3
\$\begingroup\$

I regularly use rechargeable 1.2v AA NiMH batteries, and I'd like to be able to use my multimeter to test both how much charge they have when I've finished charging them, and how much charge they have left at any one time. However, being an electronics newbie and doing some research, I'm left confused as to just how directly voltage is correlated with a battery's level of charge, due to there being lots of conflicting voices surrounding the topic.

A highly-rated EE answer here claims that voltage is directly related to a battery's level of charge, stating:

NiMH cells start at about 1.5 V right when fully charged, drop to about 1.2 V most of their discharge life, and are pretty much empty at 900 mV

Then this answer here claims the exact opposite:

It is not possible to measure or guess the capacity of a battery with a single set of instantaneous measurements, like voltage, current, and temperature.

At best you can tell how much current is going into or out of the battery a what voltage. However, there is no way to infer capacity from that. If you can control the load, you can get some idea of the internal resistance, but even that would take at least two measurements separated in time and therefore can not be done instantly. And, battery voltage and internal resistance does not tell you capacity.

The FAQ at GreenBatteries.com also claims voltage is inaccurate:

...a NiMH (or NiCd) battery stays at about 1.2 volts until it is nearly completely discharged. This makes it almost impossible to know the amount of capacity left based on its voltage alone

Elsewhere on the internet, opinion is divided at best.

So, I'm here for a final, definitive answer on the topic - are voltage and similar measurements accurate for assessing a battery's level of charge?

If they aren't, then what is?

N.B. In order to make this a definitive question for the topic, please ensure that any and all answers are well-sourced.

\$\endgroup\$
  • 1
    \$\begingroup\$ I used some Rayovac Ni-MH AAA cells around 2004 in a small flashlight. The incandescent bulb went from normal brightness or possibly a tiny bit dim to fully dead in less than 10 seconds. That's how fast the end of charge drop off was. With so many variables such as the brand of the cell, the age, the current cycle life remaining, reverse charge damage, temperature, and increased impedance from electrolyte loss (over charge damage), determining the state of charge from Voltage without knowing these other variables makes it complicated. \$\endgroup\$ – Alex Cannon Mar 26 at 15:09
6
\$\begingroup\$

The answer is, of course, neither extreme, but somewhere between the two.

You can tell the state of charge of a cell from voltage

If a cell measures 1.5v off load, it is fully charged. You should not charge this cell further. If it's 1.5v on charge, on a low current, then it's pretty much fully charged.

If a cell measures 0.9v off load, it is fully discharged. You should not discharge this cell further. In fact, you should have stopped discharging when it was 1v, on load.

By all means use a multimeter to measure the terminal voltage of a cell that's finished charging. It will tell you there's no point in charging the cell any more. It won't tell you whether the cell now contains 100%, or 80% of its original capacity, only a discharge test will tell you that.

You cannot tell the state of charge of a cell from voltage

If a cell measures 1.2v on or off load, you cannot tell where its state of charge is from 10% to 90%. If you stop or start charging or discharging, then you find the terminal voltage can wander 10s of mV in the first few seconds, then more 10s of mV over the few minutes, in a way that's not modelled by the terminal voltage being dependent on charge alone. With the flatness of the discharge curve, that dependence on history pretty much destroys any predictive power that measuring voltage alone might try to claim.

This is why practical battery metering systems use a gas-gauge IC (coulomb counter) to track the state of charge directly.

\$\endgroup\$
2
\$\begingroup\$

I'd suggest this is a reasonable documentation set on NiMh batteries.

It shows this as the charging curve:

enter image description here

Most laptops that used NiMH batteries had a temperature sensor to ensure charge cutoff ...and prevent venting batteries. That was before we had complex battery management processors that understood the charge/discharge curves much better.

You can however clearly charge the batteries with a voltage setpoint as long as you don't 100% charge them. For example, according to this data setting a fully charged point of 1.55 V per cell you would expect no problems but return only about 95% charge. That is however not true, the battery terminal voltage varies considerably with temperature.

For cells from different manufacturers you would expect some slight variation, and one general curve set for constant current charging of NiMH looks like this:

enter image description here

Here you can see that selecting a constant voltage aiming point is very challenging unless you are in the Australian desert.

So the short answer to your question is ...No you can't charge NiMH batteries to a constant voltage fully charged state. You have to use 'over the hill' voltage sensing and temperature change to take the battery to 100% fully charged.

To use your multimeter to check charge state is also problematic. It depends on cell temperature. You could build a chart of you particular cells and get some idea of charge state, but the only way to be sure is to measure energy stored during charge, and discharge.

\$\endgroup\$
  • 1
    \$\begingroup\$ OP is asking about discharging voltage, not charging voltage. You answered the wrong question (as good as your answer is). \$\endgroup\$ – Passerby Mar 12 '17 at 6:21
  • \$\begingroup\$ You are partly right ....the OP asked if he could use his multimeter to check charge state ...which includes fully charged. I'll add in the extra. \$\endgroup\$ – Jack Creasey Mar 12 '17 at 6:38
  • \$\begingroup\$ One should not cite Fig. 12 from Energizer. It compares T,U and p with arbitrary units, offset and scaling. This is either silly or intentionally misleading. \$\endgroup\$ – Jonas Stein Jan 11 '18 at 3:27

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.