I recently tested a 9V alkaline battery by measuring its open-circuit
voltage (9.0 V) and then measuring it with a car headlight lamp (R = 1
Ohm) across the terminals (4.0 V). The lamp lit up brightly and got
warm, but from the significant voltage drop I conclude that the
battery is basically dead. Correct?


From those measurements I get a Thevenin model of the circuit as
follows, where Rb is the battery's internal resistance and Rl is the
load (lamp).

- Vb + Rb
-----|||||-----/\/\/\-----
| |
| |
o o
| |
| Rl |
-----------/\/\/\---------


With the load removed, and assuming the voltmeter is an open circuit,
Vb = 9.0 V. With Rl = 1 Ohm in place and the voltage across o-o
measured as 4.0 V, the loop current is 4 A. So Rb is 1.25 Ohm.
Correct?

Is there a rule of thumb for judging a battery as "still OK" or
"dead" based on the calculated Thevenin resistance?

I measured the headlamp as 1.0 Ohm, which in a 12 V car circuit
(assuming a negligeable series resistance) should have a power of
144 W. Does that sound reasonable?

"Adam Funk" wrote in message
...
I recently tested a 9V alkaline battery by measuring its open-circuit
voltage (9.0 V) and then measuring it with a car headlight lamp (R = 1
Ohm) across the terminals (4.0 V). The lamp lit up brightly and got
warm, but from the significant voltage drop I conclude that the
battery is basically dead. Correct?


From those measurements I get a Thevenin model of the circuit as
follows, where Rb is the battery's internal resistance and Rl is the
load (lamp).

- Vb + Rb
-----|||||-----/\/\/\-----
| |
| |
o o
| |
| Rl |
-----------/\/\/\---------


With the load removed, and assuming the voltmeter is an open circuit,
Vb = 9.0 V. With Rl = 1 Ohm in place and the voltage across o-o
measured as 4.0 V, the loop current is 4 A. So Rb is 1.25 Ohm.
Correct?

Is there a rule of thumb for judging a battery as "still OK" or
"dead" based on the calculated Thevenin resistance?

I measured the headlamp as 1.0 Ohm, which in a 12 V car circuit
(assuming a negligeable series resistance) should have a power of
144 W. Does that sound reasonable?


Nice experiment and good math practice but that battery was not designed to
supply that much current. You really can not tell much about the battery
from that test. Get the test specs on the battery and load it properly and
repeat your experiment.
BTW: The resistance of the lamp filament will not stay the same as it heats
up.
Tom

Adam Funk wrote:
I recently tested a 9V alkaline battery by measuring its open-circuit
voltage (9.0 V) and then measuring it with a car headlight lamp (R = 1
Ohm) across the terminals (4.0 V). The lamp lit up brightly and got
warm, but from the significant voltage drop I conclude that the
battery is basically dead. Correct?

No, 1,25 ohm ESR (Equivalent Series Resistance) for a 9V battery is
quite good. For a 1,5V "D" cell i would consider it too high.

From those measurements I get a Thevenin model of the circuit as
follows, where Rb is the battery's internal resistance and Rl is the
load (lamp).

- Vb + Rb
-----|||||-----/\/\/\-----
| |
| |
o o
| |
| Rl |
-----------/\/\/\---------


With the load removed, and assuming the voltmeter is an open circuit,
Vb = 9.0 V. With Rl = 1 Ohm in place and the voltage across o-o
measured as 4.0 V, the loop current is 4 A. So Rb is 1.25 Ohm.
Correct?

Yep. But remember that the 1 ohm of the lamp is measured in cold state.
At 9V it is probably a lot more.

Is there a rule of thumb for judging a battery as "still OK" or
"dead" based on the calculated Thevenin resistance?

It depends of the size, technology and voltage. A large "D" cell will
have in its new state an Rb of about 0,1 ohm or even less.
A new 9V battery might give you somathing like 1 ohm.

I measured the headlamp as 1.0 Ohm, which in a 12 V car circuit
(assuming a negligeable series resistance) should have a power of
144 W. Does that sound reasonable?

No, the resistance will increase by a factor of 2 or 3 as the lamp
gets hotter. Connect it to a decent power supply (or car battery) and
measure the current. That will give you the corrent operating wattage
and resistance.

--
Joop van der Velden -
Het hele jaar door radiovlooienmarkt op:

"Adam Funk" wrote in message ...
I recently tested a 9V alkaline battery by measuring its open-circuit
voltage (9.0 V) and then measuring it with a car headlight lamp (R = 1
Ohm) across the terminals (4.0 V). The lamp lit up brightly and got
warm, but from the significant voltage drop I conclude that the
battery is basically dead. Correct?


From those measurements I get a Thevenin model of the circuit as
follows, where Rb is the battery's internal resistance and Rl is the
load (lamp).

- Vb + Rb
-----|||||-----/\/\/\-----
| |
| |
o o
| |
| Rl |
-----------/\/\/\---------


With the load removed, and assuming the voltmeter is an open circuit,
Vb = 9.0 V. With Rl = 1 Ohm in place and the voltage across o-o
measured as 4.0 V, the loop current is 4 A. So Rb is 1.25 Ohm.
Correct?

Is there a rule of thumb for judging a battery as "still OK" or
"dead" based on the calculated Thevenin resistance?

I measured the headlamp as 1.0 Ohm, which in a 12 V car circuit
(assuming a negligeable series resistance) should have a power of
144 W. Does that sound reasonable?

If this is a typical 9V battery as in:

http://data.energizer.com/PDFs/522.pdf

then the usual load is in the neighborhood of a few tens
of mA up to maybe a couple of hundred mA at greatly
shortened service life. If your load current was really
as high as 4A, then if the battery wasn't finished before
it probably is now.

Also, the car headlight will increase in resistance as
its filament gets hotter, so your load might have started
at 1 ohm, but would have risen considerably very quickly.

"Joop van der Velden"


No, the resistance will increase by a factor of 2 or 3 as the lamp
gets hotter.


** More like 10 or 12 times, actually.

Depends just how white hot the filament becomes.

Halogen projector lamps with 50 hour rated live increase by a factor of
about 16.






........ Phil

Adam Funk wrote:
I recently tested a 9V alkaline battery by measuring its open-circuit
voltage (9.0 V) and then measuring it with a car headlight lamp (R = 1
Ohm) across the terminals (4.0 V). The lamp lit up brightly and got
warm, but from the significant voltage drop I conclude that the
battery is basically dead. Correct?


From those measurements I get a Thevenin model of the circuit as
follows, where Rb is the battery's internal resistance and Rl is the
load (lamp).

- Vb + Rb
-----|||||-----/\/\/\-----
| |
| |
o o
| |
| Rl |
-----------/\/\/\---------


With the load removed, and assuming the voltmeter is an open circuit,
Vb = 9.0 V. With Rl = 1 Ohm in place and the voltage across o-o
measured as 4.0 V, the loop current is 4 A. So Rb is 1.25 Ohm.
Correct?

Is there a rule of thumb for judging a battery as "still OK" or
"dead" based on the calculated Thevenin resistance?

I measured the headlamp as 1.0 Ohm, which in a 12 V car circuit
(assuming a negligeable series resistance) should have a power of
144 W. Does that sound reasonable?

Your calculations are correct but they are on paper and your testing
arrangement is in real life with real components so:
a) R1 can and does change depending on brightness produced by ~1-10 or
20. (Read the Watts rating at 12- 13.7V -- car battery on charge).
b) Battery has I dependant on rate where the production of current has
some upper limit and then the voltage drops independant of Rb

So to test your measurment introduce in this circuit an ampermeter and
do your calculations again.

Have fun

Stanislaw
Slack user from Ulladulla.

Adam Funk ha escrito:

I recently tested a 9V alkaline battery by measuring its open-circuit
voltage (9.0 V) and then measuring it with a car headlight lamp (R = 1
Ohm) across the terminals (4.0 V). The lamp lit up brightly and got
warm, but from the significant voltage drop I conclude that the
battery is basically dead. Correct?


From those measurements I get a Thevenin model of the circuit as
follows, where Rb is the battery's internal resistance and Rl is the
load (lamp).

- Vb + Rb
-----|||||-----/\/\/\-----
| |
| |
o o
| |
| Rl |
-----------/\/\/\---------


With the load removed, and assuming the voltmeter is an open circuit,
Vb = 9.0 V. With Rl = 1 Ohm in place and the voltage across o-o
measured as 4.0 V, the loop current is 4 A. So Rb is 1.25 Ohm.
Correct?

Is there a rule of thumb for judging a battery as "still OK" or
"dead" based on the calculated Thevenin resistance?

I measured the headlamp as 1.0 Ohm, which in a 12 V car circuit
(assuming a negligeable series resistance) should have a power of
144 W. Does that sound reasonable?

Problem is that a headlamp isn´t an adequate load for a 9V battery. A
5W tail lamp would be a better load, so the test you made is esentially
meaningless, even if the math model is perfect.

On 2006-08-01, Stanislaw Flatto wrote:

So to test your measurment introduce in this circuit an ampermeter and
do your calculations again.

Have fun

I haven't got an ammeter but "fun" would be a good excuse!

On 2006-08-01, Joop van der Velden wrote:
Adam Funk wrote:
I recently tested a 9V alkaline battery by measuring its open-circuit
voltage (9.0 V) and then measuring it with a car headlight lamp (R = 1
Ohm) across the terminals (4.0 V). The lamp lit up brightly and got
warm, but from the significant voltage drop I conclude that the
battery is basically dead. Correct?

No, 1,25 ohm ESR (Equivalent Series Resistance) for a 9V battery is
quite good. For a 1,5V "D" cell i would consider it too high.

Interesting. I though a 44% voltage drop sounded like a lot, but as
you and others have pointed out, the load resistance I've used is very
low. What sort of resistance do I really need for this sort of test?


- Vb + Rb
-----|||||-----/\/\/\-----
| |
| |
o o
| |
| Rl |
-----------/\/\/\---------


With the load removed, and assuming the voltmeter is an open circuit,
Vb = 9.0 V. With Rl = 1 Ohm in place and the voltage across o-o
measured as 4.0 V, the loop current is 4 A. So Rb is 1.25 Ohm.
Correct?

Yep. But remember that the 1 ohm of the lamp is measured in cold state.
At 9V it is probably a lot more.

Right. I measured the lamp's resistance with an ohmmeter, which of
course puts very little current through it.

But I took the measurements by clipping the voltmeter (actually it's
the same meter) leads onto the battery terminals, reading the
open-circuit voltage, then pressing the lamp's terminals against the
battery terminals (the spacing was convenient --- that's where I got
the idea from) and immediately reading the loaded voltage (before the
lamp heated up).


Is there a rule of thumb for judging a battery as "still OK" or
"dead" based on the calculated Thevenin resistance?

It depends of the size, technology and voltage. A large "D" cell will
have in its new state an Rb of about 0,1 ohm or even less.
A new 9V battery might give you somathing like 1 ohm.

Thanks!

On 2006-08-01, Greg Neill wrote:

If this is a typical 9V battery as in:

http://data.energizer.com/PDFs/522.pdf

So a load in the area of 270 to 620 Ohm (the examples in that data
sheet) would be much more suitable.


then the usual load is in the neighborhood of a few tens
of mA up to maybe a couple of hundred mA at greatly
shortened service life. If your load current was really
as high as 4A, then if the battery wasn't finished before
it probably is now.

Oops!

Adam Funk wrote:
I recently tested a 9V alkaline battery by measuring its open-circuit
voltage (9.0 V) and then measuring it with a car headlight lamp (R = 1
Ohm) across the terminals (4.0 V). The lamp lit up brightly and got
warm, but from the significant voltage drop I conclude that the
battery is basically dead. Correct?

It might be dead now that you've run it at 4 amps, but 9.0 V open ckt
is perfectly fine. When the o.c. voltage drops down to 7 V or so it
might be considered dead, but that would really depend on what you're
using it for (i.e. how much is the load).

Next time, try testing it with 470 to 1k ohms. Even better would be if
you know the load resistance of your intended application.

Regards,

Mark

Tom Biasi wrote:
"Adam Funk" wrote in message
...
I recently tested a 9V alkaline battery by measuring its open-circuit
voltage (9.0 V) and then measuring it with a car headlight lamp (R = 1
Ohm) across the terminals (4.0 V). The lamp lit up brightly and got
warm, but from the significant voltage drop I conclude that the
battery is basically dead. Correct?


From those measurements I get a Thevenin model of the circuit as
follows, where Rb is the battery's internal resistance and Rl is the
load (lamp).

- Vb + Rb
-----|||||-----/\/\/\-----
| |
| |
o o
| |
| Rl |
-----------/\/\/\---------


With the load removed, and assuming the voltmeter is an open circuit,
Vb = 9.0 V. With Rl = 1 Ohm in place and the voltage across o-o
measured as 4.0 V, the loop current is 4 A. So Rb is 1.25 Ohm.
Correct?

Is there a rule of thumb for judging a battery as "still OK" or
"dead" based on the calculated Thevenin resistance?

I measured the headlamp as 1.0 Ohm, which in a 12 V car circuit
(assuming a negligeable series resistance) should have a power of
144 W. Does that sound reasonable?


Nice experiment and good math practice but that battery was not designed to
supply that much current. You really can not tell much about the battery
from that test. Get the test specs on the battery and load it properly and
repeat your experiment.
BTW: The resistance of the lamp filament will not stay the same as it heats
up.

By the way, some 9 volt batteries are made up using cells that are not
welded together, and only touching each other, which can lead to higher
ESR and unreliability. I don't know if there is another battery which
has been made like this. Beware.

GS

On 2006-08-01, Tom Biasi wrote:

Nice experiment and good math practice

How about the ASCII art?
;-)

"Adam Funk" wrote in message
...
I recently tested a 9V alkaline battery by measuring its open-circuit
voltage (9.0 V) and then measuring it with a car headlight lamp (R = 1
Ohm) across the terminals (4.0 V). The lamp lit up brightly and got
warm, but from the significant voltage drop I conclude that the
battery is basically dead. Correct?

No. A 1-ohm load is way too low a resistance to use with a 9V battery; the
excessive current that flowed (over 4 amps) will damage the battery. The
battery may be dead *now* because you killed it. Or, if the abuse was only
momentary, it may be OK.

I would have tried maybe a 100-ohm load, drawing 90 mA, which is a typical
(rather heavy) load on a 9V battery.

From those measurements I get a Thevenin model of the circuit as
follows, where Rb is the battery's internal resistance and Rl is the
load (lamp).

- Vb + Rb
-----|||||-----/\/\/\-----
| |
| |
o o
| |
| Rl |
-----------/\/\/\---------


With the load removed, and assuming the voltmeter is an open circuit,
Vb = 9.0 V. With Rl = 1 Ohm in place and the voltage across o-o
measured as 4.0 V, the loop current is 4 A. So Rb is 1.25 Ohm.
Correct?

If you're sure the load was 1.0 ohm, then the total resistance in the
circuit is 9/4 ohm, so indeed, Rth is 1.25 ohms.

That is a rather low Rth for a 9-volt battery. As much as 5 or 10 ohms
might be OK.


Is there a rule of thumb for judging a battery as "still OK" or
"dead" based on the calculated Thevenin resistance?

Indirectly, yes. At the current that the battery is designed to supply, the
voltage should be at least 90% of full voltage.

Actually, in this era of digital voltage meters, I measure Vth, not Rth.
Vth drops from well over 9.0 V for a new battery to 8.8 V for one that is
showing its age.


I measured the headlamp as 1.0 Ohm, which in a 12 V car circuit
(assuming a negligeable series resistance) should have a power of
144 W. Does that sound reasonable?

Yes. Headlamps normally draw something like 10 amps at 12 V, which would
imply about 1.2 ohms. And that is approximate.

mc wrote:

Actually, in this era of digital voltage meters, I measure Vth, not Rth.
Vth drops from well over 9.0 V for a new battery to 8.8 V for one that is
showing its age.

For the handful of batteries I've measured, I get 9.4 V when they are
brand new.

Mark

On 2006-08-01, mc wrote:

Is there a rule of thumb for judging a battery as "still OK" or
"dead" based on the calculated Thevenin resistance?

Indirectly, yes. At the current that the battery is designed to supply, the
voltage should be at least 90% of full voltage.

Actually, in this era of digital voltage meters, I measure Vth, not Rth.
Vth drops from well over 9.0 V for a new battery to 8.8 V for one that is
showing its age.

I was under the impression that the open-circuit voltage wasn't a
reliable measure of how much "juice" is left in a battery.

Adam Funk:
You are absolutely correct about open circuit testing of batteries is
useless.... many otherwise almost dead batteries will have a surprisingly
high open circuit terminal voltage......
.....BUT the test LOAD (car headlamp) that you are using is not much more
than a dead short compared to the design capabilities of the 9V battery.
These batteries were designed with practical average load currents of 20ma
and occassionally will be found in devices using 100ma or possibly a little
more.
A good and practical load that I have used for actual evaluation and
comparison with other 9volt batteries of questionable freshness is a 270 ohm
, 1 watt resistor, which gives a load current of about 33ma.... this will
"test" the battery terminal voltage with a suitable load and will not ruin
or drain the battery while doing it.
Best Regards,
Daniel Sofie
Electronics Supply & Repair
- - - - - - - -


"Adam Funk" wrote in message
...

I was under the impression that the open-circuit voltage wasn't a
reliable measure of how much "juice" is left in a battery.

Actually, in this era of digital voltage meters, I measure Vth, not Rth.
Vth drops from well over 9.0 V for a new battery to 8.8 V for one that is
showing its age.

I was under the impression that the open-circuit voltage wasn't a
reliable measure of how much "juice" is left in a battery.

It wasn't back in the days of analog meters with 10% accuracy. It is now
that we have digital voltmeters with very high input impedance and 1%
accuracy.

On Tue, 1 Aug 2006 08:01:50 -0400, "Greg Neill"
wrote:

"Adam Funk" wrote in message ...
I recently tested a 9V alkaline battery by measuring its open-circuit
voltage (9.0 V) and then measuring it with a car headlight lamp (R = 1
Ohm) across the terminals (4.0 V). The lamp lit up brightly and got
warm, but from the significant voltage drop I conclude that the
battery is basically dead. Correct?


From those measurements I get a Thevenin model of the circuit as
follows, where Rb is the battery's internal resistance and Rl is the
load (lamp).

- Vb + Rb
-----|||||-----/\/\/\-----
| |
| |
o o
| |
| Rl |
-----------/\/\/\---------


With the load removed, and assuming the voltmeter is an open circuit,
Vb = 9.0 V. With Rl = 1 Ohm in place and the voltage across o-o
measured as 4.0 V, the loop current is 4 A. So Rb is 1.25 Ohm.
Correct?

Is there a rule of thumb for judging a battery as "still OK" or
"dead" based on the calculated Thevenin resistance?

I measured the headlamp as 1.0 Ohm, which in a 12 V car circuit
(assuming a negligeable series resistance) should have a power of
144 W. Does that sound reasonable?

If this is a typical 9V battery as in:

http://data.energizer.com/PDFs/522.pdf

then the usual load is in the neighborhood of a few tens
of mA up to maybe a couple of hundred mA at greatly
shortened service life. If your load current was really
as high as 4A, then if the battery wasn't finished before
it probably is now.


Shorting this kind of battery for some seconds won't do permanent
harm.


Also, the car headlight will increase in resistance as
its filament gets hotter, so your load might have started
at 1 ohm, but would have risen considerably very quickly.


At high loads, the ESR of the battery will change with time, too.
Connect a 9v alkaline battery to an ammeter. It will start at, say, 2
amps and drop off as internal polarization kicks in; the decay and
recovery time constants are very roughly in the area of a minute. At
lower currents, ESR is pretty steady.

John

Damn, that's hard on a little 9 v. battery!!
Killed it dead.

Adam Funk wrote:
I recently tested a 9V alkaline battery by measuring its open-circuit
voltage (9.0 V) and then measuring it with a car headlight lamp (R = 1
Ohm) across the terminals (4.0 V). The lamp lit up brightly and got
warm, but from the significant voltage drop I conclude that the
battery is basically dead. Correct?


From those measurements I get a Thevenin model of the circuit as
follows, where Rb is the battery's internal resistance and Rl is the
load (lamp).

- Vb + Rb
-----|||||-----/\/\/\-----
| |
| |
o o
| |
| Rl |
-----------/\/\/\---------


With the load removed, and assuming the voltmeter is an open circuit,
Vb = 9.0 V. With Rl = 1 Ohm in place and the voltage across o-o
measured as 4.0 V, the loop current is 4 A. So Rb is 1.25 Ohm.
Correct?

Is there a rule of thumb for judging a battery as "still OK" or
"dead" based on the calculated Thevenin resistance?

I measured the headlamp as 1.0 Ohm, which in a 12 V car circuit
(assuming a negligeable series resistance) should have a power of
144 W. Does that sound reasonable?

"Adam Funk" wrote in message
...
On 2006-08-01, Tom Biasi wrote:

Nice experiment and good math practice

How about the ASCII art?
;-)

I hate ascii art :-)

On 2006-08-01, Adam Funk wrote:
On 2006-08-01, Joop van der Velden wrote:
Adam Funk wrote:
I recently tested a 9V alkaline battery by measuring its open-circuit
voltage (9.0 V) and then measuring it with a car headlight lamp (R = 1
Ohm) across the terminals (4.0 V). The lamp lit up brightly and got
warm, but from the significant voltage drop I conclude that the
battery is basically dead. Correct?

No, 1,25 ohm ESR (Equivalent Series Resistance) for a 9V battery is
quite good. For a 1,5V "D" cell i would consider it too high.

Interesting. I though a 44% voltage drop sounded like a lot, but as
you and others have pointed out, the load resistance I've used is very
low. What sort of resistance do I really need for this sort of test?

If you've got lots of spare automobile parts there,
perhaps "3W" panel indicator lamp would be more realistic
or a 22 ohm resistor.

Right. I measured the lamp's resistance with an ohmmeter, which of
course puts very little current through it.

But I took the measurements by clipping the voltmeter (actually it's
the same meter) leads onto the battery terminals, reading the
open-circuit voltage, then pressing the lamp's terminals against the
battery terminals (the spacing was convenient --- that's where I got
the idea from) and immediately reading the loaded voltage (before the
lamp heated up).

One thing you could do if the meter has a 10A range is put the meter in
series with the lamp. then you can figure out the resistance of the lamp
using ohms law.

Bye.
Jasen

On 2006-08-02, jasen wrote:

I recently tested a 9V alkaline battery by measuring its open-circuit
voltage (9.0 V) and then measuring it with a car headlight lamp (R = 1
Ohm) across the terminals (4.0 V). The lamp lit up brightly and got
warm, but from the significant voltage drop I conclude that the
battery is basically dead. Correct?
....
If you've got lots of spare automobile parts there,
perhaps "3W" panel indicator lamp would be more realistic
or a 22 ohm resistor.

Perhaps I should mention that the only reason I did this test was that
I happened to notice a used 9V battery and a headlamp lying near each
other on my workbench, and curiosity drew me in!


Right. I measured the lamp's resistance with an ohmmeter, which of
course puts very little current through it.

But I took the measurements by clipping the voltmeter (actually it's
the same meter) leads onto the battery terminals, reading the
open-circuit voltage, then pressing the lamp's terminals against the
battery terminals (the spacing was convenient --- that's where I got
the idea from) and immediately reading the loaded voltage (before the
lamp heated up).

I wrote those two paragraphs in the wrong order. I measured the
voltage drop first and then (after letting the lamp cool) measured its
resistance.

If I'd seen the 1 Ohm value first, I wouldn't have used it for the
voltage-drop test!