# Another dumb question for you guys

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?
Mike
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

That's 6 times the voltage of one discharged cell. ("12 volt" batteries are made of 6 cells in series.)

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.

When left idle a filled battery will self discharge because of its own internal resistance. Resistance implies current flow. Current flow discharge.

Depends on temperature. Do some Googling!
Take 20 amp hours out of a 100 AH battery, then measure the voltage.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

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?
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

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."
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

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.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
in a google search for "discharge curve 12v lead acid" this was the first hit :-) http://www.homepower.com/files/battvoltandsoc.pdf
wrote:

<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

Thanks for the link, even though it still doesn't explain the answer to my question. Guess I've posted a stumper!
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
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
wrote:

<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

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.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
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)
wrote:

<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

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 ,
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

Oh, RIGHT, that's *so* it! Gosh, you're so sharp. Thanks for chiming in with some NVA comments.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

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.

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.

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
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

## Site Timeline

• ### Solutions to problems in Protective Relaying Principles and Applications by J. Lewis Black...

• - last updated thread in ⏚ Electrical Engineering
• Share To

Polytechforum.com is a website by engineers for engineers. It is not affiliated with any of manufacturers or vendors discussed here. All logos and trade names are the property of their respective owners.