NiMH again

Archived from groups: rec.photo.digital (More info?)

I followed the recent thread on NiMH batteries and their charge/discharge
with interest, but am still unsure where the magic value of 0.9V comes
from. But I have a theory:

Consider the widely used series connected set of four cells. The total
voltage generated is 4.8 volts, based on an individual cell voltage of 1.2
volts. As those cells discharge, the weakest (lowest capacity) cell will be
the first to drop its voltage, and could drop to zero volts while the other
three cells remain at 1.2 volts. The total voltage is then 3.6 volts.

This 3.6 volts is the thus the maximum voltage across the four series cells
which guarantees that none of the cells is driven into electrode reversal.
Any lower than 3.6 volts and there could be cell damage.

So this 3.6 volts across four cells is considered as 0.9 volts per cell, and
the 0.9 volts becomes known as the value below which damage could occur. If
my theory is correct, 0.9 volts is an irrelevant point in the discharge
characteristic of a single cell, but 0 volts is important because the cell
voltage must not be reversed. So a single cell can be safely deep
discharged by giving it a simple resistive load and letting it run down to
0 volts.

Comments welcome, because I'm no expert and this is just a theory. If anyone
can direct me to authoratative information I'd be grateful.

Archived from groups: rec.photo.digital (More info?)

Okay. Let's put this into the real world. If you put 4 batteries into
something, and one goes dead while the others are still pumping out 1.2
volts, you've got a bad battery to begin with whether it goes into reverse
polarity or not.


"Kit" <kit@newsgroupsonlyyyy.com> wrote in message
news:431c8e8e_2@mk-nntp-2.news.uk.tiscali.com...
>I followed the recent thread on NiMH batteries and their charge/discharge
> with interest, but am still unsure where the magic value of 0.9V comes
> from. But I have a theory:
>
> Consider the widely used series connected set of four cells. The total
> voltage generated is 4.8 volts, based on an individual cell voltage of 1.2
> volts. As those cells discharge, the weakest (lowest capacity) cell will
> be
> the first to drop its voltage, and could drop to zero volts while the
> other
> three cells remain at 1.2 volts. The total voltage is then 3.6 volts.
>
> This 3.6 volts is the thus the maximum voltage across the four series
> cells
> which guarantees that none of the cells is driven into electrode reversal.
> Any lower than 3.6 volts and there could be cell damage.
>
> So this 3.6 volts across four cells is considered as 0.9 volts per cell,
> and
> the 0.9 volts becomes known as the value below which damage could occur.
> If
> my theory is correct, 0.9 volts is an irrelevant point in the discharge
> characteristic of a single cell, but 0 volts is important because the cell
> voltage must not be reversed. So a single cell can be safely deep
> discharged by giving it a simple resistive load and letting it run down to
> 0 volts.
>
> Comments welcome, because I'm no expert and this is just a theory. If
> anyone
> can direct me to authoratative information I'd be grateful.
>
>
>
>

Archived from groups: rec.photo.digital (More info?)

On Mon, 05 Sep 2005 19:40:32 +0100, Kit wrote:

> Comments welcome, because I'm no expert and this is just a theory.
> If anyone can direct me to authoratative information I'd be grateful.

Interesting theory, but wrong, unfortunately. Rework it to see if
you'd still get 0.9 volts if 2, 3, 6 or 8 cells are used. These
numbers of cells should produce values ranging from 0.6 to 1.05
volts And even if the cells are not matched very well, by the time
one cell has lost its charge the others are probably well below 1.1
volts, not still near the 1.2 volts of a fully charged cell. The
only way a manufacturer would assume a high number such as 1.2 volts
would be if they knew people would be foolish enough to throw any
batteries on hand into the camera, where one of them would initially
be almost dead. But they don't make that assumption. They always
state that batteries should not be mixed, that they all be of the
same type and only fresh batteries by used. If those warnings are
heeded, most of the batteries will be close to dead when the first
cell gives up the ghost.

Actually, most (all that I've seen) devices that discharge
batteries, whether NiCad or NiMH terminate the discharge at 1.0
volts. At this point there is hardly any energy remaining in the
cell. It may take hours to drop from 1.1 volts to 1.0, but only
seconds, or less to drop from 1.0 to 0.9 volts.

Archived from groups: rec.photo.digital (More info?)

"Glittery Gary" <OnThe@Upbtinternet.com> wrote in message
newsfi2uu$75k$1@nwrdmz02.dmz.ncs.ea.ibs-infra.bt.com...
>
[ . . . ]
>>
> yaawwwwwwnnnnnnnnnnnnnnnn

It's only boring to you because it a) is over your head, and b) has no
pictures.

Stay with comic books.

N.

Archived from groups: rec.photo.digital (More info?)

papenfussDIESPAM@juneauDOTmeDOTvt.edu wrote:

> Scouring the 'net awhile back I was
> unable to come up with any *real* reason why one couldn't discharge
> individual cells
> to 0V. The 0.9V/cell seems to be a commonly-known "rule-of-thumb," is all
> and generally prevents reversing cells, as you explained.

Yes, this is the only reason I could find.



> Of course, whether is *helps* anymore is another question. Most likely
> not. So, as long as your battery charger/discharger charges each cell
> individually, you shouldn't have to worry about it.


Agreed. However I read on this ng of people making their own dischargers
with 0.95V references etc, and suspect that a simple resistor across a cell
could do the same job, albeit without the fun of designing and building
your own box of electronics.

Having said that, I notice that, for example, the Maxim DS2711/DS2712
contains a finite state machine with one state dedicated to the gentle
charge of a cell if it has a potential of less than 1.0 volts. So avoiding
a complete discharge to 0V would remove the need to be gentle at the start
of charge, which maybe some chargers don't do.

Archived from groups: rec.photo.digital (More info?)

Nostrobino wrote:

>
>
> Somewhere I'd read that 1.0 V was the "specified voltage" at which a cell
> was considered discharged and should not be drained further. Too long ago
> for me to remember the details or source, though.
>
> N.

Well if I could find that source, and consider it authoritative, then I
could get a life again :-)

Archived from groups: rec.photo.digital (More info?)

Kit wrote:
> papenfussDIESPAM@juneauDOTmeDOTvt.edu wrote:
>
>> Scouring the 'net awhile back I was
>> unable to come up with any *real* reason why one couldn't discharge
>> individual cells
>> to 0V. The 0.9V/cell seems to be a commonly-known "rule-of-thumb,"
>> is all and generally prevents reversing cells, as you explained.
>
> Yes, this is the only reason I could find.
>
>
>
>> Of course, whether is *helps* anymore is another question. Most
>> likely not. So, as long as your battery charger/discharger charges
>> each cell individually, you shouldn't have to worry about it.
>
>
> Agreed. However I read on this ng of people making their own
> dischargers with 0.95V references etc, and suspect that a simple
> resistor across a cell could do the same job, albeit without the fun
> of designing and building your own box of electronics.

IF you add this resistor one analog quartz clock ( one running on one 1,5 V
AA battery), you will also get simple capacity meter. Using 1 ohm resistor
will give you as many hours clock will run, so many Ah battery has. Just
charge battery, set slock to 12 and go...

>
> Having said that, I notice that, for example, the Maxim DS2711/DS2712
> contains a finite state machine with one state dedicated to the gentle
> charge of a cell if it has a potential of less than 1.0 volts. So
> avoiding a complete discharge to 0V would remove the need to be
> gentle at the start of charge, which maybe some chargers don't do.

You could add a diode in series with resistor, but in this case note that
voltage across resistor will be 1.2 volts minus 0.7 = 0.5 V and so resistor
will have to be smaller. But you will get cut-off function at appr. 0.7 V.

Archived from groups: rec.photo.digital (More info?)

> Well if I could find that source, and consider it authoritative, then I
> could get a life again :-)

Check this: http://www.thomas-distributing.com/techfacts3.htm

1 Volt seems much more reasonable than 0.9 V.

Found the same curve at several locations via Google, so I guess its
authoritative?

Archived from groups: rec.photo.digital (More info?)

Charles Schuler wrote:
> > Well if I could find that source, and consider it authoritative, then I
> > could get a life again :-)
>
> Check this: http://www.thomas-distributing.com/techfacts3.htm
>
> 1 Volt seems much more reasonable than 0.9 V.
>
> Found the same curve at several locations via Google, so I guess its
> authoritative?

Looking at that curve, it seems there is little difference whether one
specifies 0.9V or 1.0V. At that point, the voltage is dropping so fast
that 0.9V is reached within a minute after 1V.

Mark

Archived from groups: rec.photo.digital (More info?)

Kit wrote:
> SleeperMan wrote:
>
>
>> You could add a diode in series with resistor, but in this case note
>> that voltage across resistor will be 1.2 volts minus 0.7 = 0.5 V and
>> so resistor will have to be smaller. But you will get cut-off
>> function at appr. 0.7 V.
>
> Are you psychic? I did this a couple of days ago, a 1N5401 and 1 ohm
> wirewound. Cell eventually settles at around 650mV. It was while I was
> doing this that I concocted my original post.

yep. Quite simple and effective. Add a clock and ...voila!

Archived from groups: rec.photo.digital (More info?)

ASAAR wrote:


> Yes, they were once built with discrete components and were fairly
> expensive. You could see the concept in what are called "switching
> power supplies", often used in computers. Now there are
> inexpensive I.C.s that do the job, essentially containing all of the
> components except for perhaps some power transistors. They're more
> efficient too, as the run-of-the-mill transformers tend to be very
> lossy (they get hot). They also require much less space and weight
> than transformers do.

A switching power regulator generally still requires some magnetic
component, either an inductor or transformer, but because they run at many
hundreds of kilohertz these days, that inductor or transformer will be low
value and small size.

Archived from groups: rec.photo.digital (More info?)

"Nostrobino" <not@home.today> wrote in message
news:LJGdnWw0r8jFToDeRVn-hg@comcast.com...
>
> "ASAAR" <caught@22.com> wrote in message
> news:b98ph11vqb62jr10s54onoer2ec8mgklcn@4ax.com...
>> On Mon, 05 Sep 2005 19:40:32 +0100, Kit wrote:
>>
>>> Comments welcome, because I'm no expert and this is just a theory.
>>> If anyone can direct me to authoratative information I'd be grateful.
>>
>> Interesting theory, but wrong, unfortunately. Rework it to see if
>> you'd still get 0.9 volts if 2, 3, 6 or 8 cells are used.
>
> Obviously not, but four or two cells are much the more common arrangement,
> aren't they? In the case of two cells it seems to me that the danger of
> reverse-charging one when it reached zero voltage would be small, since
> I'd think the half power remaining wouldn't be enough to operate the
> camera at all and it would just refuse to turn on. With one cell at 0 V
> and about three-quarters power remaining it seems that it would be much
> more of a danger.
>
> I'd never seen that 0.9 V figure mentioned in this way before, though. I
> thought these cells were considered discharged (for all practical
> purposes) when they reached 1.0 V.
>
>
>> These
>> numbers of cells should produce values ranging from 0.6 to 1.05
>> volts And even if the cells are not matched very well, by the time
>> one cell has lost its charge the others are probably well below 1.1
>> volts, not still near the 1.2 volts of a fully charged cell. The
>> only way a manufacturer would assume a high number such as 1.2 volts
>> would be if they knew people would be foolish enough to throw any
>> batteries on hand into the camera, where one of them would initially
>> be almost dead. But they don't make that assumption. They always
>> state that batteries should not be mixed, that they all be of the
>> same type and only fresh batteries by used. If those warnings are
>> heeded, most of the batteries will be close to dead when the first
>> cell gives up the ghost.
>
> I'm not so sure about that. While I've never tried to measure battery
> capacity in anything like a scientific way, my experience with NiMH cells
> has been that they appear to change characteristics quite differently from
> one another as they age. I am careful to keep the same set of cells
> together from the time they are new, insofar as this is possible.
> Typically when I put a new set of cells in a Maha charger which has four
> individual charging circuits, all four lights turn green at very nearly
> the same time; but when I put in an older set, one light often stays red
> for a long time after the other three have turned green. I'm not sure what
> this means exactly, but I presume the internal resistance of the odd cell
> has increased markedly, since the other three charge in what seems like a
> normal period of time.
>
> This is one reason I am inclined to distrust chargers that use the very
> common arrangement of only two charging circuits for four cells. If one
> cell of a pair charges more slowly because of higher internal resistance
> or some other reason, then I don't see how both cells can be properly
> charged at the same time on that same circuit. Am I missing something?
>
>
>>
>> Actually, most (all that I've seen) devices that discharge
>> batteries, whether NiCad or NiMH terminate the discharge at 1.0
>> volts. At this point there is hardly any energy remaining in the
>> cell. It may take hours to drop from 1.1 volts to 1.0, but only
>> seconds, or less to drop from 1.0 to 0.9 volts.
>
> That has always been my understanding too.
>
> N.
>

yyaaawwwnnwnnnnnnnnnnnnnnnn

Archived from groups: rec.photo.digital (More info?)

On Thu, 8 Sep 2005 04:56:48 +0000 (UTC), "Glittery Gary"
<OnThe@Upbtinternet.com> wrote:

>
>"Nostrobino" <not@home.today> wrote in message
>news:xsidnQ-UneXgbYDeRVn-qQ@comcast.com...
>>
>> "Kit" <kit@newsgroupsonlyyyy.com> wrote in message
>> news:431ddb77_2@mk-nntp-2.news.uk.tiscali.com...
>>> Nostrobino wrote:
>>>
>>>>
>>>>
>>>> Somewhere I'd read that 1.0 V was the "specified voltage" at which a
>>>> cell
>>>> was considered discharged and should not be drained further. Too long
>>>> ago
>>>> for me to remember the details or source, though.
>>>>
>>>> N.
>>>
>>> Well if I could find that source, and consider it authoritative, then I
>>> could get a life again :-)
>>
>> :-)
>>
>> And I can't begin to remember. Just tried couple of Google searches and
>> haven't yet come up with anything useful.
>>
>> N.
>>
>
>yyaawwwwnnnnnnnnnnnnnnnnn
>

djeez .. your main goal in life must be dislockating yr jaw ..
...unless you're suffering from CFS ..