\begingroup\$ So the question was something like: "why not one big battery?" Answer: because there are not many cell chemistries that can provide greater than 3V. Supercapacitors could do this. \$\endgroup\$ – user56384 May 19 '17 at 14:48
  • 2
    \$\begingroup\$ This answer feels misleading; it sounds like one battery will drain almost entirely before the next even starts draining. I know that may not have been your intent but it came across that way. Perhaps a better example would be putting in two new batteries from different packs (even if they're the same mfgr). One might have a higher starting ESR so its effective output will start out lower even though it's still new. Due to the higher starting ESR (even if only by 0.05Ω), this battery will degrade in apparent capacity sooner than the other one as its ESR continues to climb and its output drops. \$\endgroup\$ – Doktor J May 19 '17 at 19:32
  • \$\begingroup\$ @DoktorJ One will be dead and the other is still in the green. That's exactly what the TO described. I didn't say this is what will happen in most cases, but it can. Also, I was only referring to the last part of the question why the device stops working when one of the batteries is still good. The ESR aspect had already been thoroughly explained in other answers. \$\endgroup\$ – chindocaine May 19 '17 at 22:04
0
\$\begingroup\$

Every battery have some different in charge or discharge cycle. Even the most supper and costly battery have some different regardless of same batch or different batch of same model and manufactured by same company.

You might find this phenomena on not only in cheaper toy battery which normally come with pairs single pair but various source. My personal experience shows industrial grade battery use on Telecommunication​ network also digrade differently within same battery bank.

As I mentioned before, even with same model and same usage pattern every battery behaves different in atomic level when charge or discharge cycle. There is also influence of sarrounding environment of battery (middle battery on there battery bank got higher temperature exposure) creates uneven parformance and life expectancy.

So it's not a random case but most of time they show different parformance even acting together...

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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