NimH batteries

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

I get very confused about NiMh batteries. On the one hand we are told that
you don't have to have them fully discharged before recharging because they
don't have a memory. But on the other hand I understand that if you don't
have an expensive charger that knows when full charge has been reached and
automatically switches off, your batteries can overheat and deteriorate.
I think I have hit on a simple solution. I remove the batteries before they
are fully exhausted. Then I put them into a torch, switch the torch on and
leave it on until the light dies. I then give the batteries a full timed
charge as if for first charge.
Is there anything wrong with this idea please? Thanks.

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

Yes, there is one potential issue. If the batteries are not perfectly
matched in remaining charge, when you leave them in a flashlight, one
will always deplete first. The other cells will continue to force
current through the first-to-die battery, and eventually force a
"reverse" charge (wrong polarity) onto that cell. This will degrade
and destroy a cell as quickly, if not more so, than overheating.

The only "safe" way to discharge the cells, unless you can monitor the
voltages constantly, is to load/drain each one individually instead of
in a series. You can do this with battery holders shorted by resistors
or bulbs, just make sure there is one load per cell and they are not
interconnected.

Martin

Reply to martin

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

John Lee wrote:

> I get very confused about NiMh batteries. On the one hand we are told that
> you don't have to have them fully discharged before recharging because
> they don't have a memory.

Memory effect is primarily a problem for NiCads, but NiMH apparently also
like an occasional full discharge & recharge cycle.

But on the other hand I understand that if you
> don't have an expensive charger that knows when full charge has been
> reached and automatically switches off, your batteries can overheat and
> deteriorate.

Overcharging a NiMH can cause damage to it.

I think I have hit on a simple solution. I remove the
> batteries before they are fully exhausted. Then I put them into a torch,
> switch the torch on and leave it on until the light dies. I then give the
> batteries a full timed charge as if for first charge.
> Is there anything wrong with this idea please? Thanks.

If the cells are in series in the torch then the first cell to fully
discharge may suffer _reversal_ which will cause damage.

Intelligent chargers may cost a bit in intial outlay but should be cost
effective in the long term if they keep the NiMH cells in good condition.

Reply to Kit

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

On 30 Aug 2005 11:04:24 -0700, Martin wrote:

> Yes, there is one potential issue. If the batteries are not perfectly
> matched in remaining charge, when you leave them in a flashlight, one
> will always deplete first. The other cells will continue to force
> current through the first-to-die battery, and eventually force a
> "reverse" charge (wrong polarity) onto that cell. This will degrade
> and destroy a cell as quickly, if not more so, than overheating.

Far more quickly. Although it depends on what it meant by
overheating. During the last few minutes that they are charged,
cells can get very hot, too hot to comfortably hold. This may
shorten their life, but not by a great amount - they'd still be able
to undergo hundreds of recharge cycles with only a very slight
reduction of capacity. Fast chargers that aren't "smart" and use
timers are another matter. But cells can be fatally damaged after
only a single reverse charge. You're right about damage being
minimized if cells are used in matched sets, but a new set, fresh
out of the package is unlikely to be matched. At one point while
working part-time while going to school, a small part of what I did
was test and select cells to produce matched sets. This was pretty
important, as the battery packs used 8 or more cells, and they were
used under a fairly heavy load.

> The only "safe" way to discharge the cells, unless you can monitor the
> voltages constantly, is to load/drain each one individually instead of
> in a series. You can do this with battery holders shorted by resistors
> or bulbs, just make sure there is one load per cell and they are not
> interconnected.

Good advice, but there are a few decent, inexpensive chargers
that will independently discharge cells, but usually only down to
1.0 volts. At this point (as I'm sure you're aware) the cells are
essentially fully discharged. Whether this is marginally better or
worse than using a resistor or bulb to extract the last erglet I
have no idea. Do you know?

Reply to Anonymous

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

"John Lee" <jandj.lee@ntlworld.com> wrote in
news:Z31Re.1876$76.786@newsfe5-gui.ntli.net:

> I get very confused about NiMh batteries. On the one hand we are told
> that you don't have to have them fully discharged before recharging
> because they don't have a memory. But on the other hand I understand
> that if you don't have an expensive charger that knows when full
> charge has been reached and automatically switches off, your batteries
> can overheat and deteriorate. I think I have hit on a simple solution.
> I remove the batteries before they are fully exhausted. Then I put
> them into a torch, switch the torch on and leave it on until the light
> dies. I then give the batteries a full timed charge as if for first
> charge. Is there anything wrong with this idea please? Thanks.

The biggest problem is that you think that a good charger that
automatically senses when the batteries are charged is expensive, they
aren't and you should buy one.

--
Mark Heyes (New Zealand)
See my pics at www.gigatech.co.nz (last updated 16-August-05)
"The person on the other side was a young woman. Very obviously a
young woman. There was no possible way she could have been mistaken
for a young man in any language, especially Braille."
Maskerade

Reply to MarkH

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

John Lee wrote:
> I get very confused about NiMh batteries. On the one hand we are told that
> you don't have to have them fully discharged before recharging because they
> don't have a memory. But on the other hand I understand that if you don't
> have an expensive charger that knows when full charge has been reached and
> automatically switches off, your batteries can overheat and deteriorate.
> I think I have hit on a simple solution. I remove the batteries before they
> are fully exhausted. Then I put them into a torch, switch the torch on and
> leave it on until the light dies. I then give the batteries a full timed
> charge as if for first charge.
> Is there anything wrong with this idea please? Thanks.
>
>
First, it is a lousy substitute for a good charger that senses when the
batteries are charged, and second, it may damage the batteries by
forcing one into a reverse polarity situation. In short, bad idea!

--
Ron Hunter rphunter@charter.net

Reply to Anonymous

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

On Tue, 30 Aug 2005 17:55:05 +0000, John Lee wrote:

> I get very confused about NiMh batteries. On the one hand we are told that
> you don't have to have them fully discharged before recharging because they
> don't have a memory. But on the other hand I understand that if you don't
> have an expensive charger that knows when full charge has been reached and
> automatically switches off, your batteries can overheat and deteriorate.
> I think I have hit on a simple solution. I remove the batteries before they
> are fully exhausted. Then I put them into a torch, switch the torch on and
> leave it on until the light dies. I then give the batteries a full timed
> charge as if for first charge.
> Is there anything wrong with this idea please? Thanks.
Nothing.
--
Neil
Delete delete to reply by email

Reply to Anonymous

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

> I take 16 or 20 of these new batteries and put them through several
> controlled charge/discharge cycles. Then I connect all in series and
> give them a nice long trickle charge. I then disconnect them and let
> them "relax" for a day, and measure terminal voltages. Using these
> measurements, I sort them into the most-closely-matched groups of 2, 4,
> 8, or whatever I need. I then mark the batteries so that I won't
> accidentally mix them up.

Because your devices almost certainly use the batteries in series, not
parallel, matching the voltage does you little good. If you're going to
match something, match the capacities, to prevent one cell from being
overdischarged.

steve

Reply to Anonymous

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

On Wed, 31 Aug 2005 13:45:54 +0000 (UTC), Neil Ellwood wrote:

>> I think I have hit on a simple solution. I remove the batteries before they
>> are fully exhausted. Then I put them into a torch, switch the torch on and
>> leave it on until the light dies. I then give the batteries a full timed
>> charge as if for first charge.
>> Is there anything wrong with this idea please? Thanks.
>
> Nothing.

Yes, if the person monitoring the light doesn't walk away or fall
asleep. Otherwise there's a risk of damage. But if the light uses
3 or more cells, it's almost certain that at least one of them will
be "torched" if the torch isn't switched off when the light dims or
dies.

Reply to Anonymous

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

Steve Wolfe wrote:
> > I take 16 or 20 of these new batteries and put them through several
> > controlled charge/discharge cycles. Then I connect all in series and
> > give them a nice long trickle charge. I then disconnect them and let
> > them "relax" for a day, and measure terminal voltages. Using these
> > measurements, I sort them into the most-closely-matched groups of 2, 4,
> > 8, or whatever I need. I then mark the batteries so that I won't
> > accidentally mix them up.
>
> Because your devices almost certainly use the batteries in series, not
> parallel, matching the voltage does you little good. If you're going to
> match something, match the capacities, to prevent one cell from being
> overdischarged.
>
> steve

Years of direct experience have proven otherwise, it does work. The
batteries I start with have identical "capacity" by design, ie they are
identical in manufacturer, model, and date code. Differences creep in,
not by design, but more likely process tolerance. After several
charge/discharge cycles, the differences in cycle characteristics
apparently can be picked up as voltage differences after charging in
series (identical current/time for all cells).

I've thought before of actually matching true capacities, ie measuring
voltage decline as a function of energy withdrawn for each cell. In a
way, I do this by spot-checking voltage of nearly-depleted cells that
have been used in series. So far, this spot-check has verified that
what I'm already doing is working fine, ie I'm winding up with
matched-capacity cells. Nothing to gain by complicating the
proceedure.

Martin

Reply to martin

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

I've made a 4xAA conditioning discharger using 4 separate circuits
consisting of a 3A diode in series with a very low value resistor to give a
discharge of 4x1.5A. I made each resistor myself with a few turns of thin
wire till I got right current, 0R22 might do the trick though. When the
discharge has finished and the batteries are removed, their voltage is 1.0v.
No meters required, it runs hot until it's finished when it cools right down
which is when I remove the cells, but they can be left in overnight without
totally flattening the cells because of the diode forward voltage. I put
all the components inside an old timer charger with all the original parts
removed. I only use it every few months but it does seem to balance out a
set of batteries judging by how much and when they get hot at the end of
charge.

Reply to steveb

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

> I've made a 4xAA conditioning discharger using 4 separate circuits
> consisting of a 3A diode in series with a very low value resistor to give
a
> discharge of 4x1.5A. I made each resistor myself with a few turns of thin
> wire till I got right current, 0R22 might do the trick though. When the
> discharge has finished and the batteries are removed, their voltage is
1.0v.
> No meters required, it runs hot until it's finished when it cools right
down
> which is when I remove the cells, but they can be left in overnight
without
> totally flattening the cells because of the diode forward voltage. I put
> all the components inside an old timer charger with all the original parts
> removed. I only use it every few months but it does seem to balance out a
> set of batteries judging by how much and when they get hot at the end of
> charge.

Hopefully you're measuring 1.0V while the battery is still hooked up to
the circuit. If you're removing them and then measuring them after you take
them out, then the measurement is not valid, as even a *fully* depleted cell
(as long as it isn't damaged) will tend to rise back up to around 1.0V after
you disconnect it. If you're using a single diode, I would think that it
would still go lower than 1.0V if left in the circuit, because the forward
voltage of that single diode is (hopefully!) nowhere near 1 volt.

steve

Reply to Anonymous

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

Assuming the forward diode voltage is .5 volts, would two diodes in series
to give you 1.0 volt? A single diode would work is its forward voltage is
at least 0.5 volts to keep the battery from fullly discharging. Once the
voltage reached the forward diode voltage (.5 volts per diode) the diodes
would not conduct anymore and sit there until disconnected. (I am a novice
here and maybe I am missing something here, but sounds very simple.......I
am going to play around with this.)

--
Dave C.

c9ar9dar9elli@9c4.n9et

Remove the five 9's (leave the 4) for email.

"Steve Wolfe" <unt@see-signature.com> wrote in message
news:3novvbF2jl3bU2@individual.net...
>> I've made a 4xAA conditioning discharger using 4 separate circuits
>> consisting of a 3A diode in series with a very low value resistor to give
> a
>> discharge of 4x1.5A. I made each resistor myself with a few turns of
>> thin
>> wire till I got right current, 0R22 might do the trick though. When the
>> discharge has finished and the batteries are removed, their voltage is
> 1.0v.
>> No meters required, it runs hot until it's finished when it cools right
> down
>> which is when I remove the cells, but they can be left in overnight
> without
>> totally flattening the cells because of the diode forward voltage. I put
>> all the components inside an old timer charger with all the original
>> parts
>> removed. I only use it every few months but it does seem to balance out
>> a
>> set of batteries judging by how much and when they get hot at the end of
>> charge.
>
> Hopefully you're measuring 1.0V while the battery is still hooked up to
> the circuit. If you're removing them and then measuring them after you
> take
> them out, then the measurement is not valid, as even a *fully* depleted
> cell
> (as long as it isn't damaged) will tend to rise back up to around 1.0V
> after
> you disconnect it. If you're using a single diode, I would think that it
> would still go lower than 1.0V if left in the circuit, because the forward
> voltage of that single diode is (hopefully!) nowhere near 1 volt.
>
> steve
>
>

Reply to Anonymous

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

> Assuming the forward diode voltage is .5 volts, would two diodes in series
> to give you 1.0 volt? A single diode would work is its forward voltage is
> at least 0.5 volts to keep the battery from fullly discharging. Once the
> voltage reached the forward diode voltage (.5 volts per diode) the diodes
> would not conduct anymore and sit there until disconnected. (I am a
> novice here and maybe I am missing something here, but sounds very
> simple.......I am going to play around with this.)

That's exactly the idea, and most rectifier diodes have a forward voltage
of about 0.7 volts, so my guess is that in his circuit, that's what the
batteries drop to.

steve

Reply to Anonymous

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

Martin wrote:
> Steve Wolfe wrote:
>>> I take 16 or 20 of these new batteries and put them through several
>>> controlled charge/discharge cycles. Then I connect all in series and
>>> give them a nice long trickle charge. I then disconnect them and let
>>> them "relax" for a day, and measure terminal voltages. Using these
>>> measurements, I sort them into the most-closely-matched groups of 2, 4,
>>> 8, or whatever I need. I then mark the batteries so that I won't
>>> accidentally mix them up.
>> Because your devices almost certainly use the batteries in series, not
>> parallel, matching the voltage does you little good. If you're going to
>> match something, match the capacities, to prevent one cell from being
>> overdischarged.
>>
>> steve
>
> Years of direct experience have proven otherwise, it does work. The
> batteries I start with have identical "capacity" by design, ie they are
> identical in manufacturer, model, and date code. Differences creep in,
> not by design, but more likely process tolerance. After several
> charge/discharge cycles, the differences in cycle characteristics
> apparently can be picked up as voltage differences after charging in
> series (identical current/time for all cells).
>
> I've thought before of actually matching true capacities, ie measuring
> voltage decline as a function of energy withdrawn for each cell. In a
> way, I do this by spot-checking voltage of nearly-depleted cells that
> have been used in series. So far, this spot-check has verified that
> what I'm already doing is working fine, ie I'm winding up with
> matched-capacity cells. Nothing to gain by complicating the
> proceedure.
>
> Martin
>
It seems to me that checking the voltage of the discharged cells would
be more indicative of the battery health. I suspect there would be more
variation in voltage in the discharged state.

--
Ron Hunter rphunter@charter.net

Reply to Anonymous

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

"Steve Wolfe" <anx@codon.com> wrote in message
news:3nqdnvF2pj98U1@individual.net...
>> Assuming the forward diode voltage is .5 volts, would two diodes in
>> series to give you 1.0 volt? A single diode would work is its forward
>> voltage is at least 0.5 volts to keep the battery from fullly
>> discharging. Once the voltage reached the forward diode voltage (.5
>> volts per diode) the diodes would not conduct anymore and sit there until
>> disconnected. (I am a novice here and maybe I am missing something here,
>> but sounds very simple.......I am going to play around with this.)
>
> That's exactly the idea, and most rectifier diodes have a forward voltage
> of about 0.7 volts, so my guess is that in his circuit, that's what the
> batteries drop to.

I've seen this .7V as a reference too on the I-net .. as per manufacturer's
specs this is apparently too low a voltage .. recommendation is not to go
below .9V .. dunno if that is an absolutely strict value to be respected ..
but with some simple electronics it is easy to implement (I use a 723 power
regul. looping its own V ref ...)
FWIW

Reply to Anonymous

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

"Steve Wolfe" <anx@codon.com> wrote in message
news:3nqdnvF2pj98U1@individual.net...
>> Assuming the forward diode voltage is .5 volts, would two diodes in
>> series to give you 1.0 volt? A single diode would work is its forward
>> voltage is at least 0.5 volts to keep the battery from fullly
>> discharging. Once the voltage reached the forward diode voltage (.5
>> volts per diode) the diodes would not conduct anymore and sit there until
>> disconnected. (I am a novice here and maybe I am missing something here,
>> but sounds very simple.......I am going to play around with this.)
>
> That's exactly the idea, and most rectifier diodes have a forward voltage
> of about 0.7 volts, so my guess is that in his circuit, that's what the
> batteries drop to.

PS I tried this .. but the forward V drop through the diode(s) is heavily
current dependent .. In order to get some form of stable ref voltage one has
to build a constant current circuit to feed the diode .. more hassles than
just using an simple V; ref IC .. just MHO

Reply to Anonymous

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

> PS I tried this .. but the forward V drop through the diode(s) is heavily
> current dependent

It is? Perhaps when you're pushing high currents - where even small
resistances affect the voltage drop, but that's not the forward voltage,
just ohm's law in action. But as you get closer to discharged, and the
currents become more reasonable, it should get closer to the foward
voltage - and should never drop below the rated forward voltage. If you've
measured voltage drops *less* than the actual forward voltage, I'd love to
hear about it.

steve

Reply to Anonymous

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

>> Years of direct experience have proven otherwise, it does work. The
>> batteries I start with have identical "capacity" by design, ie they are
>> identical in manufacturer, model, and date code. Differences creep in,
>> not by design, but more likely process tolerance. After several
>> charge/discharge cycles, the differences in cycle characteristics
>> apparently can be picked up as voltage differences after charging in
>> series (identical current/time for all cells).
>>
>> I've thought before of actually matching true capacities, ie measuring
>> voltage decline as a function of energy withdrawn for each cell. In a
>> way, I do this by spot-checking voltage of nearly-depleted cells that
>> have been used in series. So far, this spot-check has verified that
>> what I'm already doing is working fine, ie I'm winding up with
>> matched-capacity cells. Nothing to gain by complicating the
>> proceedure.
>>
>> Martin
>>
> It seems to me that checking the voltage of the discharged cells would be
> more indicative of the battery health. I suspect there would be more
> variation in voltage in the discharged state.
>
>
> --
> Ron Hunter rphunter@charter.net

When the battery status indicator signals time to change the batteries on my
Sony camera, the manual suggests to put the camera in the slide show mode
and let it run continuously until the batteries run down to the point that
the camera shuts off. That way they are "cycled down" however the two
batteries are in series.

I haven't done that yet, but I will soon and measure the open circuit
voltage for what ever that would mean. I can't measure it under load unless
I measure it under a suitable external resistive load across each one.

Dave Cardarelli cardarelli@c4.net

Reply to Anonymous

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

"Steve Wolfe" <anx@codon.com> wrote in message
news:3nr8qoF2uoo1U1@individual.net...
> > PS I tried this .. but the forward V drop through the diode(s) is
heavily
> > current dependent
>
> It is? Perhaps when you're pushing high currents - where even small
> resistances affect the voltage drop, but that's not the forward voltage,
> just ohm's law in action. But as you get closer to discharged, and the
> currents become more reasonable, it should get closer to the foward
> voltage - and should never drop below the rated forward voltage. If
you've
> measured voltage drops *less* than the actual forward voltage, I'd love to
> hear about it.

Don't know what you guys are trying to do with a diode, but guess it's a
kind of discharge limiter for a single NiMH cell (with a resistor in
series). When I want to fully discharge a single cell, I load it with a 4.7
ohm resistor and leave it connected overnight. This provides a deep
discharge which is said to lower internal cell resistance, which helps to
reactivate cells sitting around unused for long periods. Based on my own
experience I'm convinced that no damage is done to a single cell (don't ever
do it with several cells in series) by fully discharging it. Even saw a NASA
guideline some years ago which specifies that NiCd cells not used for a
longer period shall be discharged first, then individually shorted before
storage.

Reply to Anonymous

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

On Wed, 7 Sep 2005 13:44:36 +0200, "Markus L" <uo9oew@lnubb.pbz>
wrote:

>"Steve Wolfe" <anx@codon.com> wrote in message
>news:3nr8qoF2uoo1U1@individual.net...
>> > PS I tried this .. but the forward V drop through the diode(s) is
>heavily
>> > current dependent
>>
>> It is? Perhaps when you're pushing high currents - where even small
>> resistances affect the voltage drop, but that's not the forward voltage,
>> just ohm's law in action. But as you get closer to discharged, and the
>> currents become more reasonable, it should get closer to the foward
>> voltage - and should never drop below the rated forward voltage. If
>you've
>> measured voltage drops *less* than the actual forward voltage, I'd love to
>> hear about it.
>
>Don't know what you guys are trying to do with a diode, but guess it's a
>kind of discharge limiter for a single NiMH cell (with a resistor in
>series). When I want to fully discharge a single cell, I load it with a 4.7
>ohm resistor and leave it connected overnight. This provides a deep
>discharge which is said to lower internal cell resistance, which helps to
>reactivate cells sitting around unused for long periods. Based on my own
>experience I'm convinced that no damage is done to a single cell (don't ever
>do it with several cells in series) by fully discharging it. Even saw a NASA
>guideline some years ago which specifies that NiCd cells not used for a
>longer period shall be discharged first, then individually shorted before
>storage.
>

read the other thread "NiMH again" :-)

Reply to Anonymous

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

In article <Z31Re.1876$76.786@newsfe5-gui.ntli.net>,
"John Lee" <jandj.lee@ntlworld.com> wrote:

> I get very confused about NiMh batteries. On the one hand we are told that
> you don't have to have them fully discharged before recharging because they
> don't have a memory. But on the other hand I understand that if you don't
> have an expensive charger that knows when full charge has been reached and
> automatically switches off, your batteries can overheat and deteriorate.
> I think I have hit on a simple solution. I remove the batteries before they
> are fully exhausted. Then I put them into a torch, switch the torch on and
> leave it on until the light dies. I then give the batteries a full timed
> charge as if for first charge.
> Is there anything wrong with this idea please? Thanks.

All rechargables have a limited number of full discharge / charge
cycles. Discharging a cell 50% then recharging it and then repeating
the cycle is roughly equivalent to one full discharge / charge cycle.
So depending on how discharged the cells are you are shortening their
life, possibly by quite a bit.

Get a decent charger. They aren't that expensive and they are better
even if you do a full charge. For starters the dumb chargers that run a
fixed current for a fixed amount of time are only good (and not very
good) for a given capacity cell.

--
Clark Martin
Redwood City, CA, USA Macintosh / Internet Consulting

"I'm a designated driver on the Information Super Highway"

Reply to Anonymous

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

I need an opinion on what I am trying regarding discharging NiMH cells:

I took one cell with a good charge on it, put a 3 amp diode forward biased
across the cell with no load resistor. The readings I measured with a
digital volt meter are:

Starting voltage across the cell unloaded is 1.28 volts.

With the diode forward biased across the cell, discharge current is 1.19
amps.

Drop across the diode is 0.838 volts.

Battery voltage with the diode conducting is 0.820.

After one hour, battery is hardly warm and diode is very warm, and hot
enough to hold. I figure about 100 degrees F or so.

So, I would expect that the battery will discharge down until about 0.838 or
so volts and then stop discharging expect for diode leakage.

Does all this sound like a workable situation? If so, I would do this with
spent batteries from my camera before recharging.

Would like opinions on this.

Dave

--
Dave C.

c9ar9dar9elli@9c4.n9et

Remove the five 9's (leave the 4) for email.

"SteveB" <sbrads@nildramDOTcoDOTuk> wrote in message
news:HZSdnai6BNVm3YreRVn-vw@pipex.net...
> I've made a 4xAA conditioning discharger using 4 separate circuits
> consisting of a 3A diode in series with a very low value resistor to give
> a discharge of 4x1.5A. I made each resistor myself with a few turns of
> thin wire till I got right current, 0R22 might do the trick though. When
> the discharge has finished and the batteries are removed, their voltage is
> 1.0v. No meters required, it runs hot until it's finished when it cools
> right down which is when I remove the cells, but they can be left in
> overnight without totally flattening the cells because of the diode
> forward voltage. I put all the components inside an old timer charger
> with all the original parts removed. I only use it every few months but
> it does seem to balance out a set of batteries judging by how much and
> when they get hot at the end of charge.
>

Reply to Anonymous

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

Dave C. wrote:

> I need an opinion on what I am trying regarding discharging NiMH cells:
>
> I took one cell with a good charge on it, put a 3 amp diode forward biased
> across the cell with no load resistor. The readings I measured with a
> digital volt meter are:
>
> Starting voltage across the cell unloaded is 1.28 volts.
>
> With the diode forward biased across the cell, discharge current is 1.19
> amps.
>
> Drop across the diode is 0.838 volts.
>
> Battery voltage with the diode conducting is 0.820.
>
> After one hour, battery is hardly warm and diode is very warm, and hot
> enough to hold. I figure about 100 degrees F or so.
>
> So, I would expect that the battery will discharge down until about 0.838
> or so volts and then stop discharging expect for diode leakage.
>
> Does all this sound like a workable situation? If so, I would do this
> with spent batteries from my camera before recharging.
>
> Would like opinions on this.
>
> Dave
>

Based on recent discussions on this ng, and information

at http://www.dvanhorn.org/NIMH/Index.php

QUOTE
Deep discharging cells is no problem whatsoever, and is somewhat theraputic.
/QUOTE

I would suggest discharging individual cells using a low value resistive
load (one or two ohms).

Your diode load apparently only draws an initial current of 1.19A, but
sticking just a diode across a cell is rather uncontrolled as far as
current level is concerned.

Reply to Kit

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

On Fri, 9 Sep 2005 11:09:46 -0400, Dave C. wrote:

> With the diode forward biased across the cell, discharge current is 1.19
> amps.
>
> Drop across the diode is 0.838 volts.
>
> Battery voltage with the diode conducting is 0.820.
>
> After one hour, battery is hardly warm and diode is very warm, and hot
> enough to hold. I figure about 100 degrees F or so.
>
> So, I would expect that the battery will discharge down until about 0.838 or
> so volts and then stop discharging expect for diode leakage.
>
> Does all this sound like a workable situation? If so, I would do this with
> spent batteries from my camera before recharging.

After one hour the cell (assuming that it's high capacity, over
2000mah) should be only about 1/2 depleted. Were the given voltages
taken at the beginning or after 1 hour? Also, the 0.838 voltage
drop across the diode depends on the current. As the battery nears
depletion, the voltage should drop well below 0.8 volts. Try adding
a 2.2k ohm resistor (or greater) to the circuit and see what kind of
voltage drop you'll see across the diode. Using a diode rated at
only 3 amps would have it operate dangerously close to its limits,
especially if it's not solidly mounted to something that can act as
an effective heat sink. If you use the diode to drain a cell that
has a much lower internal resistance, such as an NiMH C or D cell or
a good NiCad AA, you may notice the odor of burning silicon.
Lastly, draining batteries as quickly as this may shorten their
lives by a good deal more than an single charge/discharge cycle. If
it's not done frequently though, it probably wouldn't be worth
worrying about.

Reply to Anonymous

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

"Dave C." <c9ar9dar9elli@9c4.n9et> wrote in message
news:4321a5c4.0@paperboy.c4.net...
>I need an opinion on what I am trying regarding discharging NiMH cells:
>
> I took one cell with a good charge on it, put a 3 amp diode forward biased
> across the cell with no load resistor. The readings I measured with a
> digital volt meter are:
>
> Starting voltage across the cell unloaded is 1.28 volts.
>
> With the diode forward biased across the cell, discharge current is 1.19
> amps.
>
> Drop across the diode is 0.838 volts.
>
> Battery voltage with the diode conducting is 0.820.
>
> After one hour, battery is hardly warm and diode is very warm, and hot
> enough to hold. I figure about 100 degrees F or so.
>
> So, I would expect that the battery will discharge down until about 0.838
> or so volts and then stop discharging expect for diode leakage.
>
> Does all this sound like a workable situation? If so, I would do this
> with spent batteries from my camera before recharging.
>
> Would like opinions on this.

I'm sorry but this is lousy electronics .. :-)

What do you consider a "3 amp diode" ? .. you mean the max forward current
is 3 amp ?

You will need to check the forward characteristic of this diode as that
forward voltage drop does seem odd to me unless you are still not in full
conduction which I think is the case given forward current and drop voltage
.... so you may be just outside the conducting "knee voltage" for that
specific diode. There's a big chance that as soon as the current start
dropping the diode goes into non-conducting state below the knee-voltage and
the discharge will stop
Do not forget that using some fast switching diodes in this type of setup
will virtually short circuit the battery as there is no current limiting
charge resistor ..
all in all, to me, this is a completely unuseable setup ..
success or failure will be dictated by random luck parameters ..

Reply to Anonymous

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

ASAAR wrote:
> On Fri, 9 Sep 2005 11:09:46 -0400, Dave C. wrote:
>
>
>>With the diode forward biased across the cell, discharge current is 1.19
>>amps.
>>
>>Drop across the diode is 0.838 volts.
>>
>>Battery voltage with the diode conducting is 0.820.
>>
>>After one hour, battery is hardly warm and diode is very warm, and hot
>>enough to hold. I figure about 100 degrees F or so.
>>
>>So, I would expect that the battery will discharge down until about 0.838 or
>>so volts and then stop discharging expect for diode leakage.
>>
>>Does all this sound like a workable situation? If so, I would do this with
>>spent batteries from my camera before recharging.
>
>
> After one hour the cell (assuming that it's high capacity, over
> 2000mah) should be only about 1/2 depleted. Were the given voltages
> taken at the beginning or after 1 hour? Also, the 0.838 voltage
> drop across the diode depends on the current. As the battery nears
> depletion, the voltage should drop well below 0.8 volts. Try adding
> a 2.2k ohm resistor (or greater) to the circuit and see what kind of
> voltage drop you'll see across the diode. Using a diode rated at
> only 3 amps would have it operate dangerously close to its limits,
> especially if it's not solidly mounted to something that can act as
> an effective heat sink. If you use the diode to drain a cell that
> has a much lower internal resistance, such as an NiMH C or D cell or
> a good NiCad AA, you may notice the odor of burning silicon.
> Lastly, draining batteries as quickly as this may shorten their
> lives by a good deal more than an single charge/discharge cycle. If
> it's not done frequently though, it probably wouldn't be worth
> worrying about.

Hi...

I can't imagine any reason to use a diode, and wouldn't
even dream of doing so...

Terrible terrible, idea. Along with many other ideas not
to do it, a diode when it fails will 99 out of a hundred
times fail shorted.

Imagine the results if one were to inadvertantly put in a
"full" battery, and have it then dump its full capacity
into an (almost) dead short.

I totally disagree with the concept of discharging them,
but appreciate that others disagree, so if you must
deplete them at least do it into some reasonable dummy
load.

Take care.

Ken
(old retired electrical guy)

Reply to Anonymous

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

On Fri, 09 Sep 2005 18:59:17 GMT, Ken Weitzel wrote:

> ASAAR wrote:
>> On Fri, 9 Sep 2005 11:09:46 -0400, Dave C. wrote:
> . . .
> I totally disagree with the concept of discharging them,
> but appreciate that others disagree, so if you must
> deplete them at least do it into some reasonable dummy
> load.

There's nothing wrong will occasionally discharging batteries if
done appropriately. And by appropriately I mean discharging them at
a slower rate that the cells are normally charged (taking at least
several hours) and stopping at about the 1.0 or 0.9 volt point.
This is the way the discharge circuits of battery chargers and
cameras are designed. I agree in general with most of what you said
though, so I'm wondering if your reply was really intended to be to
the same person/people my reply was to?

Reply to Anonymous

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

ASAAR wrote:
> On Fri, 09 Sep 2005 18:59:17 GMT, Ken Weitzel wrote:
>
>
>>ASAAR wrote:
>>
>>>On Fri, 9 Sep 2005 11:09:46 -0400, Dave C. wrote:
>>
>>. . .
>>I totally disagree with the concept of discharging them,
>>but appreciate that others disagree, so if you must
>>deplete them at least do it into some reasonable dummy
>>load.
>
>
> There's nothing wrong will occasionally discharging batteries if
> done appropriately. And by appropriately I mean discharging them at
> a slower rate that the cells are normally charged (taking at least
> several hours) and stopping at about the 1.0 or 0.9 volt point.
> This is the way the discharge circuits of battery chargers and
> cameras are designed. I agree in general with most of what you said
> though, so I'm wondering if your reply was really intended to be to
> the same person/people my reply was to?

Hi...

Yeppers, sorry for causing confusion

Though you and I do disagree (that's OK ) - nothing wrong
with discharging them. Nothing right with it, either

Take care.

Ken

Reply to Anonymous

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

Regarding discharging batteries, I recently bought a Sony DSC-W5 which uses
two AAs. And I bought four new 2500 mah batteries. The battery status in
the camera is pretty good and has 3 or 4 level of status including
"batteries nearly dead." I figure once every two or three months they could
use a discharging. So I thought I'd figure out a circuit that might be
useful to drain the last few percent from the battery pair before recharging
(in my MAHA charger). Since most of the energy will be gone from the
batteries, the discharge cycle probably would probably be not too long.

My testing was done with an older set of 1800mah batteries that were to a
full charge about three months ago. I did add a .5 ohm power resister is
series with the 3 amp diode, and current value was about 0.80 amps. I
monitored the battery voltage, and it seemed to settle down to about 0.75
volts and the current went down to about 0.050 amps at which time I
disconnected the battery.

My manual suggested to occasionally discharge them by setting the camera on
slide show mode with a half dozen or so pictures and let it continuously run
in this mode until the camera shuts off.

All in all, this might be a better option rather than fooling around with
"off line" discharging. My charger does not do a discharge before charging,
but I have read somewhere that that feature is in some chargers.
--
Dave C.

c9ar9dar9elli@9c4.n9et

Remove the five 9's (leave the 4) for email.

"ASAAR" <caught@22.com> wrote in message
news:qln3i19s4p7tpf4i6olajg44fefkbl2a79@4ax.com...
> On Fri, 09 Sep 2005 18:59:17 GMT, Ken Weitzel wrote:
>
>> ASAAR wrote:
>>> On Fri, 9 Sep 2005 11:09:46 -0400, Dave C. wrote:
>> . . .
>> I totally disagree with the concept of discharging them,
>> but appreciate that others disagree, so if you must
>> deplete them at least do it into some reasonable dummy
>> load.
>
> There's nothing wrong will occasionally discharging batteries if
> done appropriately. And by appropriately I mean discharging them at
> a slower rate that the cells are normally charged (taking at least
> several hours) and stopping at about the 1.0 or 0.9 volt point.
> This is the way the discharge circuits of battery chargers and
> cameras are designed. I agree in general with most of what you said
> though, so I'm wondering if your reply was really intended to be to
> the same person/people my reply was to?
>

Reply to Anonymous

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

"Dave C." <c9ar9dar9elli@9c4.n9et> wrote in message
news:43223ffb.0@paperboy.c4.net...
> My testing was done with an older set of 1800mah batteries that were to a
> full charge about three months ago. I did add a .5 ohm power resister is
> series with the 3 amp diode, and current value was about 0.80 amps. I
> monitored the battery voltage, and it seemed to settle down to about 0.75
> volts and the current went down to about 0.050 amps at which time I
> disconnected the battery.

you should have followed this thread more carefully :-)
Manufacturers suggest not to drop voltages below 0.9V ..for the sanity of
the batteries and BTW your discharge cicuit is not learning anything except
you are discharging the battery up to just before cutoff of the diode.. IMHO
the 0.75V at 0.05 amps is the value reading for the diode just before it
goes in cut-off.

Reply to Anonymous

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