# Charge 2 x 6V on 12V Charger

Started by December 18, 2011
```On Wed, 21 Dec 2011 09:27:00 +1100, "Phil Allison" <phil_a@tpg.com.au>
wrote:

>
>"Jim Thompson"
>>
>> Connect the two 6V batteries in parallel until equalized, then charge
>> in series.
>
>
>** The OP does not have a 6 volt SLA charger -  he has a 12 volt one.
>
>Most SLA chargers work as a current limited PSU with a fixed voltage limit
>of 13.7 volts for a nominal 12 volt battery.  13.7 volts is carefully chosen
>so 6 fully charged cells will draw only a small current -  the terminal
>voltage must be raised to 15 or 16 volts to overcharge the battery.
>
>When a discharged battery is connected, charging commences at the current
>limit and as voltage rises it gradually tapers off until at full at charge
>( ie 13.7 volts ) it is only a trickle.  The limit current value depends on
>the amp hour rating of the cells so charging is completed in about 2 to 4
>hours.
>
>So, if you connect the wrong voltage battery ( 6V instead of 12V ) high
>current will flow until the battery explodes  - the better designs recognise
>when the voltage across the battery is dangerously low and refuse to apply
>charge.
>
>The interesting question is what happens if one or more cells in a battery
>are in a much lower state of charge when connected to the charger.
>
>If it is only one, charging will drop to a trickle with that cell not fully
>charged. This is because 5 fully charged cells require 13 volts to be
>applied to pass significant charging current and the remaining cell requires
>about 2 volts. The total needed of 15 volts is simply not available.
>
>A similar scenario applies for 2 out of 6.
>
>However, if it is 3 out of six charging will still continue at a slow rate
>until all 6 cells are topped up.
>
>
>
>...   Phil

---
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.

--
JF
```
```Jim Thompson wrote:

(...)

> Connect the two 6V batteries in parallel until equalized, then charge
> in series.

Most of the time, we would 'get away' with
doing that.  Ideally, one would measure
the charge on each battery first, to assure
they were within a few tens of millivolts
of each other before strapping them together.

If not, a C/1 resistor in series would
limit the current between them until they
were safely within 'paralleling' range.

--Winston
```
```Phil Allison wrote:
> "Winston"
>>
>> 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
>
> ** Should be " charge " -  right ?

The arithmetic works the same way for the
first few milliseconds of charging or
discharging at 10 A. R=E/I, in theory.

At a 10 A charge however, the user can
be fooled by the higher 'surface charge'
voltage measured on the cell, thus
overestimating it's 'equivalent resistance'.

There is also the danger of underestimating
the 'equivalent resistance' due to the
current flow, especially with smaller capacity
batteries.  Hey, it is just a thought
experiment meant to reveal the danger of
connecting mismatched batteries.

--Winston
```
```On Tue, 20 Dec 2011 16:46:47 -0600, John Fields
<jfields@austininstruments.com> wrote:

>On Wed, 21 Dec 2011 09:27:00 +1100, "Phil Allison" <phil_a@tpg.com.au>
>wrote:
>
>>
>>"Jim Thompson"
>>>
>>> Connect the two 6V batteries in parallel until equalized, then charge
>>> in series.
>>
>>
>>** The OP does not have a 6 volt SLA charger -  he has a 12 volt one.
>>
>>Most SLA chargers work as a current limited PSU with a fixed voltage limit
>>of 13.7 volts for a nominal 12 volt battery.  13.7 volts is carefully chosen
>>so 6 fully charged cells will draw only a small current -  the terminal
>>voltage must be raised to 15 or 16 volts to overcharge the battery.
>>
>>When a discharged battery is connected, charging commences at the current
>>limit and as voltage rises it gradually tapers off until at full at charge
>>( ie 13.7 volts ) it is only a trickle.  The limit current value depends on
>>the amp hour rating of the cells so charging is completed in about 2 to 4
>>hours.
>>
>>So, if you connect the wrong voltage battery ( 6V instead of 12V ) high
>>current will flow until the battery explodes  - the better designs recognise
>>when the voltage across the battery is dangerously low and refuse to apply
>>charge.
>>
>>The interesting question is what happens if one or more cells in a battery
>>are in a much lower state of charge when connected to the charger.
>>
>>If it is only one, charging will drop to a trickle with that cell not fully
>>charged. This is because 5 fully charged cells require 13 volts to be
>>applied to pass significant charging current and the remaining cell requires
>>about 2 volts. The total needed of 15 volts is simply not available.
>>
>>A similar scenario applies for 2 out of 6.
>>
>>However, if it is 3 out of six charging will still continue at a slow rate
>>until all 6 cells are topped up.
>>
>>
>>
>>...   Phil
>
>---
>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.

Yep.  But Phyllis was too dense to grasp the leveling scheme ;-)

...Jim Thompson
--
| James E.Thompson, CTO                            |    mens     |
| 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.
```
```On Tue, 20 Dec 2011 14:51:09 -0800, Winston <Winston@BigBrother.net>
wrote:

>Jim Thompson wrote:
>
>(...)
>
>> Connect the two 6V batteries in parallel until equalized, then charge
>> in series.
>
>Most of the time, we would 'get away' with
>doing that.  Ideally, one would measure
>the charge on each battery first, to assure
>they were within a few tens of millivolts
>of each other before strapping them together.
>
>If not, a C/1 resistor in series would
>limit the current between them until they
>were safely within 'paralleling' range.
>
>--Winston

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 :-)

...Jim Thompson
--
| James E.Thompson, CTO                            |    mens     |
| 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.
```
```"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.

...   Phil

```
```"Jim Thompson"

> Yep.  But Phyllis was too dense to grasp the leveling scheme ;-)

** Cos it is bullshit.

```
```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.

Dummy, one will charge the other until they equalize.  The down side is that
large currents are possible.
```
```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?

--
JF
```
```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
--
| James E.Thompson, CTO                            |    mens     |