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can i use multimeter as a battery tester?

M

Martin Lynch

Jan 1, 1970
0
Can I use a multimeter to test your standard AA or AAA
batteries?

Not sure if I should be checking for voltage or current.

Seems like checking voltage doesnt give me accurate
results, as a "dead" battery will often still read
1.2 volts or so.

And problem with current is my 1600 mAh batteries seem
to be rated to high for the multimeter.

What does a "battery tester" that is designed specifically
for this actually measure?
 
E

EEng

Jan 1, 1970
0
Can I use a multimeter to test your standard AA or AAA
batteries?

Not sure if I should be checking for voltage or current.
Both

Seems like checking voltage doesnt give me accurate
results, as a "dead" battery will often still read
1.2 volts or so.

Put a known load on the battery and measure it then.
And problem with current is my 1600 mAh batteries seem
to be rated to high for the multimeter.

What does a "battery tester" that is designed specifically
for this actually measure?

it measures voltage and current via a fixed load internal to the
meter.
 
J

Joel Kolstad

Jan 1, 1970
0
EEng said:

I think 'EEng' knows what he's talking about, but to make it clearer: You're
looking to measure the voltage of the cell while a known load (more or less
a known curren) is placed across it.

It measure voltages while placing a resistor across the battery. The
battery tester I have from Radio Shack even has different positions for "AA"
vs. "AAA" and denotes that the "AA" test is done at... 250mA, maybe? --
dunno off-hand -- and the "AAA" test is done at something more like 50mA.
I.e., the AA test would use a ~1.5V/250mA=6 ohm resistor whereas the 50mA
test uses ~1.5V/30 ohms.

You can model a cell as an ideal voltage source plus an internal battery
resistance and just say that the internal resistance grows as the cell's
useful energy is depleted. Hence, if you draw _no_ current whatsoever (a
digital multimeter is a very good approximation of this!), even with a large
internal resistance you get something close to the full cell voltage.
However, as soon as you place a load across the cell, you have a voltage
divider and the output voltage will drop (or plummet, in the case of a cell
near the end of its life).

---Joel Kolstad

....such a nice study break from singular value decomposition...
 
P

Paul Burridge

Jan 1, 1970
0
You can model a cell as an ideal voltage source plus an internal battery
resistance and just say that the internal resistance grows as the cell's
useful energy is depleted.

This sounds like a clever analogy, but does it really accurately model
what happens when a cell runs down?
 
J

John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that Paul Burridge
[email protected]>) about 'can i use multimeter as a battery tester?', on Wed,
26 Nov 2003:
This sounds like a clever analogy, but does it really accurately model
what happens when a cell runs down?

Pretty accurately, if it's discharged normally. If it dries up, due to
old age, the source voltage tends to go down as well as the resistance
rising a great deal.
 
M

Martin Lynch

Jan 1, 1970
0
Put a known load on the battery and measure it then.

This is a stupid question, but... what is a simple way to put
a load on it? Sounds to me like you guys are saying that
a multimeter has virtually zero current, so I assume
you mean to send a current thru the battery somehow... ?
 
W

Walter Harley

Jan 1, 1970
0
Paul Burridge said:
This sounds like a clever analogy, but does it really accurately model
what happens when a cell runs down?

Maybe if you're discharging it at a steady discharge rate?

But I recently spent some time, and some batteries, studying how partly-dead
batteries behave. In my experiments, what I saw was that a
partly-discharged battery has a lower-than-fresh voltage even into a very
high load (straight into a multimeter, e.g.); and then when a load is
applied, its voltage drops more than a fresh battery would (and continues to
quickly sag over the next few seconds).

So, it behaves as though both the ideal voltage is sagging *and* the
internal resistance is increasing. Furthermore, the internal resistance
behaves rather like a positive thermal coefficient thermistor: that is, as
it heats up the resistance increases. I don't actually know if that is a
heat-related phenomenon (didn't try, e.g., freeze spray or heat gun on the
batteries), but it acts right: the recovery time after load is related to
the amount of time it was under load, and so forth.
 
R

Reg Edwards

Jan 1, 1970
0
When I was about 12 years of age I almost electrocuted myself by testing a
120-volt dry battery with a spoon in one hand and a fork in the other.

There was no warning message on the container.
 
J

John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that Walter Harley
Furthermore, the internal resistance
behaves rather like a positive thermal coefficient thermistor: that is,
as it heats up the resistance increases. I don't actually know if that
is a heat-related phenomenon (didn't try, e.g., freeze spray or heat gun
on the batteries), but it acts right: the recovery time after load is
related to the amount of time it was under load, and so forth.

The source voltage falls with increasing temperature. Maybe 10 mV/K.
 
W

Walter Harley

Jan 1, 1970
0
John Woodgate said:
I read in sci.electronics.design that Walter Harley


The source voltage falls with increasing temperature. Maybe 10 mV/K.

Must be something else I was seeing, then. Take a nominal 9v battery,
discharged to near the end of its useful life, measuring 7.2v into a
multimeter. Put a load on it of, say, 47R (sorry, don't have my notes in
front of me). What I saw was the voltage drops immediately to 6v,
suggesting the instantaneous internal impedance is about 10R. Then, over a
period of seconds, it drops gradually further, down to perhaps 2V after a
minute of load. The voltage falls not smoothly but in a series of steep
cascades with less-steep plateaus between them. Take the load off, and the
voltage gradually climbs back up to near 7.2v; that is, the battery wasn't
completely depleted, but it seemingly built up more internal resistance.

I replicated this with several different batteries; the effect is less
pronounced the fresher the battery is.
 
R

Reg Edwards

Jan 1, 1970
0
All types of flat dry batteries can be revived for a short time by heating
to 50 or 60 degrees.

The high internal resistance falls fast with increasing temperature.

It can't be done more than one or twice.

Try it with a torch battery and bulb.
 
J

John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that Walter Harley
Must be something else I was seeing, then. Take a nominal 9v battery,
discharged to near the end of its useful life, measuring 7.2v into a
multimeter. Put a load on it of, say, 47R (sorry, don't have my notes
in front of me). What I saw was the voltage drops immediately to 6v,
suggesting the instantaneous internal impedance is about 10R. Then,
over a period of seconds, it drops gradually further, down to perhaps 2V
after a minute of load. The voltage falls not smoothly but in a series
of steep cascades with less-steep plateaus between them.


You don't say what sort of battery that was, but the effect is almost
certainly 'polarization' - a build-up of (usually) hydrogen gas
molecules on the surface of the negative electrode. This is due to the
oxidant ('depolarizer'), which is supposed to convert the hydrogen to
water or something else harmless, losing efficiency, or speed of
reaction, as it wears out.
Take the load
off, and the voltage gradually climbs back up to near 7.2v; that is, the
battery wasn't completely depleted, but it seemingly built up more
internal resistance.

This is due to the depolarizer gradually gobbling up the hydrogen.
I replicated this with several different batteries; the effect is less
pronounced the fresher the battery is.

Yes. fresh depolarizer works faster.
 
J

John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that Reg Edwards
et.com>) about 'can i use multimeter as a battery tester?', on Thu, 27
Nov 2003:
All types of flat dry batteries can be revived for a short time by heating
to 50 or 60 degrees.

The high internal resistance falls fast with increasing temperature.

It can't be done more than one or twice.

Try it with a torch battery and bulb.
The high temperature also makes the depolarizer work better, thus
cleaning up hydrogen and/or other reduction products from around the
negative electrode.
 
N

normanstrong

Jan 1, 1970
0
Martin Lynch said:
This is a stupid question, but... what is a simple way to put
a load on it? Sounds to me like you guys are saying that
a multimeter has virtually zero current, so I assume
you mean to send a current thru the battery somehow... ?

I can tell you how I did it:

I cut the banana plugs off of the end of the test leads and substitued
a pair of Pomona hermaphrodite banana plugs. I have several dual
banana plugs with various load resistors. I can plug in whichever one
is appropriate for the battery in question. I use 33 ohms for a
single cell and 470 ohms for a 9v battery.

Norm Strong
 
P

Paul Hovnanian P.E.

Jan 1, 1970
0
Reg said:
When I was about 12 years of age I almost electrocuted myself by testing a
120-volt dry battery with a spoon in one hand and a fork in the other.

There was no warning message on the container.

Every spoon and fork should be clearly labeled 'Not suitable for testing
electrical power sources'. Where are the product liability lawyers when
we need them.
 
P

Paul Hovnanian P.E.

Jan 1, 1970
0
Martin said:
This is a stupid question, but... what is a simple way to put
a load on it? Sounds to me like you guys are saying that
a multimeter has virtually zero current,

Yes. In the voltage mode. In the current mode, they are nearly a dead
short and are not a suitable test load for battery life.
so I assume you mean to send a current thru the battery somehow... ?

Yes. And then measure the batteries terminal voltage while you do.
 
W

Walter Harley

Jan 1, 1970
0
Paul Hovnanian P.E. said:
Every spoon and fork should be clearly labeled 'Not suitable for testing
electrical power sources'. Where are the product liability lawyers when
we need them.

Au contraire: the spoon and fork worked perfectly; the test was successful
and reasonable results were obtained.

However, our bodies should all be tattooed with a warning message, along the
lines of "CAUTION: This container may be permanently damaged by doing stupid
things to or with it."
 
R

Reg Edwards

Jan 1, 1970
0
Every spoon and fork should be clearly labeled 'Not suitable for testing
Au contraire: the spoon and fork worked perfectly; the test was successful
and reasonable results were obtained.

However, our bodies should all be tattooed with a warning message, along the
lines of "CAUTION: This container may be permanently damaged by doing stupid
things to or with it."
============================

Dear Walter, your observations, although made about circumstances which
existed around 70 years ago, are still relevant. If I had been a weakly
child the experience may very well have abruptly ended my electical
engineering education. As it transpired the sudden shock greatly stimulated
my interest in amps, volts and ohms. Now, at the age of 78, having had a
career (still going strong) involving these 3 mysterious quantities, I can
be grateful for the happy coincidence of a fork, spoon and 120 volts all
being together on the kitchen table.

It was some time later I appreciated the purpose of the extra 9 volts tacked
on to the end of 120.

I'm sure many others can thank accidents for setting themselves off into a
technical career.
 
M

Martin Lynch

Jan 1, 1970
0
Yes. In the voltage mode. In the current mode, they are nearly a dead
short and are not a suitable test load for battery life.


Yes. And then measure the batteries terminal voltage while you do.

Ok time for next stupid question. How do I create the current. Do I
need to get another multimeter and put it to the current setting?

What about the resistor? Is it internal to the battery or do I have
to hook one up externally?

I'm not an electronics guy, as you may have guessed.
 
R

Robert Monsen

Jan 1, 1970
0
Martin Lynch said:
Ok time for next stupid question. How do I create the current. Do I
need to get another multimeter and put it to the current setting?

What about the resistor? Is it internal to the battery or do I have
to hook one up externally?

I'm not an electronics guy, as you may have guessed.

You go buy a 100 ohm resistor at radio shack. Then, you hook one side of the
resistor to one side of the battery, and the other side of the resistor to
the other side of the battery.

Now, measure the voltage at the battery terminals using the voltage
measurement mode of your multimeter.

If its much less than the rated voltage, then the battery is dead and should
be disposed of in an environmentally safe way.

You can also just run it in the circuit you are intending to use it in, and
measure it while its on.

Regards,
Bob Monsen
 
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