James Hahn wrote:

"Meindert Sprang" wrote

I start to charge the set in series, so the current through both
batteries is exactly the same.

No. You are failing to understand that the charging process is a chemical
reaction that converts energy into a differnt form.. The current flowing
at the positive and negative terminal of the whole string of cells is the
same, obviously,

Yes, indeed, obviously. And equally obviously the current flowing at the
positive and negative terminal of each individual cell in the string is
also the same. Surely you would not be daft enough to deny that!

but the current that flows 'through' any cell is a
complex result of the conversion processes occurring inside the cells of
which it is a string.

What exactly happens at the microscopic level may well be complex and
diverse, but the current passing through any surface you'd care to cut
through any cell will be the same.

Each cell can be considered an energy sink. The amount of energy
imparted to each cell will differ, either within a single battery or
within a series of batteries.

This is not in dispute.

The current flowing across any cell, or any group of cells, is most
definately not equal, and there is no reason it should be.

There is every reason it should be. It makes no difference whether
the load seen by a charging device is a string of resistors or of
battery cells, the current at every point in the circuit must be
equal at all times.

There is, however, no reason why the amount of energy absorbed by
each cell during charging should be the same.

One battery will reach the full state before the other but is still being
charged with full current because the other battery hasn't reached the
voltage that corresponds with full charge.

One battery wil reach the full state before the other, like one cell will
reach the full state before some others. It's not a problem.

It is a problem, because the fuller battery's internal voltage will
rise sooner than that of the not-so-full one, and so it will continue
to take a higher share of the monitored charging voltage applied to the
string. This not only means the less full one will take longer to
reach full charge but also that it's being charged at a suboptimal
voltage and so may never reach proper full charge, while the fuller
battery is being overcharged leading to a shorter life.

It is not
being charged with full current because the 'load' that the system
presents to the charging process is reduced by the reduced charging
requirements of the cell (or cells) that have reached or are approaching
full charge.

"Ronald Raygun" wrote in message
k...
James Hahn wrote:

"Meindert Sprang" wrote

I start to charge the set in series, so the current through both
batteries is exactly the same.

No. You are failing to understand that the charging process is a
chemical
reaction that converts energy into a differnt form.. The current
flowing
at the positive and negative terminal of the whole string of cells is
the
same, obviously,

Yes, indeed, obviously. And equally obviously the current flowing at the
positive and negative terminal of each individual cell in the string is
also the same. Surely you would not be daft enough to deny that!

but the current that flows 'through' any cell is a
complex result of the conversion processes occurring inside the cells of
which it is a string.

What exactly happens at the microscopic level may well be complex and
diverse, but the current passing through any surface you'd care to cut
through any cell will be the same.

Each cell can be considered an energy sink. The amount of energy
imparted to each cell will differ, either within a single battery or
within a series of batteries.

This is not in dispute.

The current flowing across any cell, or any group of cells, is most
definately not equal, and there is no reason it should be.

There is every reason it should be. It makes no difference whether
the load seen by a charging device is a string of resistors or of
battery cells, the current at every point in the circuit must be
equal at all times.

There is, however, no reason why the amount of energy absorbed by
each cell during charging should be the same.

One battery will reach the full state before the other but is still
being
charged with full current because the other battery hasn't reached the
voltage that corresponds with full charge.

One battery wil reach the full state before the other, like one cell
will
reach the full state before some others. It's not a problem.

It is a problem, because the fuller battery's internal voltage will
rise sooner than that of the not-so-full one, and so it will continue
to take a higher share of the monitored charging voltage applied to the
string. This not only means the less full one will take longer to
reach full charge but also that it's being charged at a suboptimal
voltage and so may never reach proper full charge, while the fuller
battery is being overcharged leading to a shorter life.

It is not
being charged with full current because the 'load' that the system
presents to the charging process is reduced by the reduced charging
requirements of the cell (or cells) that have reached or are approaching
full charge.

So how does all this techno babble answer the OP's question I wonder? :-)

ChrisR wrote:

So how does all this techno babble answer the OP's question I wonder? :-)

Well, it doesn't directly. The OP *should* know by now however, that CN
and all his drivel should be studiously ignored. But, to be on-topic, I
would suggest the following:

1. If the 12V device draws minimal current, and/or is to be in use only
sporadically, then tap one battery and don't worry about it. If you're
using a couple of amp-hours per day, and recharge fairly frequently, the
diffential in the batteries will be, IMO, negligible.

2. If the 12V device (VHF IIRC) is a significant load, you can always
build a simple voltage divider (the VHF should be input voltage tolerant
enough to handle the voltage sag during transmission) and run from both
batteries. The overall battery draw will, of course, be somewhat larger
since you'll dissipate heat in the dropping resistor. If multiple
devices are used, however, this approach quickly becomes problematic.

Keith

On Thu, 21 Apr 2005 09:22:27 -0700, Keith Hughes
wrote:


2. If the 12V device (VHF IIRC) is a significant load, you can always
build a simple voltage divider (the VHF should be input voltage tolerant
enough to handle the voltage sag during transmission) and run from both
batteries. The overall battery draw will, of course, be somewhat larger
since you'll dissipate heat in the dropping resistor. If multiple
devices are used, however, this approach quickly becomes problematic.

Keith

A series voltage dropping resistor may work for some devices where the
current draw is constant, but it _will not_ work for a VHF radio. The
radio will only draw a few hundred mA on receive when squelched,
somewhat more when actually receiving, and a few amps when
transmitting - there is no way a simple resistor can keep the supply
voltage to the radio within acceptable bounds with that current
variation.



--
Peter Bennett VE7CEI
email: peterbb4 (at) interchange.ubc.ca
GPS and NMEA info and programs: http://vancouver-webpages.com/peter/index.html
Newsgroup new user info: http://vancouver-webpages.com/nnq

Peter Bennett wrote:

On Thu, 21 Apr 2005 09:22:27 -0700, Keith Hughes
wrote:


2. If the 12V device (VHF IIRC) is a significant load, you can always
build a simple voltage divider (the VHF should be input voltage tolerant
enough to handle the voltage sag during transmission) and run from both
batteries. The overall battery draw will, of course, be somewhat larger
since you'll dissipate heat in the dropping resistor. If multiple
devices are used, however, this approach quickly becomes problematic.

A series voltage dropping resistor may work for some devices where the
current draw is constant, but it _will not_ work for a VHF radio. The
radio will only draw a few hundred mA on receive when squelched,
somewhat more when actually receiving, and a few amps when
transmitting - there is no way a simple resistor can keep the supply
voltage to the radio within acceptable bounds with that current
variation.

Keith mentioned a voltage divider. That's not a simple resistor.

Even a voltage divider is unlikely to be very satisfactory either,
since in order to keep the voltage within acceptable range for the
VHF, the resistance values will probably need to be so low, you'd
be dissipating as much power in the divider all the time as the
VHF on full-power transmit.

One possibility might be to use a zener diode in place of a simple
series resistor.

At the end of the day, having different-voltage devices aboard is
always going to be a pig's breakfast unless you have proper
fully independent systems, with two alternators, one for charging
the 24V batteries, and one for the 12V ones, or a charge controller
capable of charging a 12V battery from the 24V alternator.

On Thu, 21 Apr 2005 21:46:09 GMT, Ronald Raygun
wrote:

Peter Bennett wrote:

On Thu, 21 Apr 2005 09:22:27 -0700, Keith Hughes
wrote:


2. If the 12V device (VHF IIRC) is a significant load, you can always
build a simple voltage divider (the VHF should be input voltage tolerant
enough to handle the voltage sag during transmission) and run from both
batteries. The overall battery draw will, of course, be somewhat larger
since you'll dissipate heat in the dropping resistor. If multiple
devices are used, however, this approach quickly becomes problematic.

A series voltage dropping resistor may work for some devices where the
current draw is constant, but it _will not_ work for a VHF radio. The
radio will only draw a few hundred mA on receive when squelched,
somewhat more when actually receiving, and a few amps when
transmitting - there is no way a simple resistor can keep the supply
voltage to the radio within acceptable bounds with that current
variation.

Keith mentioned a voltage divider. That's not a simple resistor.

Even a voltage divider is unlikely to be very satisfactory either,
since in order to keep the voltage within acceptable range for the
VHF, the resistance values will probably need to be so low, you'd
be dissipating as much power in the divider all the time as the
VHF on full-power transmit.

One possibility might be to use a zener diode in place of a simple
series resistor.

At the end of the day, having different-voltage devices aboard is
always going to be a pig's breakfast unless you have proper
fully independent systems, with two alternators, one for charging
the 24V batteries, and one for the 12V ones, or a charge controller
capable of charging a 12V battery from the 24V alternator.


Come on guys, get back to the proper way to do the job and the
simplest. The 24 to 12 volt converter.

Resistor dividers will draw more power than the radio. A zener has to
be very large for the job at hand. Again it will consume as much power
as the radio.

regards
Gary

Gary Schafer wrote:
On Thu, 21 Apr 2005 21:46:09 GMT, Ronald Raygun
wrote:


Peter Bennett wrote:


On Thu, 21 Apr 2005 09:22:27 -0700, Keith Hughes
wrote:



2. If the 12V device (VHF IIRC) is a significant load, you can always
build a simple voltage divider (the VHF should be input voltage tolerant
enough to handle the voltage sag during transmission) and run from both
batteries. The overall battery draw will, of course, be somewhat larger
since you'll dissipate heat in the dropping resistor. If multiple
devices are used, however, this approach quickly becomes problematic.

A series voltage dropping resistor may work for some devices where the
current draw is constant, but it _will not_ work for a VHF radio. The
radio will only draw a few hundred mA on receive when squelched,
somewhat more when actually receiving, and a few amps when
transmitting - there is no way a simple resistor can keep the supply
voltage to the radio within acceptable bounds with that current
variation.

I did not recall, for sure, what the device was. That's why I hedged.
Whatever the device, you'd have to look at the full load/idle ratio and
see if a voltage divider makes sense. In the OP's case, I'm not sure.

Keith mentioned a voltage divider. That's not a simple resistor.

Even a voltage divider is unlikely to be very satisfactory either,
since in order to keep the voltage within acceptable range for the
VHF, the resistance values will probably need to be so low, you'd
be dissipating as much power in the divider all the time as the
VHF on full-power transmit.

One possibility might be to use a zener diode in place of a simple
series resistor.

At the end of the day, having different-voltage devices aboard is
always going to be a pig's breakfast unless you have proper
fully independent systems, with two alternators, one for charging
the 24V batteries, and one for the 12V ones, or a charge controller
capable of charging a 12V battery from the 24V alternator.


True, but do-able. And relatively cheaply compared to off-the-shelf
solutionsa.

Come on guys, get back to the proper way to do the job and the
simplest. The 24 to 12 volt converter.

I would certainly agree. Shouldn't be too hard to find a 12VDC output,
12-12VDC regulated power supply.


Resistor dividers will draw more power than the radio. A zener has to
be very large for the job at hand. Again it will consume as much power
as the radio.

Hmmm...where have I hear that before??? :-)

Keith

In article ,
Gary Schafer wrote:

Come on guys, get back to the proper way to do the job and the
simplest. The 24 to 12 volt converter.

Resistor dividers will draw more power than the radio. A zener has to
be very large for the job at hand. Again it will consume as much power
as the radio.

regards
Gary

As I posted very early on, one could use a 24-36Vdc/12Vdc Switching
Converter of an appropriate current rating, to charge a small 12Vdc
Group 12, AGM, or Gellcell battery, and then run a 12Vdc buss for all
the loads aboard that require that range of power. This is entirely
consistant with The Fishing Vessel Safety Act, and SOLAS Requirments
for Reserve Power for Required Communications Systems. A lot of the
older Coastal Freighters af less than 1600 Tons that trade in the
gulf of Mexico, and in the North Pacific and Gulf of Alaska, were
32Vdc powered vessels, and the SOLAS requirments were meet using this
type of system.


Bruce in alaska
--
add a 2 before @

On Thu, 21 Apr 2005 11:07:23 UTC, Ronald Raygun
wrote:

: What exactly happens at the microscopic level may well be complex and
: diverse, but the current passing through any surface you'd care to cut
: through any cell will be the same.

Is that strictly true? When a cell is charging, with electrochemical
processes taking place at the plates, does the charging current still
exist between them? It wouldn't in a capacitor, though I suppose the
displacemet current would even up the score a bit there.

Ian


--

Ian Johnston wrote:

On Thu, 21 Apr 2005 11:07:23 UTC, Ronald Raygun
wrote:

: What exactly happens at the microscopic level may well be complex and
: diverse, but the current passing through any surface you'd care to cut
: through any cell will be the same.

Is that strictly true?

Yes, I very much believe so, and know of no reason why it should not be.

When a cell is charging, with electrochemical
processes taking place at the plates, does the charging current still
exist between them? It wouldn't in a capacitor, though I suppose the
displacemet current would even up the score a bit there.

That's because in a capacitor the dielectric replaces the electrolyte,
and the energy is stored in the form of an electric field created by
the physical separation of "real" (electric) charge.

In a battery no electric charge is stored at all (to speak of) and
energy is stored by chemical changes to plates and electrolyte.

Out of every two SO4-- ions in any drop of electrolyte, on average one
travels to each plate, but all four corresponding H+ ions travel only
to the PbO2 plate. As I tried to explain yesterday by considering the
drops laid end to end in series, this means that for each two electrons
travelling left to right on the outside, the charge balance in each drop
of electrolyte is neutral (-2 out to the left, -2 out to the right, +4
out to the right) and at each inter-drop interface this means charge
travels either -2 left or +2 right, plus self-balancing through traffic.

Hence charge travels on the inside as well as on the outside.
Where charge travels, that's current.