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Some quick questions about cells and batteries...

P

phaeton

Jan 1, 1970
0
Just some quick questions to see if I've got the concepts correctly.
Any responses, flames or pointers welcome:

1) With similar types, you can generally expect comparative battery
life to be proportional to the physical volume of material- i.e. twelve
D-cells can continuously supply 18V into a load longer than two 9V
batteries could, right?

2) Charging a sealed lead-acid battery is about as complicated as
connecting it to another current source and slowly pulling electrons
through it? In other words, *could* you safely charge a 12V motorcycle
or car or similar battery with an AC-to-DC wall adaptor and a simple
voltage divider circuit?

3) The same can be said about NiMH or NiCd batteries.

4) Do connecting cells in series add voltages but keep current the
same, connecting cells in parallel add current but keep voltages the
same?

5) Putting cells in parallel also increases the overall physical volume
of material. Thus, if you connect 10 ordinary D-cells in parallel you
still get ~1.5v but it will last roughly 10 times longer than a single
ordinary D-cell driving the same load. True, False or Maybe?

Thanks for any and all!
 
P

phaeton

Jan 1, 1970
0
Thanks Bob.

I'm aware of the 'voltage depletion' of batteries, and in fact that's
what brings me to this. I built a small LM386-based guitar amplifier
that i've been powering with a 9V battery. All is well if the battery
is fairly healthy, but after about 30 minutes or so i've 'worn it down'
to about 7V. That's about where (i assume) it starts affecting
transistor bias in the chipamp. I get distortion and in a short time
it will simply bias itself off. (No it's not overheating. it is a
diminished battery).

If i turn it off and let it sit for awhile (i assume) the chemical
reaction has time to bolster the battery back up to about 9V or so and
I can go at it again, for another 30 minutes. So I was figuring that
if I were to power this with multiple batteries, i would be able to
expect more time out of it- and an hour is about as long as I want to
listen to a 500mW amp. ;)

(more later, gotta work)
 
B

Bob Myers

Jan 1, 1970
0
phaeton said:
Just some quick questions to see if I've got the concepts correctly.
Any responses, flames or pointers welcome:

1) With similar types, you can generally expect comparative battery
life to be proportional to the physical volume of material- i.e. twelve
D-cells can continuously supply 18V into a load longer than two 9V
batteries could, right?

In general, yes, but...where you went wrong in the above is
that "continuously supply 18V into a load" bit. You're not
WAY wrong, but I am worried that you might have this
notion that batteries are truly decent voltage sources (i.e., they
will maintain their "labeled" voltage until they're out of
juice, and then just shut off) when in fact they're not all THAT
great. Batteries/cells will in general show a slow but steady
decrease in output voltage at a given load, and for that
matter even the initial voltage of that curve is subject to
some variation due to their internal resistance. With
that qualification, though, the above isn't too bad. But for
example - NiCd cells are supposedly 1.2V each, but that's
only at full charge; for most of the discharge cycle, you'll
see closer to 1.1V or under. NiCds have a pretty flat curve,
but it doesn't just stay at 1.2V for the full cycle and then shut
down. (This, by the way, brings up a problem with simple
battery-operated equipment which uses voltage, and ONLY
voltage, to determine the remaining charge in a battery - a
problem which contributed to the myth of NiCd "memory."
The real problem was that NiCd cells suffer from a "voltage
depression" phenomenon - leave them on the charger a long
time, and they OVERcharge and the cell voltage actually
drops off from what it "should" be. All of the expected
energy or charge is still in there, it's just available at a lower
voltage - but this lead some equipment to declare these types
as "out of charge" when in fact most of the stored charge was
still there!)
2) Charging a sealed lead-acid battery is about as complicated as
connecting it to another current source and slowly pulling electrons
through it? In other words, *could* you safely charge a 12V motorcycle
or car or similar battery with an AC-to-DC wall adaptor and a simple
voltage divider circuit?

3) The same can be said about NiMH or NiCd batteries.

The way most batteries/cells are "supposed" to be charged
is via a current source limited to a given percentage of the
specified maximum discharge current (0.1C is a good starting
point for a lot of technologies), and maintaining that current
limit UNTIL a given cell voltage is reached, at which point it
is generally a good idea to switch the current level to a
"trickle" (0.01C, perhaps, but in both of these it's a really good
idea to read the specifics for the technology/type in question)
simply to maintain the thing in a state of "full charge." Too much
charging current and you run the risk of overheating and/or
venting (which is generally a Bad Thing); too little isn't
bad, generally, from a reliability point of view, but most often
you DO want the thing to charge within some reasonable
time period. "Quick chargers" pile more charge in at
once (i.e., higher charging current), but have to very carefully
monitor the state of the battery to avoid any problems.
4) Do connecting cells in series add voltages but keep current the
same, connecting cells in parallel add current but keep voltages the
same?

Yes, again to a very rough approximation. Connecting cells
in series does result in a greater voltage (this is what a simple
"battery" generally is - e.g., a 9V battery is a stack of 6 1.5V
cells), but you also, don't forget, are also putting the resistances
of those cells in series as well. These are NOT little voltage
sources, quite.
5) Putting cells in parallel also increases the overall physical volume
of material. Thus, if you connect 10 ordinary D-cells in parallel you
still get ~1.5v but it will last roughly 10 times longer than a single
ordinary D-cell driving the same load. True, False or Maybe?

Yes, this one is pretty close. Of course, putting the cells in
series ALSO "increases the overall physical volume of material,"
so you really DO have more energy available in that case as well.
It's just a matter of how (and how effectively) you can extract
that energy and do something useful with it.
 
D

Dan H

Jan 1, 1970
0
phaeton said:
Just some quick questions to see if I've got the concepts correctly.
Any responses, flames or pointers welcome:

1) With similar types, you can generally expect comparative battery
life to be proportional to the physical volume of material- i.e. twelve
D-cells can continuously supply 18V into a load longer than two 9V
batteries could, right?

TRUE



2) Charging a sealed lead-acid battery is about as complicated as
connecting it to another current source and slowly pulling electrons
through it? In other words, *could* you safely charge a 12V motorcycle
or car or similar battery with an AC-to-DC wall adaptor and a simple
voltage divider circuit?

3) The same can be said about NiMH or NiCd batteries.

CHARGING BATTERIES IS MORE COMPLICATED. DIFFERENT TYPES REQUIRE
DIFFERENT METHODS
4) Do connecting cells in series add voltages but keep current the
same, connecting cells in parallel add current but keep voltages the
same?
TRUE

5) Putting cells in parallel also increases the overall physical volume
of material. Thus, if you connect 10 ordinary D-cells in parallel you
still get ~1.5v but it will last roughly 10 times longer than a single
ordinary D-cell driving the same load. True, False or Maybe?
TRUE


Thanks for any and all!

DAN H
 
P

Phil Allison

Jan 1, 1970
0
"phaeton"
2) Charging a sealed lead-acid battery is about as complicated as
connecting it to another current source and slowly pulling electrons
through it?


** Slowly being the operative word.

In other words, *could* you safely charge a 12V motorcycle
or car or similar battery with an AC-to-DC wall adaptor and a simple
voltage divider circuit?


** Again - only very slowly.

Takes days that way.

3) The same can be said about NiMH or NiCd batteries.


** Slow as hell, but.




.......... Phil
 
W

Walter Harley

Jan 1, 1970
0
phaeton said:
Thanks Bob.

I'm aware of the 'voltage depletion' of batteries, and in fact that's
what brings me to this. I built a small LM386-based guitar amplifier
that i've been powering with a 9V battery. All is well if the battery
is fairly healthy, but after about 30 minutes or so i've 'worn it down'
to about 7V. That's about where (i assume) it starts affecting
transistor bias in the chipamp. I get distortion and in a short time
it will simply bias itself off. (No it's not overheating. it is a
diminished battery).

If i turn it off and let it sit for awhile (i assume) the chemical
reaction has time to bolster the battery back up to about 9V or so and
I can go at it again, for another 30 minutes. So I was figuring that
if I were to power this with multiple batteries, i would be able to
expect more time out of it- and an hour is about as long as I want to
listen to a 500mW amp. ;)

Yes, adding current capacity (by putting more batteries in parallel, or
using bigger batteries) will help.

A concept to understand is "energy capacity", measured in milliamp-hours
(mAh). Roughly speaking, a battery rated at (say) 600mAh will be able to
produce 600mA for one hour, or 1mA for 600 hours, before its voltage starts
falling rapidly. Take a look at a battery manufacturer site, like
www.duracell.com, to see ratings. But understand that it's a very rough
measure, because it varies depending on how much current you draw. The
manufacturer sites give lots of graphs showing how the battery will
discharge over time with various different loads.

Most mfr's 9V alkaline batteries are rated 625mAh, so if you're wearing it
down in 30 minutes, you're drawing roughly 1A. That is counterintuitive,
because I don't think you can manage to squeeze 1A through an LM386 without
it burning up. I wonder if there is a problem in your circuit somewhere?

Also, as was already pointed out, batteries have internal resistance; the
smaller the battery (physically) the more resistance. When you draw current
from the battery, that internal resistance causes the voltage at the battery
terminals to drop (Ohm's Law), and it also causes the battery itself to
heat. That means that you need to use a battery that is large enough (= has
small enough internal impedance) to supply the needed current without
wasting too much energy in internal heating. A 9V battery definitely can't
supply 1A for very long. That may be part of what's causing the "recovery"
phenomenon you mention.
 
N

NoThanks

Jan 1, 1970
0
phaeton said:
Just some quick questions to see if I've got the concepts correctly.
Any responses, flames or pointers welcome:
1) With similar types, you can generally expect comparative battery
life to be proportional to the physical volume of material- i.e. twelve
D-cells can continuously supply 18V into a load longer than two 9V
batteries could, right?

Assuming equivalent construction and equivalent chemistries, I suppose
thats roughly correct.
2) Charging a sealed lead-acid battery is about as complicated as
connecting it to another current source and slowly pulling electrons
through it? In other words, *could* you safely charge a 12V motorcycle
or car or similar battery with an AC-to-DC wall adaptor and a simple
voltage divider circuit?

Bad idea. Very bad idea. Every battery chemistry requires its own unique
manner of charging (and sensing when a charge is finished). Lead acid
batteries should be typically charged at a constant current until they
reach a certain voltage, and then charged at constant voltage there-after.
There's a current limit and a voltage limit when charging lead acid
batteries, at no time should either be exceeded.

Two problems with what you describe. First, a wall wart with a resistor
does not make for a current source, it makes for a voltage source with
source resistance. Second, the voltage of an unregulated wallwart can
vary significantly with load. When left open circuit, a 12V wallwart
might actually be putting out 17V. Lead acid battery resistance increases
as it becomes charged, so a fully charged battery will appear nearly open
circuit, and your voltage dropping resistor will cease to work well. This
will put 17V across the battery which is quite bad for it.
3) The same can be said about NiMH or NiCd batteries.

NiMH and NiCd batteries can probably be safely charged at very low
currents (1/20th of their capacity?), but if you charge them rapidly you
better have a much more sophisticated charging scheme or you will risk
damaging the cells and/or creating a fire.
4) Do connecting cells in series add voltages but keep current the
same, connecting cells in parallel add current but keep voltages the
same?

Connecting cells in series cause the voltages to increase, but current
capability will be proportional to the worst of the cells. In theory
connecting cells in parallel should retain the same voltage but have the
sum of the current (and charge capacity). There is a big caveat here...
Putting cells in parallel is not a good idea, because if they do not
discharge equally, one cell will be trying to charge the other. If one
cell outright fails (such as a dead short or resistive short -- not
uncommon) this can lead to very unfortunate consequences.
5) Putting cells in parallel also increases the overall physical volume
of material. Thus, if you connect 10 ordinary D-cells in parallel you
still get ~1.5v but it will last roughly 10 times longer than a single
ordinary D-cell driving the same load. True, False or Maybe?

In theory, yes, if the cells are perfectly matched, you'd get 10x. In
reality, it may well be less (or if one cell fails prematurely, all may
die).
 
P

phaeton

Jan 1, 1970
0
4) Do connecting cells in series add voltages but keep current the
same, connecting cells in parallel add current but keep voltages the
same?

-Connecting cells in series cause the voltages to increase, but current
-capability will be proportional to the worst of the cells. In theory
-connecting cells in parallel should retain the same voltage but have
the
-sum of the current (and charge capacity). There is a big caveat
here... -
-Putting cells in parallel is not a good idea, because if they do not
-discharge equally, one cell will be trying to charge the other. If
one
-cell outright fails (such as a dead short or resistive short -- not
-uncommon) this can lead to very unfortunate consequences.
5) Putting cells in parallel also increases the overall physical volume
of material. Thus, if you connect 10 ordinary D-cells in parallel you
still get ~1.5v but it will last roughly 10 times longer than a single
ordinary D-cell driving the same load. True, False or Maybe?

-In theory, yes, if the cells are perfectly matched, you'd get 10x. In
-reality, it may well be less (or if one cell fails prematurely, all
may
-die).

Would any clever diode networks alleviate you from one dead cell taking
down the rest?
 
A

Alan B

Jan 1, 1970
0
Bad idea. Very bad idea. Every battery chemistry requires its own unique
manner of charging (and sensing when a charge is finished). Lead acid
batteries should be typically charged at a constant current until they
reach a certain voltage,

A process called "equalization." I differ here with your idea that the
current will be constant; rather I think the proper technique is to keep
the voltage high, while the current will gradually drop off. Again, as you
say, different vendors may suggest different methods.
and then charged at constant voltage there-after.

A process called "float."
There's a current limit and a voltage limit when charging lead acid
batteries, at no time should either be exceeded.

Precisely. Much more complicated than "connecting to another current
source and pulling electrons through." A true statement in an extremely
general way, but lead and acid are rather harmful substances on their own,
and when combined create quite a volatile mixture that requires proper care
at all times.
 
A

Alan B

Jan 1, 1970
0
-In theory, yes, if the cells are perfectly matched, you'd get 10x. In
-reality, it may well be less (or if one cell fails prematurely, all
may
-die).

Would any clever diode networks alleviate you from one dead cell taking
down the rest?

Adding diodes adds voltage drop. With a typical cell (of any chemical
construction) being <=~2.0VDC, there isn't much room for voltage drop.
Better to monitor battery voltage and take quick action if it falls out of
spec.
 
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