Wall Warts

[Puddin' Man]

I've a Skil Mod 2375 3/8" cordless drill (9.6v) thats nearly an antique.

The old wall wart outputs dc 11v, 250 mA when functional. It now has a slight
bulge and outputs 0v.

I rumble thru my box of WW's, find one at rated output of dc 12v, 300 mA,
splice the wires, plug it in to see if it will charge. It will, but it heats
up a bunch, enough to burn my fingers a little.

Are there any guidelines for substituting these things? I'd like to keep the drill,
but can hardly afford to burn the house down.

Thx,
P

"Law Without Equity Is No Law At All. It Is A Form Of Jungle Rule."

 

[whit3rd]

On Monday, May 23, 2011 1:54:46 PM UTC-7, Puddin' Man wrote:

I've a Skil Mod 2375 3/8" cordless drill (9.6v) thats nearly an antique.

The old wall wart outputs dc 11v, 250 mA when functional. It now has a slight
bulge and outputs 0v.

I rumble thru my box of WW's, find one at rated output of dc 12v, 300 mA,

To charge a 9.6V battery (which would be eight 1.2V NiCd sub-C cells in series)
I'd consider 150 mA for eight hours. So, connect a series resistor
to the 12V supply that chokes the charge current down to that level
(measure with ammeter into a part-charged battery).

The purpose here is to DEFEAT THE REGULATION of that 12V power supply,
it oughtn't drive its maximum current just because its load is intended to
be 9.6V instead of 12V...

At a guess, the original power was unfiltered but rectified, and the replacement
is rectified, filtered, and regulated. The filtering doesn't hurt, but that
voltage regulation is your enemy here.

 

[Trevor Wilson]

whit3rd wrote:

On Monday, May 23, 2011 1:54:46 PM UTC-7, Puddin' Man wrote:
I've a Skil Mod 2375 3/8" cordless drill (9.6v) thats nearly an
antique.

The old wall wart outputs dc 11v, 250 mA when functional. It now has
a slight bulge and outputs 0v.

I rumble thru my box of WW's, find one at rated output of dc 12v,
300 mA,

To charge a 9.6V battery (which would be eight 1.2V NiCd sub-C cells
in series) I'd consider 150 mA for eight hours. So, connect a series
resistor
to the 12V supply that chokes the charge current down to that level
(measure with ammeter into a part-charged battery).

The purpose here is to DEFEAT THE REGULATION of that 12V power supply,
it oughtn't drive its maximum current just because its load is
intended to
be 9.6V instead of 12V...

At a guess, the original power was unfiltered but rectified, and the
replacement is rectified, filtered, and regulated. The filtering
doesn't hurt, but that voltage regulation is your enemy here.

**Agreed with the above, but for a few cents more, you could use an LM317 as
a current source (see the Natsemi application notes), set for the
appropriate current. You may need an unregulated 12 Volt wall wart, or,
perhaps, a 15 Volt one.


--
Trevor Wilson
www.rageaudio.com.au

 

[Phil Allison]

"Puddin' Man"

I've a Skil Mod 2375 3/8" cordless drill (9.6v) thats nearly an antique.

The old wall wart outputs dc 11v, 250 mA when functional. It now has a
slight
bulge and outputs 0v.

** Wall warts break up into THREE distinct groups:

1. AC output.

2. DC output.

3. Battery chargers.

You cannot exchange units from one group into another.

Those in group 3 are the most variable in design and are purpose built to
charge a specific type and number of cells.

The fact they all look much the same and often have the same plugs attached
is very misleading.


..... Phil

 

[Wild_Bill]

The reply from whit is a good suggestion, but I suspect that the cells in
the drill are shorted, which will damage a charging power supply. It seems
that the old supply became overheated and failed, which will happen if the
cells are shorted.
I believe this is the reason for the substitute wall wart getting hot.

To find out if the cells are shorted, the cells need to be accessed, and
individually checked with an ohm meter or an instrument capable of measuring
battery cell impedance or ESR.
Also, a reading for individual cells with voltmeter reading of zero volts
generally indicates shorted cells.

Replacing all of the cells is the most effective solution, and cells with
tabs can be soldered together in the original order, and restore full (or
even better) operation of the drill (or most cordless tools).

If you have the ability to make good solder connections, replacement is
usually less costly than paying someone else to do the work.

There are numerous online sellers with reasonable prices for replacement
cells.

--
Cheers,
WB
..............


"Puddin' Man" <puddingDOTman@gmail.com> wrote in message
news:6bilt6tedes72j7jmhgvbctu41evoou66g@4ax.com...

I've a Skil Mod 2375 3/8" cordless drill (9.6v) thats nearly an antique.

The old wall wart outputs dc 11v, 250 mA when functional. It now has a
slight
bulge and outputs 0v.

I rumble thru my box of WW's, find one at rated output of dc 12v, 300 mA,
splice the wires, plug it in to see if it will charge. It will, but it
heats
up a bunch, enough to burn my fingers a little.

Are there any guidelines for substituting these things? I'd like to keep
the drill,
but can hardly afford to burn the house down.

Thx,
P

"Law Without Equity Is No Law At All. It Is A Form Of Jungle Rule."

 

[Puddin' Man]

On Mon, 23 May 2011 20:24:18 -0400, "Wild_Bill" <wb_wildbill@XSPAMyahoo.com> wrote:

The reply from whit is a good suggestion, but I suspect that the cells in
the drill are shorted, which will damage a charging power supply. It seems
that the old supply became overheated and failed, which will happen if the
cells are shorted.
I believe this is the reason for the substitute wall wart getting hot.

Certainly possible, for all I know.

I shoulda mentioned, I'm not an electronics guy, just a tinker. Limited
(make that -very- limited) skills, eqpt.

I might've charged with the new supply 20 min. before I noticed the heat.
The drill worked a little thereafter.

To find out if the cells are shorted, the cells need to be accessed, and
individually checked with an ohm meter or an instrument capable of measuring
battery cell impedance or ESR.
Also, a reading for individual cells with voltmeter reading of zero volts
generally indicates shorted cells.

The drill still works a little. I measure about 4v aggregate from the
battery pack.

Replacing all of the cells is the most effective solution, and cells with
tabs can be soldered together in the original order, and restore full (or
even better) operation of the drill (or most cordless tools).

The usual case, not practical. Wierd cell shapes, very old unit.

If you have the ability to make good solder connections, replacement is
usually less costly than paying someone else to do the work.

There are numerous online sellers with reasonable prices for replacement
cells.

If I can't tinker-fix it, it goes in the trash. Any replacement parts
either won't fit or are too expensive or both. I hate to throw
potentially useful tools away, but ...

Thanks,
P

"Law Without Equity Is No Law At All. It Is A Form Of Jungle Rule."

 

[Jeff Liebermann]

On Mon, 23 May 2011 15:54:46 -0500, Puddin' Man
<puddingDOTman@gmail.com> wrote:

I've a Skil Mod 2375 3/8" cordless drill (9.6v) thats nearly an antique.

The charger was involved in a safety recall. I have a similar model
with the same problem. The xformer will overheat when trying to
charge a dead battery. If you call the 800 number, you might get a
free charger.
<http://www.all-cordless.com/skilrecalls.html>
However, if it's a later replacement, then all you did is blow the
thermal fuse inside the charger. You can crack it open and replace
it, but my guess(tm) is you also have some very dead batteries.

Are there any guidelines for substituting these things? I'd like to keep the drill,
but can hardly afford to burn the house down.

Not without knowing where the charge controller (probably just a
resistor) is located. If it's inside the wall wart, you'll need to
add an identical resistor. If it's inside the drill, you can probably
get away with your 12VDC substitute. Measure the current drain at the
charger output. My guess(tm) is you should have a mess of AA size
NiCd batteries inside the drill, which are usually rated at 750ma-hr.
Using the 0.1C rule, that would be a charging current of about 75ma
max.

--
Jeff Liebermann jeffl@cruzio.com
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558

 

[spamtrap1888]

On May 24, 10:55 am, Puddin' Man <puddingDOT...@gmail.com> wrote:

On Mon, 23 May 2011 20:24:18 -0400, "Wild_Bill" <wb_wildb...@XSPAMyahoo.com> wrote:

I shoulda mentioned, I'm not an electronics guy, just a tinker. Limited
(make that -very- limited) skills, eqpt.

I might've charged with the new supply 20 min. before I noticed the heat.
The drill worked a little thereafter.

To find out if the cells are shorted, the cells need to be accessed, and
individually checked with an ohm meter or an instrument capable of measuring
battery cell impedance or ESR.
Also, a reading for individual cells with voltmeter reading of zero volts
generally indicates shorted cells.

The drill still works a little. I measure about 4v aggregate from the
battery pack.

Replacing all of the cells is the most effective solution, and cells with
tabs can be soldered together in the original order, and restore full (or
even better) operation of the drill (or most cordless tools).

The usual case, not practical. Wierd cell shapes, very old unit.


If I can't tinker-fix it, it goes in the trash. Any replacement parts
either won't fit or are too expensive or both. I hate to throw
potentially useful tools away, but ...

If the pack didn't consist of 8 AA-sized NiCds I would be muchly
surprised. Anybody who can plug new AAs into a remote control, and
knows which end of a soldering pencil to hold, likely has the skill
level to rebuild a battery pack.

 

[Puddin' Man]

On Tue, 24 May 2011 11:35:26 -0700, Jeff Liebermann <jeffl@cruzio.com> wrote:

On Mon, 23 May 2011 15:54:46 -0500, Puddin' Man
puddingDOTman@gmail.com> wrote:

I've a Skil Mod 2375 3/8" cordless drill (9.6v) thats nearly an antique.

The charger was involved in a safety recall. I have a similar model
with the same problem. The xformer will overheat when trying to
charge a dead battery. If you call the 800 number, you might get a
free charger.
http://www.all-cordless.com/skilrecalls.html
However, if it's a later replacement, then all you did is blow the
thermal fuse inside the charger. You can crack it open and replace
it, but my guess(tm) is you also have some very dead batteries.

Thanks. That was worth checking into.

They have no replacement parts for the 2375. They say that they'll
send paperwork so, if I send 'em the charger, they'll
send more paperwork so, if I buy a new Skil drill, they'll refund
$25 or somesuch -if- I send 'em the UPC code -and- the orig.
receipt, etc, etc ad nauseum. I don't think they really wanna
honor the recall.

Are there any guidelines for substituting these things? I'd like to keep the drill,
but can hardly afford to burn the house down.

Not without knowing where the charge controller (probably just a
resistor) is located. If it's inside the wall wart, you'll need to
add an identical resistor. If it's inside the drill, you can probably
get away with your 12VDC substitute. Measure the current drain at the
charger output.

Lost me there. Measure the *drain* at the charger output? But I'm
pretty sure it is inside the buggered charger.

My guess(tm) is you should have a mess of AA size
NiCd batteries inside the drill, which are usually rated at 750ma-hr.
Using the 0.1C rule, that would be a charging current of about 75ma
max.

It's evidently all original: charger is 92950, power pack is 92955,
and is a *sealed* unit. I see no way to take it apart.

Am I missing something here?

Thx,
P

"Law Without Equity Is No Law At All. It Is A Form Of Jungle Rule."

 

[Jeff Liebermann]

On Tue, 24 May 2011 17:03:17 -0500, Puddin' Man
<puddingDOTman@gmail.com> wrote:

I don't think they really wanna
honor the recall.

<http://www.cpsc.gov/cpscpub/prerel/prhtml03/03082.html>
Looks like the $25 rebate is all that they are offering.

Maybe calling them back and asking for the name of the company
attorney and whether this recall is defacto admission of their
responsibility if your garage burns down, might get their attention.

Are there any guidelines for substituting these things? I'd like to keep the drill,
but can hardly afford to burn the house down.

Not without knowing where the charge controller (probably just a
resistor) is located. If it's inside the wall wart, you'll need to
add an identical resistor. If it's inside the drill, you can probably
get away with your 12VDC substitute. Measure the current drain at the
charger output.

Lost me there. Measure the *drain* at the charger output? But I'm
pretty sure it is inside the buggered charger.

I guess it might be difficult to measure the current if the charger is
dead.

The problem is that we don't really know how the charging system works
without tearing it apart. If the blown charger has a resistor in
series with the output, there's no way to know at this point without
cracking it open, or comparing it with another working unit.

Start by using an ohms-guesser on the charger base. Cut the cord. Is
there DC continuity between the two wires and the contacts on the
charger base? If there's a measurable resistance, then the charger
base has either a resistor, or complexicated charge controller in the
base. I would therefore guess(tm) that the wall wart is just a simple
wall wart with nothing more complex than a thermal fuse inside.
However, if there is DC continuity between the leads and the battery
terminal connections, then the charge controller is inside the wall
wart, and a simple replacement isn't going to work.

Since the power supply is obviously blown and useless, tearing it
apart should reveal if there's anything inside. You'll probably find
a thermal fuse. If it's blown, just replace it with a similar thermal
fuse, glue the case back together, and continue charging. However, if
there's a series resistor inside (which is what I suspect), then use
the 12V adapter, add a similar resistor in series, measure the
charging current, and see if it's reasonable 0.1C. If not, adjust the
resistor value for 0.1C charging current.

My guess(tm) is you should have a mess of AA size
NiCd batteries inside the drill, which are usually rated at 750ma-hr.
Using the 0.1C rule, that would be a charging current of about 75ma
max.

It's evidently all original: charger is 92950, power pack is 92955,
and is a *sealed* unit. I see no way to take it apart.

Am I missing something here?

Most chargers and charging bases are solvent welded together. They
can usually be cracked open with brute force. Place a masons chisel
along the glue line, and beat on it with a hammer. The glue line
should crack as the plastic bends. If not, just use a hack saw to saw
along the glue line. Do the repair, and then glue the case back
together.

--
Jeff Liebermann jeffl@cruzio.com
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558

 

[Jeff Liebermann]

On Tue, 24 May 2011 22:36:26 -0700, Jeff Liebermann <jeffl@cruzio.com>
wrote:

This is much too easy...

Skil 9.6v 92950 Battery Charger Repair 9.6 volt- 92955
<http://cgi.ebay.com/200454895811>
$13 plus $6 shipping.

--
Jeff Liebermann jeffl@cruzio.com
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558

 

[spamtrap1888]

On May 24, 3:03 pm, Puddin' Man <puddingDOT...@gmail.com> wrote:

On Tue, 24 May 2011 11:35:26 -0700, Jeff Liebermann <je...@cruzio.com> wrote:

My guess(tm) is you should have a mess of AA size
NiCd batteries inside the drill, which are usually rated at 750ma-hr.
Using the 0.1C rule, that would be a charging current of about 75ma
max.

It's evidently all original: charger is 92950, power pack is 92955,
and is a *sealed* unit. I see no way to take it apart.

Am I missing something here?

The battery packs I've repaired consisted of two shells epoxied
together. Does this one not look like it could be knifed apart?

 

[Geoffrey S. Mendelson]

Jeff Liebermann wrote:

Most chargers and charging bases are solvent welded together. They
can usually be cracked open with brute force. Place a masons chisel
along the glue line, and beat on it with a hammer. The glue line
should crack as the plastic bends. If not, just use a hack saw to saw
along the glue line. Do the repair, and then glue the case back
together.

Overnight in a freezer helps. It makes the plastic brittle. If you have a deep
freezer (0F or colder) use it instead of a refigerator freezer (around 10F).

Geoff.

--
Geoffrey S. Mendelson N3OWJ/4X1GM
It's amazing how many people have no clue what the word "contiguous" means. :-(

 

[Archon]

On 5/23/2011 4:54 PM, Puddin' Man wrote:

I've a Skil Mod 2375 3/8" cordless drill (9.6v) thats nearly an antique.

The old wall wart outputs dc 11v, 250 mA when functional. It now has a slight
bulge and outputs 0v.

I rumble thru my box of WW's, find one at rated output of dc 12v, 300 mA,
splice the wires, plug it in to see if it will charge. It will, but it heats
up a bunch, enough to burn my fingers a little.

Are there any guidelines for substituting these things? I'd like to keep the drill,
but can hardly afford to burn the house down.

Thx,
P

"Law Without Equity Is No Law At All. It Is A Form Of Jungle Rule."

Recall on the charger, though not much good if the battery is knackered.

http://store.boschpowertoolsource.com/skil9296vwar1.html

JC

 

[hrhofmann@att.net]

On May 25, 1:02 am, Jeff Liebermann <je...@cruzio.com> wrote:

On Tue, 24 May 2011 22:36:26 -0700, Jeff Liebermann <je...@cruzio.com
wrote:

This is much too easy...

Skil 9.6v 92950 Battery Charger Repair 9.6 volt- 92955
http://cgi.ebay.com/200454895811
$13 plus $6 shipping.

--
Jeff Liebermann     je...@cruzio.com
150 Felker St #D    http://www.LearnByDestroying.com
Santa Cruz CA 95060http://802.11junk.com
Skype: JeffLiebermann     AE6KS    831-336-2558

But the OP needs to know that none of the cells are shorted before he
goes any further, that requires trying to charge the cells for a few
minutes and then measuring the voltage at the drill. The voltage
should be 8x1.2V = 9.6V or slightly higher. If the voltage is not at
that level, then one or more cells in the drill are shorted and no
charger is going to fix that. You can blast the shorted cell by
putting it across a very large charged capacitor if you take the drill
apart and measure each cell individually, but even then the cell is
likely to redevelop a short unless it is kept on a charger almost
continually.

 

[Puddin' Man]

On Tue, 24 May 2011 22:36:26 -0700, Jeff Liebermann <jeffl@cruzio.com> wrote:

Start by using an ohms-guesser on the charger base. Cut the cord. Is
there DC continuity between the two wires and the contacts on the
charger base? If there's a measurable resistance, then the charger
base has either a resistor, or complexicated charge controller in the
base.

The + side measures 3 ohms.

I would therefore guess(tm) that the wall wart is just a simple
wall wart with nothing more complex than a thermal fuse inside.
However, if there is DC continuity between the leads and the battery
terminal connections, then the charge controller is inside the wall
wart, and a simple replacement isn't going to work.

Since the power supply is obviously blown and useless, tearing it
apart should reveal if there's anything inside. You'll probably find
a thermal fuse. If it's blown, just replace it with a similar thermal
fuse, glue the case back together, and continue charging. However, if
there's a series resistor inside (which is what I suspect), then use
the 12V adapter, add a similar resistor in series, measure the
charging current, and see if it's reasonable 0.1C. If not, adjust the
resistor value for 0.1C charging current.

My guess(tm) is you should have a mess of AA size
NiCd batteries inside the drill, which are usually rated at 750ma-hr.
Using the 0.1C rule, that would be a charging current of about 75ma
max.

It's evidently all original: charger is 92950, power pack is 92955,
and is a *sealed* unit. I see no way to take it apart.

Am I missing something here?

Most chargers and charging bases are solvent welded together. They
can usually be cracked open with brute force. Place a masons chisel
along the glue line, and beat on it with a hammer. The glue line
should crack as the plastic bends. If not, just use a hack saw to saw
along the glue line. Do the repair, and then glue the case back
together.

Thanks for this opportunity for taking revenge on the offending WW.

After freezing, it cracked open easily, but I see no fuse.
Apologies for lack of quality of pix:

http://imageshack.us/photo/my-images/231/skilwallwart1.jpg/
http://imageshack.us/photo/my-images/831/skilwallwart2.jpg/

Thanks,
P

"Law Without Equity Is No Law At All. It Is A Form Of Jungle Rule."

 

[Jeff Liebermann]

On Wed, 25 May 2011 14:07:41 -0500, Puddin' Man
<puddingDOTman@gmail.com> wrote:

The + side measures 3 ohms.

Ok. There may be a diode inside the base. Try reversing the leads on
the ohms-guesser and see if it acts like a diode.

Thanks for this opportunity for taking revenge on the offending WW.

Learn by Destroying(tm). If you haven't destroyed it and fixed it,
you don't understand it.

After freezing, it cracked open easily, but I see no fuse.
Apologies for lack of quality of pix:

http://imageshack.us/photo/my-images/231/skilwallwart1.jpg/
http://imageshack.us/photo/my-images/831/skilwallwart2.jpg/

Yech. I can't see anything in there. Fix the focus. Is there any
manner of diode, resitor, charge controller, PCB, pilot lamp (current
regulator), or thermal fuse inside? If not, it's an ordinary AC
xformer. Also, no points for covering the xformer with the burned
insulating material, so I can't see anything.

If AC only, then there's the usual problem of what voltage to get.
Peak voltage on a single diode is 1.414 * AC voltage. If there's no
series resistor, and just a diode in the base unit, then my guess(tm)
is that you should be looking for a 7.5VAC wall wart. Again, you'll
need to measure the charging current and check for 0.1C current to be
sure you got it right. Of course, make sure you don't have a shorted
battery pack.

--
# Jeff Liebermann 150 Felker St #D Santa Cruz CA 95060
# 831-336-2558
# http://802.11junk.com jeffl@cruzio.com
# http://www.LearnByDestroying.com AE6KS

 

[Mike Tomlinson]

En el artículo <6bilt6tedes72j7jmhgvbctu41evoou66g@4ax.com>, Puddin' Man
<puddingDOTman@gmail.com> escribió:

Are there any guidelines for substituting these things?

You've done the right thing, substituting one of a slightly higher spec.
The fact that you've had one burn out and its replacement gets hot
suggests there may be something wrong with the charging circuit in the
drill or the battery.

--
(\__/)
(='.'=)
(")_(")

 

[Puddin' Man]

On Wed, 25 May 2011 15:57:21 -0700, Jeff Liebermann <jeffl@cruzio.com> wrote:

On Wed, 25 May 2011 14:07:41 -0500, Puddin' Man
puddingDOTman@gmail.com> wrote:

The + side measures 3 ohms.

Ok. There may be a diode inside the base. Try reversing the leads on
the ohms-guesser and see if it acts like a diode.

Measures infinite ohms when reversed.

...

After freezing, it cracked open easily, but I see no fuse.
Apologies for lack of quality of pix:

http://imageshack.us/photo/my-images/231/skilwallwart1.jpg/
http://imageshack.us/photo/my-images/831/skilwallwart2.jpg/

Yech. I can't see anything in there. Fix the focus. Is there any
manner of diode, resitor, charge controller, PCB, pilot lamp (current
regulator), or thermal fuse inside? If not, it's an ordinary AC
xformer. Also, no points for covering the xformer with the burned
insulating material, so I can't see anything.

There's nothing else in there.

If AC only, then there's the usual problem of what voltage to get.
Peak voltage on a single diode is 1.414 * AC voltage. If there's no
series resistor, and just a diode in the base unit, then my guess(tm)
is that you should be looking for a 7.5VAC wall wart. Again, you'll
need to measure the charging current and check for 0.1C current to be
sure you got it right.

I'm 'fraid you'll have to elaborate on this last. And, what is "C" in
..1C?

Of course, make sure you don't have a shorted
battery pack.

So I gotta crack the battery pack open. And test each cell individually?
How?

Thx,
P

"Law Without Equity Is No Law At All. It Is A Form Of Jungle Rule."

 

[spamtrap1888]

On May 26, 10:51 am, Puddin' Man <puddingDOT...@gmail.com> wrote:

Of course, make sure you don't have a shorted
battery pack.

So I gotta crack the battery pack open. And test each cell individually?
How?

A 9.6 V NiCd battery pack should consist of 8 NiCd penlight cells,
connected in series.