Testing batteries - LaCrosse charger OK?

[burpee]

What's better than using a LaCrosse charger - to read battery levels?

Is there favorite size resistor I can AAs with?

How about a resistor for 7.4V Li Ion packs?

What if I "load" a battery with a 10watt 2ohm resistor?

What is the poor man's battery loading device if their meter doesn't do it?

 

[Battery Guy]

What's better than using a LaCrosse charger - to read battery levels?

Is there favorite size resistor I can AAs with?

How about a resistor for 7.4V Li Ion packs?

What if I "load" a battery with a 10watt 2ohm resistor?

What is the poor man's battery loading device if their meter doesn't do it?

Not sure what you are trying to "read", state-of-charge or total cell capacity?

I don't know if it is better than the LaCrosse, but I really like my new Maha C9000 charger. I am very pleasantly surprised by the quality, functionality and ease of use. Quite a nice unit for $50.

Cheers,
Battery Guy

 

[fishinfool]

+1 Maha C9000 is simply the best. :thumbsup:

 

[burpee]

What if I "load" a battery with a 10watt 2ohm resistor?

What I wanted to know is if the terminal voltage you read on a battery in a charger is representative of it's actually capacity?

In other words, if I pull batteries out of a charger as soon as they read 1.40V is that indicative of the battery being 95% charged - given the fact that they usually will read "full" if taken out and replaced in a charger when at ~1.42- 1.43V?

I thought the right way to get a real indication of charge requires putting a load on the battery while reading a terminal voltage. And I also thought that a 10watt, 1ohm resistor ought to suffice.

 

[Battery Guy]

What I wanted to know is if the terminal voltage you read on a battery in a charger is representative of it's actually capacity?

In other words, if I pull batteries out of a charger as soon as they read 1.40V is that indicative of the battery being 95% charged - given the fact that they usually will read "full" if taken out and replaced in a charger when at ~1.42- 1.43V?

I thought the right way to get a real indication of charge requires putting a load on the battery while reading a terminal voltage. And I also thought that a 10watt, 1ohm resistor ought to suffice.

Hey Burpee,

Unless I misunderstand you, the fundamental problem is that the voltage of the cell on the charger is the charging voltage, whereas the voltage of the cell under a 1 ohm resistor load is a discharging voltage. So they are not going to be even close to being the same.

But the bottom line is that it is extremely difficult to determine the state-of-charge of a NiMH cell based only on voltage, even if you take the trouble of measuring voltage under load. The only way to do this is to make a calibration curve for a specific cell type by charging (or discharging) to a known state-of-charge, then measuring the OCV and CCV under a given load. There is simply no other way to do this for a NiMH, at least not without some pretty expensive testing equipment.

Quite frankly, the only time you know the state-of-charge of your NiMH cells is when they come off your charger at 100%.

Sorry that I am not more help.

Cheers,
BG

 

[Lynx_Arc]

Like said earlier, as you use and get familiar with the batteries you have you will learn how much charge they have left at certain voltages. Typically putting a load on nimh to test them is unneeded if you let them rest before testing the voltage on nimh don't sag much and recover like primaries do. Personally I measure the voltage of nimh and here is my "estimate" of how much charge is left.
1.40-1.50v + fully charged/just off the charger
1.35-1.39 90-95% full after resting
1.28-1.34 50-89%
1.22-1.27 15-49%
1.10-1.21 depleted to 14%
<1.09v dead
This is no way exact, and some cells like eneloops would show higher voltages while other cells would be lower. Under light loads this would be decent "estimates" but under heavy loads you can throw it out the door because when battery capacity gets to half under a heavy load you may find yourself with only 25% runtime left as some regulated boost circuits increase the power used to compensate for declining voltage which sags the voltage more and increases the power sapped.