Charge 2 x 6V on 12V Charger

Started by December 18, 2011
```"John Fields"
>>
>>** Cannot work.
>>
>>The terminal voltage will obviously be the same immediately, but very
>>little
>>charging will go on.
>
> ---
> Let's say we have two lead-acid batteries,

** ROTFL !!

A " thought experiment " involving SLAs !!

Einstein would be pissing himself.

...  Phil

```
```On Wed, 21 Dec 2011 12:13:34 +1100, "Phil Allison" <phil_a@tpg.com.au> wrote:

>
>"John Fields"
>>>
>>>** Cannot work.
>>>
>>>The terminal voltage will obviously be the same immediately, but very
>>>little
>>>charging will go on.
>>
>> ---
>> Let's say we have two lead-acid batteries,
>
>
>** ROTFL !!
>
> A " thought experiment " involving SLAs !!
>
> Einstein would be pissing himself.
>
A " thought experiment " involving Phyllis !!

Einstein would be pissing himself.
```
```<krw@att.bizzzzzzzzzzzz> = psychopath
>>
>>> I think he meant to connect them in parallel, sans charger, until they
>>> leveled off to the same voltage, and then to connect them in series
>>> and charge them.
>>>
>>
>>** Cannot work.
>>
>>The terminal voltage will obviously be the same immediately, but very
>>little
>>charging will go on.
>
> Dummy, one will charge the other until they equalize.

** Capacitors do something like that.

But not SLAs.

....  Phil

```
```Winston wrote:
> ehsjr wrote:
>
>> Winston wrote:
>>
>>> NT wrote:
>
>
> (...)
>
>>> Two 6 V SLAs.
>>>
>>> One is charged to nearly 100% or say 6.3 V.
>>> We place it in series with one that is charged
>>> to about 35% or say 6.05 V
>>>
>>> At 10 A discharge we can think of the top battery as
>>> a resistor measuring 0.63 ohm and the bottom
>>> battery equivalent to a 0.60 ohm resistor.
>>>
>>> The top battery will have 51% of the charge voltage
>>> across it and the bottom battery will have only 49%.
>>>
>>> If we begin charging them at say 10 A, the battery
>>> charged to nearly 100% will get about 3 W more
>>> charge power than the relatively discharged battery.
>>
>>
>> No it won't. The battery that is discharged the most gets the
>> most charge power.
>
>
> In a parallel circuit, yes.
> These batteries are in series, though.
>
>> The lower the charge on a battery, the higher
>> its internal resistance.
>
>
> I don't believe that is true. I don't think
> ESR changes much with state of charge.
>

Battery voltage, measured at the battery posts with a DMM, drops
significantly when a heavy load is applied, demonstrating the
effect of internal resistance.

That voltage drop is significantly lower with a fully charged
battery than it is with a discharged battery.  It is why you
can't start your car when the battery is discharged too far,
yet you can read ~12 volts on the DMM when the starter load
is not present.

Whatever "much" means in your statement "I don't think
ESR changes much with state of charge.", the change in ESR
is the difference in being able to start your car with a
charged battery, and not being able to start it with a
too far discharged battery.

> I think that my 'equivalent resistance'
> of a relatively discharged battery looks
> lower than the 'equivalent resistance'
> of a fully charged battery.  'Equivalent
> resistance' is just ESR plus (the battery
> voltage divided by the charge current).

If that were true, there would be no voltage drop when a
load is applied to the battery, and the voltage drop would
be less with a discharged battery than it would be with
a charged battery when. Exactly the opposite is true.
Electrically, a battery is an ideal voltage source with a series
internal resistance.  A discharged battery has a higher internal
resistance than a charged battery.

>
> Watch the ammeter next time you charge a
> battery and see if charge current starts
> out low and then increases as the battery
> charges, or starts high and then decreases
> with increasing battery charge.

The current starts high and decreases as the battery charges.
That's no mystery when you understand a battery in terms of
an ideal voltage source in series with an internal resistance.

The battery attempts to push current into the charger, and
the charger attempts to push current into the battery. The
charger "wins" because it is at a higher voltage. Current
flowing into the battery from the charger lowers the internal
resistance in the battery. The current that the battery is able
to exert against the charging current increases, due to the
lowered internal resistance. As the battery current in opposition
to the charger current increases, the current from the charger
to the battery decreases.

>
> (Then you say:)
>
>> The .63 ohm equivalent top battery will
>> receive higher power than the .60 ohm equivalent bottom battery.
>> P = I^2 * R, so the top battery gets more.
>
>
> I agree with your third statement but not your
> first or second.
>
> :)
>
>> Seems moot anyway, as the op specified batteries of the same
>> brand, capacity and usage.
>
>
> Note that the OP did not mention state of charge.
> One would not place them in series (or parallel)
> without being sure they were at the same charge level.

Agreed.

My guess is that they are already in series in whatever
device they are part of. Setting my guess aside, there
are devices that use 2 6V SLAs in series and charge them
in series, so it has been done as a practical matter.
Specifically I know of emergency lighting that is configured
that way.

>
> I have two 5 Ah SLA batteries of the same lot number
> from the same manufacturer.  They were both used
> in burglar alarms. One of the batteries measures
> 6.36 V and the other 6.05 V because only the second
> alarm was turned on.  Can I expect these batteries
> to perform to spec if I use them in series?  No.

Agreed.  And in the emergency lighting devices I mentioned
above with two identical SLAs installed at the same time,
one battery usually fails before the other.  I maintain a
bunch of those lights at church.  The batteries last about
5 years, and I replace the pair when they fail. I find
in testing that I can replace one and the other is good.
Putting the bad one back in with the new one, the light still
fails.  (However, I get rid of both old batteries.)

Ed

>
>> If one wants to discuss batteries of
>> different usage/discharge level/specs/whatever, then it is
>> best to independently charge, where the charge can be tailored to
>> the individual battery.
>
>
> I agree. That is the most prudent approach.
>
> It is also moral and correct to match the battery's
> state of charge, then charge them in series or
> parallel using the proper sequence for the
> chemistry in question.
>
> --Winston
```
```On Wed, 21 Dec 2011 12:18:41 +1100, "Phil Allison" <phil_a@tpg.com.au> wrote:

>
><krw@att.bizzzzzzzzzzzz> = psychopath
>>>
>>>> I think he meant to connect them in parallel, sans charger, until they
>>>> leveled off to the same voltage, and then to connect them in series
>>>> and charge them.
>>>>
>>>
>>>** Cannot work.
>>>
>>>The terminal voltage will obviously be the same immediately, but very
>>>little
>>>charging will go on.
>>
>> Dummy, one will charge the other until they equalize.
>
>
>** Capacitors do something like that.
>
>     But not SLAs.

Conservation of energy and Ohm's law don't work in your universe, huh,
Phyllis.
```
```On Tue, 20 Dec 2011 17:58:36 -0700, Jim Thompson
<To-Email-Use-The-Envelope-Icon@On-My-Web-Site.com> wrote:

>On Tue, 20 Dec 2011 18:32:08 -0600, John Fields
><jfields@austininstruments.com> wrote:
>
>>On Wed, 21 Dec 2011 10:34:39 +1100, "Phil Allison" <phil_a@tpg.com.au>
>>wrote:
>>
>>>
>>>"John Fields"
>>>
>>>> I think he meant to connect them in parallel, sans charger, until they
>>>> leveled off to the same voltage, and then to connect them in series
>>>> and charge them.
>>>>
>>>
>>>** Cannot work.
>>>
>>>The terminal voltage will obviously be the same immediately, but very little
>>>charging will go on.
>>
>>---
>>Let's say we have two lead-acid batteries, one of which has been
>>discharged to 12V, and the other to 10V.
>>
>>Let's also say that they both have internal resistances of 0.1 ohm:
>>
>>     12V      11V      10V
>>    /        /        /
>>.  +-[0.1R]-+-[0.1R]-+
>>.  |+       |        |+
>>.[BA1]      Vt     [BA2]
>>.  |        |        |
>>.  +--------+--------+
>>
>>Since the batteries are in parallel, Vt will of course be the same for
>>both batteries, but because of the internal resistances of the
>>batteries and the difference in their internal voltages, there will be
>>an EMF of 2 volts impressed across the sum of the internal
>>resistances, 0.2 ohms, which will cause charge to be transferred from
>>BA1 to BA2 at the rate of 10 coulombs per second, which is 10 amperes.
>>
>>That's quite a bit more than "very little charging", yes?
>
>Is Phyllis really that ignorant?  Or is she just trying to challenge
>Larkin for the title ?:-)
>
>                                        ...Jim Thompson

---
Hey, Jim, why not bury the hatchet?

We all make mistakes and the more gently we're corrected the less
likely we are to go Columbine.

--
JF
```
```On Tue, 20 Dec 2011 20:21:56 -0600, John Fields
<jfields@austininstruments.com> wrote:

>On Tue, 20 Dec 2011 17:58:36 -0700, Jim Thompson
><To-Email-Use-The-Envelope-Icon@On-My-Web-Site.com> wrote:
>
>>On Tue, 20 Dec 2011 18:32:08 -0600, John Fields
>><jfields@austininstruments.com> wrote:
>>
>>>On Wed, 21 Dec 2011 10:34:39 +1100, "Phil Allison" <phil_a@tpg.com.au>
>>>wrote:
>>>
>>>>
>>>>"John Fields"
>>>>
>>>>> I think he meant to connect them in parallel, sans charger, until they
>>>>> leveled off to the same voltage, and then to connect them in series
>>>>> and charge them.
>>>>>
>>>>
>>>>** Cannot work.
>>>>
>>>>The terminal voltage will obviously be the same immediately, but very little
>>>>charging will go on.
>>>
>>>---
>>>Let's say we have two lead-acid batteries, one of which has been
>>>discharged to 12V, and the other to 10V.
>>>
>>>Let's also say that they both have internal resistances of 0.1 ohm:
>>>
>>>     12V      11V      10V
>>>    /        /        /
>>>.  +-[0.1R]-+-[0.1R]-+
>>>.  |+       |        |+
>>>.[BA1]      Vt     [BA2]
>>>.  |        |        |
>>>.  +--------+--------+
>>>
>>>Since the batteries are in parallel, Vt will of course be the same for
>>>both batteries, but because of the internal resistances of the
>>>batteries and the difference in their internal voltages, there will be
>>>an EMF of 2 volts impressed across the sum of the internal
>>>resistances, 0.2 ohms, which will cause charge to be transferred from
>>>BA1 to BA2 at the rate of 10 coulombs per second, which is 10 amperes.
>>>
>>>That's quite a bit more than "very little charging", yes?
>>
>>Is Phyllis really that ignorant?  Or is she just trying to challenge
>>Larkin for the title ?:-)
>>
>>                                        ...Jim Thompson
>
>---
>Hey, Jim, why not bury the hatchet?
>
>We all make mistakes and the more gently we're corrected the less
>likely we are to go Columbine.

I _would_ like a statement from Larkin that he's withdrawing from
cluck-cluck-cluck, then I'll bury the hatchet... like the butcher in
"Boardwalk" ;-)

...Jim Thompson
--
| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    |
| Phoenix, Arizona  85048    Skype: Contacts Only  |             |
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  |
| E-mail Icon at http://www.analog-innovations.com |    1962     |

I love to cook with wine.     Sometimes I even put it in the food.
```
```Jim Thompson wrote:

(...)

> An amusing thought, make a smart charger that bypasses some charging
> current around the "better" unit until leveling is accomplished.
>
> Solution left as an exercise for the student :-)

You are joshing, but I wonder if the electrical
characteristics of a relatively discharged
battery vs those of a charged battery could be
used to that end.  Ferinstance, is there a
pulse waveform that would be ignored by the
charged battery but would tend to charge the
'flatter' battery connected to it in series?

Hrm.

--Winston
```
```On Dec 20, 9:35=A0pm, Winston <Wins...@BigBrother.net> wrote:
> NT wrote:
> > On Dec 20, 2:53 pm, Winston<Wins...@BigBrother.net> =A0wrote:
> >> NT wrote:
>
> (...)
>
> >> Which is unfortunate because, in our thought
> >> experiment, the top battery requires no more
> >> I or T and the bottom battery requires nearly
> >> a full dose of both I and T if it is to perform
> >> to spec and indeed, survive over the long haul.
>
> > /If/ the batteries are discharged asymmetrically, one will eventually
> > need water adding. That's all. Normal maintenance.
>
> I don't believe the 'bottom' battery would survive.
> Stored in a consistently discharged state, it would
> sulfate and fail.

Since it would get i x t at every charge, it would not be stored in a
consistently discharged state.

NT
```
```ehsjr wrote:
> Winston wrote:

(...)

> Whatever "much" means in your statement "I don't think
> ESR changes much with state of charge."

In a chart referenced below, ESR does increase by 71%
from full charge to well flattened.  At it's highest,
though, it is only 15 milliohms!  More about this later.

(...)

> the change in ESR
> is the difference in being able to start your car with a
> charged battery, and not being able to start it with a
> too far discharged battery.
>
>> I think that my 'equivalent resistance'
>> of a relatively discharged battery looks
>> lower than the 'equivalent resistance'
>> of a fully charged battery. 'Equivalent
>> resistance' is just ESR plus (the battery
>> voltage divided by the charge current).
>
> If that were true, there would be no voltage drop when a
> load is applied to the battery, and the voltage drop would
> be less with a discharged battery than it would be with
> a charged battery when. Exactly the opposite is true.
> Electrically, a battery is an ideal voltage source with a series
> internal resistance. A discharged battery has a higher internal
> resistance than a charged battery.

http://batteryuniversity.com/learn/article/how_does_internal_resistance_affect_performance

We see a chart that shows a wheelchair SLA internal
resistance as a function of state - of - charge.
(Chart titled "Changing Resistance with SoC".)

I see that the internal resistance *does*
increase as the charge depletes.  I also see that
it changes from 8.75 milliohms at full charge
up to 15 milliohms with a well flattened battery.

Ohms law says that 11.8 V and 15 milliohms would
still allow the battery to source 120 amps into
a load requiring 10 V. Clearly this battery is not
going to do anything of the sort, so we have to
look for an additional series resistance if we
are going to create a good arithmetic model.

That is what I hoped to do by adding my
"Equivalent Resistance" to ESR.

(...)

> As the battery current in opposition
> to the charger current increases, the current from the charger
> to the battery decreases.

That is a way of looking at it, I guess.

I could substitute an electrolytic capacitor for
the battery and see that charge current starts
high and then diminishes, too.  It is because
of the diminishing relative voltage, not the
increasing current of the capacitor opposing that
of the charger, though.  I conjecture the same
is true for the battery.

You appear to agree because you say:
"The charger "wins" because it is at a higher voltage."

(...)

> My guess is that they are already in series in whatever
> device they are part of.

I didn't get that from what the OP stated.

> Setting my guess aside, there
> are devices that use 2 6V SLAs in series and charge them
> in series, so it has been done as a practical matter.
> Specifically I know of emergency lighting that is configured
> that way.

If capacity and state of charge are matched,
one can use SLA batteries in series or parallel.

>> I have two 5 Ah SLA batteries of the same lot number
>> from the same manufacturer. They were both used
>> in burglar alarms. One of the batteries measures
>> 6.36 V and the other 6.05 V because only the second
>> alarm was turned on. Can I expect these batteries
>> to perform to spec if I use them in series? No.
>
> Agreed. And in the emergency lighting devices I mentioned
> above with two identical SLAs installed at the same time,
> one battery usually fails before the other.

This tends to support my earlier statement about no
two batteries aging in exactly the same way and the
attendant 'death spiral' when one of the series
batteries can no longer charge at the same rate as
it's partner.

> I maintain a
> bunch of those lights at church. The batteries last about
> 5 years, and I replace the pair when they fail. I find
> in testing that I can replace one and the other is good.
> Putting the bad one back in with the new one, the light still
> fails. (However, I get rid of both old batteries.)

Could one double the service life of these batteries
by re-hydrating them say, annually?

http://www.instructables.com/id/Refilling-SLAs-Sealed-Lead-Acid-battery/

Thought-provoking, wot?  :)

--Winston
```