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Home > Archive > Electrical Engineering > March 2007 > Another dumb question for you guys
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Another dumb question for you guys
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| Zerex71 2007-03-06, 9:25 am |
| Group,
I am reading up on batteries and vehicle power systems for an
application I'm working on at work. I've learned that a standard 12V
battery is considered "discharged" when its output voltage level
actually drops to 10.5 V. My questions to you are:
1. Why 10.5 V? What's so special at that level?
2. What occurs electrochemically during the discharge process that
causes the voltage to drop?
3. Why do batteries discharge even when they are not used (assume
sealed, maintenance-free for the purposes of discussion)?
4. In reading about deep cycle batteries and discharge cycles in
general, when the battery has reached the 20% DOD (depth of
discharge), how does that correspond to the 10.5V level?
Many thanks in advance.
Mike
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| contrex 2007-03-06, 1:25 pm |
| On 6 Mar, 15:07, "Zerex71" <Zere...@msn.com> wrote:
> Group,
>
> I am reading up on batteries and vehicle power systems for an
> application I'm working on at work. I've learned that a standard 12V
> battery is considered "discharged" when its output voltage level
> actually drops to 10.5 V. My questions to you are:
>
> 1. Why 10.5 V? What's so special at that level?
That's 6 times the voltage of one discharged cell. ("12 volt"
batteries are made of 6 cells in series.)
> 2. What occurs electrochemically during the discharge process that
> causes the voltage to drop?
Each cell contains (in the charged state) electrodes of lead metal and
lead oxide (PbO2) in an electrolyte of about 37% sulfuric acid. In the
discharged state both electrodes turn into lead sulfate and the
electrolyte loses its dissolved sulfuric acid and becomes primarily
water.
> 3. Why do batteries discharge even when they are not used (assume
> sealed, maintenance-free for the purposes of discussion)?
When left idle a filled battery will self discharge because of its own
internal resistance. Resistance implies current flow. Current flow =
discharge.
> 4. In reading about deep cycle batteries and discharge cycles in
> general, when the battery has reached the 20% DOD (depth of
> discharge), how does that correspond to the 10.5V level?
Depends on temperature. Do some Googling!
Take 20 amp hours out of a 100 AH battery, then measure the voltage.
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| Zerex71 2007-03-06, 1:25 pm |
| On Mar 6, 11:47 am, "contrex" <mike.j.har...@gmail.com> wrote:
> On 6 Mar, 15:07, "Zerex71" <Zere...@msn.com> wrote:
>
>
>
>
> That's 6 times the voltage of one discharged cell. ("12 volt"
> batteries are made of 6 cells in series.)
That's not exactly what I was asking...what is special about 10.5V?
Why isn't fully discharged
consider 9V, 6V or even 0V?
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| contrex 2007-03-06, 1:25 pm |
| On 6 Mar, 17:28, "Zerex71" <Zere...@msn.com> wrote:
> On Mar 6, 11:47 am, "contrex" <mike.j.har...@gmail.com> wrote:
>
>
>
>
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> That's not exactly what I was asking...what is special about 10.5V?
> Why isn't fully discharged
> consider 9V, 6V or even 0V?
I was about to tell you, but then I thought "You're getting this free
information because you're too lazy to use Google, and you find fault
with it, so fuck off and find out for yourself."
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| Zerex71 2007-03-06, 1:25 pm |
| On Mar 6, 12:57 pm, "contrex" <mike.j.har...@gmail.com> wrote:
> On 6 Mar, 17:28, "Zerex71" <Zere...@msn.com> wrote:
>
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> I was about to tell you, but then I thought "You're getting this free
> information because you're too lazy to use Google, and you find fault
> with it, so fuck off and find out for yourself."
So your real answer is, "I'm an XXXXXXX who doesn't know the real
answer."
Am I reading that right?
It's not a matter of me Googling. I've been reading websites and I
have not run across
the answer about why 10.5V is considered fully charged? Why not
10.7? If you don't
know, just say so; no need to be a prick about it.
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| cr500r 2007-03-06, 1:25 pm |
| in a google search for "discharge curve 12v lead acid" this was the first
hit :-)
http://www.homepower.com/files/battvoltandsoc.pdf
"Zerex71" <Zerex71@msn.com> wrote in message
news:1173204210.131962.309790@n33g2000cwc.googlegroups.com...
> On Mar 6, 12:57 pm, "contrex" <mike.j.har...@gmail.com> wrote:
>
> So your real answer is, "I'm an XXXXXXX who doesn't know the real
> answer."
> Am I reading that right?
>
> It's not a matter of me Googling. I've been reading websites and I
> have not run across
> the answer about why 10.5V is considered fully charged? Why not
> 10.7? If you don't
> know, just say so; no need to be a prick about it.
>
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| Zerex71 2007-03-06, 5:25 pm |
| On Mar 6, 2:16 pm, "cr500r" <jeffnope_nos...@hotmail.com> wrote:[color=darkred]
> in a google search for "discharge curve 12v lead acid" this was the first
> hit :-)http://www.homepower.com/files/battvoltandsoc.pdf
>
> "Zerex71" <Zere...@msn.com> wrote in message
>
> news:1173204210.131962.309790@n33g2000cwc.googlegroups.com...
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Thanks for the link, even though it still doesn't explain the answer
to my question.
Guess I've posted a stumper!
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| cr500r 2007-03-06, 5:25 pm |
| In a way it answers the 10.5v question.
If you look at the discharge curves, you can see that almost all the energy
is gone from the battery at 10.5V, it's the tail end of the charge curve.
and if you read the text it has the bit about the chemical reaction of
sulfuric acid being turned into water during the discharge. It seems at
about 10.5v there is almost no sulfuric acid left in the battery to "push"
the electrons, so below that terminal voltage you won't get much current
flow to the load, in short: the battery is empty of charge at about 10.5v
due to the nature of the chemical reaction.
Hope that clears that up.
-Jeff
"Zerex71" <Zerex71@msn.com> wrote in message
news:1173209126.123872.25930@8g2000cwh.googlegroups.com...
> On Mar 6, 2:16 pm, "cr500r" <jeffnope_nos...@hotmail.com> wrote:
>
> Thanks for the link, even though it still doesn't explain the answer
> to my question.
> Guess I've posted a stumper!
>
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| Zerex71 2007-03-06, 5:25 pm |
| On Mar 6, 3:30 pm, "cr500r" <jeffnope_nos...@hotmail.com> wrote:[color=darkred]
> In a way it answers the 10.5v question.
> If you look at the discharge curves, you can see that almost all the energy
> is gone from the battery at 10.5V, it's the tail end of the charge curve.
> and if you read the text it has the bit about the chemical reaction of
> sulfuric acid being turned into water during the discharge. It seems at
> about 10.5v there is almost no sulfuric acid left in the battery to "push"
> the electrons, so below that terminal voltage you won't get much current
> flow to the load, in short: the battery is empty of charge at about 10.5v
> due to the nature of the chemical reaction.
> Hope that clears that up.
> -Jeff
>
> "Zerex71" <Zere...@msn.com> wrote in message
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> news:1173209126.123872.25930@8g2000cwh.googlegroups.com...
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Okay, that makes more sense. I'm sure it had something to do with
that but I suppose
I am looking for some physics or math-based proof that works out to
the emf dropping to
10.5V for the six-cell cluster. In other words, I *see* that it's
considered 0% charged/100%
discharged, but I was wondering about where the number came from.
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| Lionel 2007-03-06, 5:25 pm |
| On 6 Mar 2007 11:25:29 -0800, "Zerex71" <Zerex71@msn.com> wrote:
>On Mar 6, 2:16 pm, "cr500r" <jeffnope_nos...@hotmail.com> wrote:
>
>Thanks for the link, even though it still doesn't explain the answer
>to my question.
>Guess I've posted a stumper!
That, or it's just a really dumb question, & people are getting a kick
out of watching you spinning your wheels.
--
W "Some people are alive only because it is illegal to kill them."
. | ,. w ,
\|/ \|/ Perna condita delenda est
---^----^---------------------------------------------------------------
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| cr500r 2007-03-06, 5:25 pm |
| Sometimes that kind of theoretical answer doesn't exist, and I've got a
feeling this may be one of them.
The answer is: it's based on empirical evidence
I also know that a real world battery will still provide some current at
10.5v, it probably won't start your car, but you can't consider it 0%
charged until the terminal voltage is ZERO, but for practical applications
it can be said to be "discharged" (in some cases of course)
"Zerex71" <Zerex71@msn.com> wrote in message
news:1173213685.159269.108570@c51g2000cwc.googlegroups.com...
> On Mar 6, 3:30 pm, "cr500r" <jeffnope_nos...@hotmail.com> wrote:
>
> Okay, that makes more sense. I'm sure it had something to do with
> that but I suppose
> I am looking for some physics or math-based proof that works out to
> the emf dropping to
> 10.5V for the six-cell cluster. In other words, I *see* that it's
> considered 0% charged/100%
> discharged, but I was wondering about where the number came from.
>
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| Zerex71 2007-03-06, 5:25 pm |
| On Mar 6, 3:45 pm, Lionel <use...@imagenoir.com> wrote:
> On 6 Mar 2007 11:25:29 -0800, "Zerex71" <Zere...@msn.com> wrote:
>
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> That, or it's just a really dumb question, & people are getting a kick
> out of watching you spinning your wheels.
>
> --
> W "Some people are alive only because it is illegal to kill them."
> . | ,. w ,
> \|/ \|/ Perna condita delenda est
> ---^----^---------------------------------------------------------------
Oh, RIGHT, that's *so* it! Gosh, you're so sharp. Thanks for chiming
in with some NVA comments.
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| daestrom 2007-03-11, 1:25 pm |
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"Zerex71" <Zerex71@msn.com> wrote in message
news:1173193663.154512.145510@c51g2000cwc.googlegroups.com...
> Group,
>
> I am reading up on batteries and vehicle power systems for an
> application I'm working on at work. I've learned that a standard 12V
> battery is considered "discharged" when its output voltage level
> actually drops to 10.5 V. My questions to you are:
>
> 1. Why 10.5 V? What's so special at that level?
> 2. What occurs electrochemically during the discharge process that
> causes the voltage to drop?
Others have given you some ideas about these. Why does voltage drop at all
when there is less acid in the cell and more lead sulphate on the plates?
Sorry, guess it has to do with the electro-chemical potential. There is
still some pure lead and lead-dioxide in the plates, so it would seem the
open-circuit voltage would still be the same.
Of course, if there is any load on the battery, then the fact that there is
less plate surface for the acid to react with, and weaker concentration of
acid means the internal resistance to current flow is higher, so a given
load will cause more internal voltage drop from the ideal electro-chemical
potential.
> 3. Why do batteries discharge even when they are not used (assume
> sealed, maintenance-free for the purposes of discussion)?
Well, it's *not* 'internal resistance'. That term is used in battery
technology to refer to the effective resistance in series with the cell.
When a cell is loaded, the voltage at the terminals drops immediately. This
is due to the 'internal resistance'. The resistance of current flow through
the plate materials (lead isn't the greatest conductor) and the electrolyte
(acid). As a cell is discharged, the internal resistance goes up because
the conductivity of the electrolyte drops (less acid, more pure water), and
the contact resistance of the plate surface to the electrolyte goes up (lead
sulphate is a poor conductor).
What *does* cause self discharge is a *parallel* resistance across the cell.
In effect, the cell has a tiny amount of load on it all the time due to this
internal parallel resistance. This can take many forms: A layer of damp
corrosion products on the top of the battery between terminal posts, a layer
of crystaline products underneath the lid, a tiny 'whisker' of lead material
that grew from a plate surface and contacts a plate of opposite polarity.
There is *also* a phenomenon where an impurity in the electrolyte or plate
surface can cause self-discharge. Basically, a bit of some other metal such
as copper or iron on the plate surface will create a lead-copper or
lead-iron cell. You have two dissimilar materials immersed in the
electrolyte, and the other 'end' of these materials is connected together
where the bit of material touches the plate. Voila, a tiny cell that comes
already shorted out. It gradually consumes the electrolyte until the bit of
material is consumed. Recharge the cell and the charging current will
'electroplate' the material dissolved in the acid back onto a plate surface
and you start all over again.
A sure fire way to ruin a lead-acid battery is to put 'hard' water into it.
The dissolved minerals in tap-water will do just as I described above.
> 4. In reading about deep cycle batteries and discharge cycles in
> general, when the battery has reached the 20% DOD (depth of
> discharge), how does that correspond to the 10.5V level?
>
The chemistry of modern lead-acid batteries has come a long way. Sponge
lead, used for the plate material, by itself is not 'structurally sound'.
It would fall apart from vibration very easily. So the material is pressed
into a 'grid' of a lead alloy that is structurally stiffer/stronger. But
now you have an alloy, with a dissimilar material and you have some of the
problems I described above. Years of experimentation with different alloys
has been done to find 'the right combination' that is strong and resists
vibration and shedding (the phenomenon of plate material falling off the
plate to the bottom of the cell), but doesn't cause too many side-effects
(like high rates of self discharge).
Lead-antimony alloy has been used for years in deep cycle batteries. It
allows very high currents. But it has a high self-discharge rate (batteries
need frequent charging / topping-off) and uses a lot of water. Lead-calcium
alloy came into prominence in the 70's. It has a much lower self-discharge
rate, and doesn't cause as much water loss. These are the so-called
'maintenance free' batteries. Because they use little water, they often
don't even have service caps (and that keeps folks from putting tap-water
into them :-)
But for any type of battery, it's best to follow the manufacturer's curve
and data. They have done the emprical testing to come up with that data.
OBTW, you need to know the current load for the 10.5 V to be meaningful. As
I mentioned above, the internal resistance that is in series with the cell
will cause an internal voltage drop. A car battery's voltage will drop very
low when cranking the engine (a high current draw), even a fully charged
one. But as soon as the load is removed, the voltage comes right back up.
Just watch the brightness of the headlights while starting a car.
Some manufacturers will supply you with a 'fully discharged' or
'terminating' voltage for different discharge rates. These voltages are a
combination of the actual voltage developed electrochemically and the
voltage drop across the internal resistances.
Hope this helps some...
daestrom
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