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

On Thu, 21 Apr 2005 02:23:40 GMT, "James Hahn"
wrote:

"Meindert Sprang" wrote in message
...
snip

The charge of a battery is the product of current x time.
True, but irrelevant

Both batteries are in series and one load is connected to the set,
operating
at 24V. Another load is connected across only one battery, operating at
12V.
So it is evident that one battery is discharged more than the other.
Yes, but you don't need the load - batteries will always have different
charge levels in individual cells, whether in the same case or not. It's
easy to measure the differences.

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

Since one battery is discharged more than the other and
the current throug both is the same, one battery must be charged longer
that
the other.
No. All cells are charged for as long as the charge is applied. It makes
no sense to say that one is charged longer than the other. Perhaps you
meant to say that some will continue to have the full potential applied when
they are not accepting additional energy - this is normal in any charging
process and causes no problem for the battery. It is, in fact, exactly what
happens when any battery is fully charged and a maintaining charge is being
applied. It is no different for the battery as a whole as it is for the
individual cells.

Exactly how am I going to achive that with the same current
through both batteries?
You do not achieve that, and you don't need to.

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 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. The current
that can be absorbed by a bank of cells depends on the charge state of the
cells in the string, not the charge state of the most discharged cell.
That's why I said before that the system 'monitors' the state of the string
of cells, not individual cells. If the charge rate was controlled by the
most discharged cell then modern lead acid batteries, with typical
manufacturing differences, wouldn't last any time at all.



Current is always equal at any point in a series circuit. All cells
are in series therefore current is the same through each cell.

What does change is the resistance of each cell as it charges. The
more charge a cell accumulates the higher it's resistance becomes.
That makes the current through it decrease. But as the current
decreases through that cell so does it decrease by an equal amount
through all the other cells.

As resistance increases in a particular cell so does the voltage
across it increase.

Total current decreases as cells charge if charge voltage is held
constant because cell resistance increases. Resistance can increase
more in one cell than in another but total current will be still be
equal in all cells.

It comes down to good old ohms law. Applied voltage divided by the sum
of the resistance of each cell equals total current.

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

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


--

On Thu, 21 Apr 2005 22:33:12 UTC, Gary Schafer
wrote:

: It comes down to good old ohms law. Applied voltage divided by the sum
: of the resistance of each cell equals total current.

Only if you define resistance that way. For non-linear circuit
elements it's a pretty meaningless thing to say.

Ian

"Ian Johnston" wrote in message
news:cCUlhtvFIYkV-pn2-1yXN8Ihw3NKL@localhost...
On Thu, 21 Apr 2005 22:33:12 UTC, Gary Schafer
wrote:

: It comes down to good old ohms law. Applied voltage divided by the sum
: of the resistance of each cell equals total current.

Only if you define resistance that way. For non-linear circuit
elements it's a pretty meaningless thing to say.

Ian

Absolutely. And if we dare mention in this thread that internal resistance
of a battery is dI/dV (rather than I/V), then god knows what chaotic
argument that will create amongst the cognoscenti who think that a battery
is a big bucket full of electrons!

"Pete Styles" wrote in message
...

"Ian Johnston" wrote in message
news:cCUlhtvFIYkV-pn2-1yXN8Ihw3NKL@localhost...
On Thu, 21 Apr 2005 22:33:12 UTC, Gary Schafer
wrote:

: It comes down to good old ohms law. Applied voltage divided by the sum
: of the resistance of each cell equals total current.

Only if you define resistance that way. For non-linear circuit
elements it's a pretty meaningless thing to say.

Ian

Absolutely. And if we dare mention in this thread that internal
resistance of a battery is dI/dV (rather than I/V), then god knows what
chaotic argument that will create amongst the cognoscenti who think that a
battery is a big bucket full of electrons!

WHOOPS - meant dV/dI and V/I, sorry!

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.