Watson's Weston Meter Mystery

[Mjolinor]

"Dave VanHorn" <dvanhorn@cedar.net> wrote in message
news:bZudnaNeau9IG-PdRVn_iw@comcast.com...

I pay less than $2 per cell on Ebay.

Most people don't shop on Ebay, they buy their batteries at the grocery
store or worse yet, Radio Scrap. And they're _not_ cheap there!

Half the people you meet are below average intelligence.

Hmmm, half the people you meet are average intelligence or below.

 

[Mjolinor]

"Mjolinor" <mjolinor@hotmail.com> wrote in message
news:amwfc.80$O07.17@newsfe1-win...

"Dave VanHorn" <dvanhorn@cedar.net> wrote in message
news:bZudnaNeau9IG-PdRVn_iw@comcast.com...

I pay less than $2 per cell on Ebay.

Most people don't shop on Ebay, they buy their batteries at the
grocery
store or worse yet, Radio Scrap. And they're _not_ cheap there!

Half the people you meet are below average intelligence.



Hmmm, half the people you meet are average intelligence or below.

I don't think that's right either.

What about:- Just under half the people you meet are below average
intelligence.

Half the average intelligence people you meet plus (half the people you meet
minus those of average intelligence) are below average intelligence.

Is that any good?

 

[William P.N. Smith]

"Mjolinor" <mjolinor@hotmail.com> wrote:

Half the average intelligence people you meet plus (half the people you meet
minus those of average intelligence) are below average intelligence.

Half the people you meet are below the average intelligence of people
you meet. [I don't hang out in places (*) that represent a
statistical sample of intelligences]

(*) except for newsgroups, of course. 8*)

--
William Smith
ComputerSmiths Consulting, Inc. www.compusmiths.com

 

[jakdedert]

Watson A.Name - "Watt Sun, the Dark Remover" wrote:

"Dave VanHorn" <dvanhorn@cedar.net> wrote in message
news:04GdnTeKUMrw7-XdRVn-gQ@comcast.com...

The problem with Inova is they use a much more expensive lithium
battery.
See URL http://www.inovalight.com/site.html?X5-ov

I pay less than $2 per cell on Ebay.

Most people don't shop on Ebay, they buy their batteries at the
grocery store or worse yet, Radio Scrap. And they're _not_ cheap
there!

Go to your local Batteries Plus store and get fresh Duracell Procell AA's
for around half a buck apiece.

jak

 

[Watson A.Name \"Watt Sun]

Watson A.Name - "Watt Sun, the Dark Remover" wrote:

"jakdedert" <jdedert@bellsouth.net> wrote in message
news:hYcfc.42797$Lh2.5187@bignews1.bellsouth.net...

[snip]

Looks like a lot of the auto mfgrs are making custom reflectors for the
LEDs in their taillights. They don't act like regular incandescents, ya
know.

jak

I found this URL that says that regular 5mm white LEDs degrade to 70%
brightness after only 3000 hours. This confirms my empirical
experiments with some LEDs that I've had running since before xmas.
See page 2, paragraph 4.
http://www.lumileds.com/newsandevents/releases/May_06_2003_Warn_white.pdf

 

[Jim Yanik]

"Watson A.Name - \"Watt Sun, the Dark Remover\"" <NOSPAM@dslextreme.com>
wrote in news:107s96077lc1n2c@corp.supernews.com:

"Dave VanHorn" <dvanhorn@cedar.net> wrote in message
news:04GdnTeKUMrw7-XdRVn-gQ@comcast.com...

The problem with Inova is they use a much more expensive lithium
battery.
See URL http://www.inovalight.com/site.html?X5-ov

I pay less than $2 per cell on Ebay.

Most people don't shop on Ebay, they buy their batteries at the grocery
store or worse yet, Radio Scrap. And they're _not_ cheap there!

If you can find an $8.00 battery for $1.66 from an Internet source,why
would you buy from the more expensive place?

--
Jim Yanik
jyanik-at-kua.net

 

[Jim Yanik]

"Watson A.Name - \"Watt Sun, the Dark Remover\"" <NOSPAM@dslextreme.com>
wrote in news:107sui1la473acc@corp.supernews.com:

William P.N. Smith> wrote in message
news:ksos709p65qntlek9rs64rjpd4cshhs4th@4ax.com...
"Watson A.Name - \"Watt Sun, the Dark Remover\""
NOSPAM@dslextreme.com> wrote:
What's bothering me is that very few flashlight makers have
progressed
to designing their lights to use rechargeable cells. If more AA cell
sized flashlights would be designed to use Ni-MH cells, the world
would
have a lot less batteries to trash, recycle, etc.

And a lot more dead flashlights sitting in drawers, cars, and other
places. Sure, if they get a lot of use, rechargable batteries are
nice, but they'll self-discharge to uselessness in a month or two. I
dunno how many flashlights I've got, but if I had to worry about
keeping their batteries charged I'd end up doing nothing else...

I don't know about you, but it doesn't seem to be a problem for the
several cordless phones, cell phones and cordless tools I have. They
come with a cradle to put them in when not in use. Another way is like
the ones we have at work: there is a fold-out plug on the side of the
flashlight and it stays plugged into the wall when not in use.

This isn't rocket science, you know. The cost or a rechargeable
cordless phone, screwdriver or whatever might be in the $20 to $30
range, less on sale. The leap from a 900 MHz full duplex transceiver to
a simple light beam is a leap backwards in technology.

--
William Smith
ComputerSmiths Consulting, Inc. www.compusmiths.com

Those cordless phone batteries are intended for frequent use.
Flashlights mostly sit around for long periods before being used.
Those lithiums have a 10 year shelf life.

--
Jim Yanik
jyanik-at-kua.net

 

[R Adsett]

In article <77ij5c.80k.ln@cubalibre.marini>, NO_SPAM@marini-spa.com
says...

Dear Friends,
I'm a student, and I have to write a SW to connect an ATMEL CPU + CAN
INTERFACE to a CATERPILLAR CPU via CAN-BUS.
The protocol is SAE J1939.
My problem is I don't find anything on the WEB about the specifications.
Can anybody help me?
Try http://www.kvaser.com/ and follow the education link to white papers

for this paper.

Diesel Engine Control, CAN Kingdom and J1939

That should give you a starting point.

Robert

 

[Graham W]

<William P.N. Smith> wrote in message
news:vt7t70p0tc04mavqvmbrme1u78p3r0ung0@4ax.com...

"Mjolinor" <mjolinor@hotmail.com> wrote:
Half the average intelligence people you meet plus (half the people you
meet
minus those of average intelligence) are below average intelligence.

Half the people you meet are below the average intelligence of people
you meet. [I don't hang out in places (*) that represent a
statistical sample of intelligences]

(*) except for newsgroups, of course. 8*)

Creep!

But did you know that most people have more fingers than the average?


--
Graham W http://www.gcw.org.uk/ PGM-FI page updated, Graphics Tutorial
WIMBORNE http://www.wessex-astro-society.freeserve.co.uk/ Wessex
Dorset UK Astro Society's Web pages, Info, Meeting Dates, Sites & Maps
Change 'news' to 'sewn' in my Reply address to avoid my spam filter.

 

[jakdedert]

Watson A.Name "Watt Sun - the Dark Remover" wrote:

Watson A.Name - "Watt Sun, the Dark Remover" wrote:
"jakdedert" <jdedert@bellsouth.net> wrote in message
news:hYcfc.42797$Lh2.5187@bignews1.bellsouth.net...

[snip]

Looks like a lot of the auto mfgrs are making custom reflectors for
the LEDs in their taillights. They don't act like regular
incandescents, ya know.

jak

I found this URL that says that regular 5mm white LEDs degrade to 70%
brightness after only 3000 hours. This confirms my empirical
experiments with some LEDs that I've had running since before xmas.
See page 2, paragraph 4.
http://www.lumileds.com/newsandevents/releases/May_06_2003_Warn_white.pdf

Three thousand hours is a long time in a flashlight...typically several
bulbs worth, assuming the bulbs don't fail from shock first. Compared to
the little halogens in a mini-maglite, 70% is still several times brighter
(assuming equal output when new) than end-of-life out put for the halogen.
Someone else pointed out something I had forgotten about halogens: namely
that the silvering of the envelope accelerates under lower than rated
voltage, due to the fact that the filament does not regenerate itself as it
does in a full-voltage condition. Of course, batteries--especially
alkalines and carbon/zincs--spend most of their useful life at less than
'fresh' voltage....

jak

 

[Watson A.Name \"Watt Sun]

jakdedert wrote:

Watson A.Name - "Watt Sun, the Dark Remover" wrote:

"Dave VanHorn" <dvanhorn@cedar.net> wrote in message
news:04GdnTeKUMrw7-XdRVn-gQ@comcast.com...

The problem with Inova is they use a much more expensive
lithium battery. See URL
http://www.inovalight.com/site.html?X5-ov

I pay less than $2 per cell on Ebay.

Most people don't shop on Ebay, they buy their batteries at the
grocery store or worse yet, Radio Scrap. And they're _not_ cheap
there!

Go to your local Batteries Plus store and get fresh Duracell Procell
AA's for around half a buck apiece.

Never heerd of it. I'll have to see if there a store around here. I
doubt it, tho.

Nope. I checked their website, the closest store is 40 miles away, cost
me ten bucks for gas to get there and back. Forget it.

I can go to Fry's and get a brick of AA alkalines for $.25 to $.30 each.
Cheaper online, at such places as cheapbatteries.com.

> jak

 

[Robert Baer]

Jorgen Lund-Nielsen wrote:

George Francis schrieb:

Hi All:

I need a small vacuum pump for use in a photo vacuum frame. Need not
be able to draw large vacuum (relatively small frame) but preferably
120 AC. Any suggestions for an alternate vacuum device will also be
most appreciated. Thanks, George

Maybe an aquarium air pump (fish tank) could do the job?
Those i knew are membrane pumps (diaphragm pump) and are able
to buila a nice vacuum.

HTH

Jorgen
dj0ud

*THAT* is exactly what i was going to recommend.

 

[wylbur37]

"Watson A.Name - \"Watt Sun, the Dark Remover\"" <NOSPAM@dslextreme.com> wrote in message news:<107sv0dl29r3730@corp.supernews.com>...

"wylbur37" <wylbur37nospam@yahoo.com> wrote in message
news:8028c236.0404150300.6cf40740@posting.google.com...
"Watson A.Name - \"Watt Sun, the Dark Remover\""
NOSPAM@dslextreme.com> wrote in message
news:<107s90arisj914@corp.supernews.com>...

I've been running white LEDs at 50 to 60 mA, and I'm finding them
dying
in a matter of days to weeks. The 20 mA that they are rated for is
okay, but 30 mA is iffy and and above that will start to overheat
them,
and they lose their long lifetime.

How would you reduce the current of the power source to make it
compatible
with the rated current of the LED?
After all, the standard formula for the desired resistance is ...

R = (V8 - Vf)/If

R = the resistor
V8 = dc supply voltage
Vf = rated forward voltage of the LED
If = rated forward current of the LED at specified forward voltage

Notice that nowhere in the formula is the current of the power source.
So if the voltage of the source is already equal to the rated voltage
of the LED, how would you reduce the current?

That last paragraph, above, has got me bumfuzzled. I'm not sure what
you mean.

I'd like to take the LED from the key-fob flashlight and make it into
a booklight (so I can read a book or use a computer in the dark so as
not to disturb others in the room).

However, since I plan to have this LED on continuously for several
hours at a time, it's not practical or economical to run it on the
original lithium disk batteries (because the batteries would quickly
be exhausted).

So I figured on using an AC adapter that puts out the same 6V that the
two batteries in series had put out. But if the AC adapter puts out
100mA of current (as it says on the unit) and if the LED is rated at
20mA (as someone had suggested), then wouldn't I be overloading the
LED? (After all, you said in your original posting that running your
LEDs at a higher current than their rating would prematurely burn
them out).

So, assuming that's true, the idea is to find some way to reduce the
current that the LED is exposed to from 100mA to 20mA when running it
on the AC adapter.

 

[Michael A. Covington]

To reduce the current in an LED, use a larger resistor.

Ohm's Law is:

R = E / I

where R is resistance, E is voltage, and I is current.

In the case of an LED circuit, E is the voltage drop across the resistor.


Example:

A good old-fashioned red LED whose internal voltage drop is 1.8 volts.

A 6-volt power source.

You want 20 mA through the LED.

R = (6 - 1.8) / 0.020 = 210 ohms.



We should also check the power rating the resistor, since in some cases a
1/8-watt resistor will not be big enough. (The power rating of the resistor
is a maximum, so anything larger than the actual power dissipation will be
safe.)

P = E * I = (6 - 1.8) * 0.020 = 0.084 watt

This is well under 1/8 watt, so a common 1/8-watt resistor will do fine.



Do similar calculations for any LED and any voltage source.

Or, if you like to experiment, simply hook up a milliammeter in series with
the power source. Start with a resistor that you know is too large (like
1000 ohms) and measure the current flow. Change resistors until you get
what you want.


Crucially, unlike a light bulb, and LED does not limit its own current. The
voltage drop across an LED is nearly constant regardless of the current.
That's why I treated it as constant in the calculation above. That's also
why an LED requires a resistor. If you connect it directly to a voltage
source, it will generally burn out, and if it doesn't, its current will vary
tremendously with small fluctuations in the voltage source.


This must be one of the most frequently asked questions in all of
electronics. I haven't seen anyone post a good answer recently.



--
----------------------------------------------------------------------------
---------
Michael A. Covington, Associate Director
Artificial Intelligence Center, The University of Georgia
http://www.ai.uga.edu/~mc

 

[George Francis]

Robert Baer <robertbaer@earthlink.net> wrote in message news:<407F8F71.BB2FF60F@earthlink.net>...

Jorgen Lund-Nielsen wrote:

George Francis schrieb:

Hi All:

I need a small vacuum pump for use in a photo vacuum frame. Need not
be able to draw large vacuum (relatively small frame) but preferably
120 AC. Any suggestions for an alternate vacuum device will also be
most appreciated. Thanks, George

Maybe an aquarium air pump (fish tank) could do the job?
Those i knew are membrane pumps (diaphragm pump) and are able
to buila a nice vacuum.

HTH

Jorgen
dj0ud

*THAT* is exactly what i was going to recommend.

Thanks for the suggestions......the pump may do the job. Thanks to
Robt. Baer also. George

 

[Watson A.Name - \"Watt Su]

"wylbur37" <wylbur37nospam@yahoo.com> wrote in message
news:8028c236.0404160156.6e06bd1a@posting.google.com...

[snip]

Notice that nowhere in the formula is the current of the power
source.
So if the voltage of the source is already equal to the rated
voltage
of the LED, how would you reduce the current?

That last paragraph, above, has got me bumfuzzled. I'm not sure
what
you mean.

I'd like to take the LED from the key-fob flashlight and make it into
a booklight (so I can read a book or use a computer in the dark so as
not to disturb others in the room).

However, since I plan to have this LED on continuously for several
hours at a time, it's not practical or economical to run it on the
original lithium disk batteries (because the batteries would quickly
be exhausted).

So I figured on using an AC adapter that puts out the same 6V that the
two batteries in series had put out. But if the AC adapter puts out
100mA of current (as it says on the unit) and if the LED is rated at
20mA (as someone had suggested), then wouldn't I be overloading the
LED? (After all, you said in your original posting that running your
LEDs at a higher current than their rating would prematurely burn
them out).

So, assuming that's true, the idea is to find some way to reduce the
current that the LED is exposed to from 100mA to 20mA when running it
on the AC adapter.

I did exactly the same thing, I put several white LEDs in a wall wart AC
adapter to let them run for the last 5 or 6 months. I used a 6VDC, 100
mA adapter, which had a voltage of about 7.7 to 8VDC with no load on it.
I used 150 ohm resistors in series with the LEDs, but if you have only
one LED, then use a 180 or 220 ohm resistor. For a 220 ohm resistor,
you should measure 4.4VDC across it at 20 mA. For a 180 ohm resistor,
you should measure 3.6VDC across it at 20 mA. A bit higher or lower
should be okay. To find the current, just divide the voltage you
meassure by the resistance.

View the following with courier font.

220 ohm
+ o----/\/\/\--------+
|
Wall Wart --- --> white LED
6VDC 100 mA \ / -->
--- Cathode is flattened point
|
= o------------------+

 

[Watson A.Name - \"Watt Su]

"Michael A. Covington" <look@www.covingtoninnovations.com.for.address>
wrote in message news:_dmdnT2ww9y8S-Ld4p2dnA@speedfactory.net...

To reduce the current in an LED, use a larger resistor.

Ohm's Law is:

R = E / I

where R is resistance, E is voltage, and I is current.

In the case of an LED circuit, E is the voltage drop across the
resistor.


Example:

A good old-fashioned red LED whose internal voltage drop is 1.8 volts.

A 6-volt power source.

You want 20 mA through the LED.

R = (6 - 1.8) / 0.020 = 210 ohms.

We should also check the power rating the resistor, since in some
cases a
1/8-watt resistor will not be big enough. (The power rating of the
resistor
is a maximum, so anything larger than the actual power dissipation
will be
safe.)

P = E * I = (6 - 1.8) * 0.020 = 0.084 watt

This is well under 1/8 watt, so a common 1/8-watt resistor will do
fine.

Do similar calculations for any LED and any voltage source.

Or, if you like to experiment, simply hook up a milliammeter in series
with
the power source. Start with a resistor that you know is too large
(like
1000 ohms) and measure the current flow. Change resistors until you
get
what you want.

Or to save having to change resistors, just put a 1000 ohm pot in there,
and adjust to get the correct current. Then measure the pot and use the
next higher standard value.

BTW, I dunno about you, Mike, but most resistors (at least the thru hole
type, anyway) that I see are 1/4W, not 1/8W.

Crucially, unlike a light bulb, and LED does not limit its own
current. The
voltage drop across an LED is nearly constant regardless of the
current.
That's why I treated it as constant in the calculation above. That's
also
why an LED requires a resistor. If you connect it directly to a
voltage
source, it will generally burn out, and if it doesn't, its current
will vary
tremendously with small fluctuations in the voltage source.


This must be one of the most frequently asked questions in all of
electronics. I haven't seen anyone post a good answer recently.

Hey, yours is a good one ;-)

--
---------
Michael A. Covington, Associate Director
Artificial Intelligence Center, The University of Georgia
http://www.ai.uga.edu/~mc

 

[H. R. Bob Hofmann]

wylbur37nospam@yahoo.com (wylbur37) wrote in message news:<8028c236.0404150246.30a3546f@posting.google.com>...

don@manx.misty.com (Don Klipstein) wrote in message news:<slrnc7rpg6.106.don@manx.misty.com>...
In article <8028c236.0404140322.bb1f288@posting.google.com>, wylbur37 wrote:
I recently bought one of those cute key-fob LED flashlights at K-Mart.
The brand name is Coleman, and the flashlight is flat and shaped like
one of their typical Coleman lanterns. It cost $3.49 and runs on two
CR2016 (3V lithium) disk batteries (6V total). There's a resistor
inside in series that has a value of 51 or 52 ohms. (The bands on the
resistor were green, brown, black and gold, although I wasn't sure if
the second band was brown or red).

Anyway, I'm trying to identify the specs (voltage, amperage, mcd,
etc.) of the LED used in this flashlight (in case I want to transplant
it for use in another application).

I can say that the resistor is probably 51 and not 52 ohms since 51 is a
standard value and 52 is not.

As for the LED: Based on the LEDMuseum page, I am guessing that this is
a plain old ordinary common (by today's standards) "super red" or "super
bright red" or "ultrabright red" GaAlAsP LED with a typical forward
voltage drop at 20 mA around 1.85-1.9 volts, maximum rated current of 30
mA, brightness probably 3,000 mcd (possibly less), and peak wavelength
probably around 660 nm.

No, it's not a red LED, it's a *WHITE* LED.
Sorry I neglected to mention that.\

Start with V=IR, R=V/I

The wall wart will put out the voltage stated, at a current up to the
value given (150ma I think you said). The voltage drop across the
series resistor and
the LED will have to add up to the output voltage of the wallwart. The
LED drop is more or less constant voltage, so as you increase the
resistance of the series resistor, the current trhough the total
resistor and LED goes down,. So you need to measure the output
voltage of the wallwart feeding a pure resistance at let's say 30 ma,
that may be a little higher than the rated voltage when the output
current is a full 150 ma (say you get 7V). Then put the LED in series
with the resistor and see how bright it is, it should be fairly dim
because there is less current flowing. Measure the voltage across the
resistor (say 4V) Now, divide the voltage across the resistor (4V) by
the current you want thru the LED, say 0.030 amperes (30 ma), 4/0.03 =
133 ohms.
That is the nominal valuue you want for the resistor.

Remember, voltage is like water pressure, current is like the size of
the pipe used to deliver the water.

H. R. (Bob) Hofmann

 

[Soeren]

wylbur37nospam@yahoo.com (wylbur37) wrote in
news:8028c236.0404160156.6e06bd1a@posting.google.com:

So I figured on using an AC adapter that puts out the same 6V that
the two batteries in series had put out.

Just remember there is a resistor in the circuit (in between the lithium
cells).

(It is *not* the voltage of the batteries you need for calculating the
resistor for use with an adapter, it is the LEDs voltage drop - measure it
or calculate with ~3V6)

But if the AC adapter puts
out 100mA of current (as it says on the unit)

The adapter doesn't "put out", the circuit it feeds draws from it. The
100mA is a rating of the maximum current you can draw (without overloading
the adapter).

and if the LED is rated
at 20mA (as someone had suggested), then wouldn't I be overloading
the LED? (After all, you said in your original posting that running
your LEDs at a higher current than their rating would prematurely
burn them out).

You control the current of the LED by choosing the right size resistor as
others have allready explained.

So, assuming that's true, the idea is to find some way to reduce the
current that the LED is exposed to from 100mA to 20mA when running it
on the AC adapter.

You need to replicate the resistor (51 Ohm) in the lantern, but you just
need a higher value.
In the lantern, when the battery cells is full, the current draw is around
47mA (given a LED voltage drop of 3V6). While this is acceptable in a
lantern used only for a few seconds (or minutes) occasionally, any long
term use at this current will degrade and eventually kill the LED too
fast. go for a max. current of 20..25mA.

The resistor will then be:
(6-3.6)/0.025 = 96 Ohm
(6-3.6) / 0.020 = 120 Ohm
Standard values (E12) are 100 Ohm and 120 Ohm.

The LED will not give the same amount of light as in the lantern, of
course, but it will last longer.


--
Regards,
Soeren

* If it puzzles you dear... Reverse engineer *

 

On Wed, 14 Apr 2004 23:08:33 -0700, "Watson A.Name - \"Watt Sun, the
Dark Remover\"" <NOSPAM@dslextreme.com> wrote:

"Jeff Wiseman" <wisemanja@earthlink.net> wrote in message
news:4078D308.F95B11F5@earthlink.net...


Dave VanHorn wrote:

The problem with Inova is they use a much more expensive lithium
battery.
See URL http://www.inovalight.com/site.html?X5-ov

I pay less than $2 per cell on Ebay.


Check out Surefire's website (www.surefire.com). You can get them
there by the box for as low as $1.25 each.

I can buy a brick of AA alkalines for $.25 - $.30 each not on sale, and
even less when they're on sale. Hard to beat that for inexpensive
batteries, and everything seems to use AA cells.

What's bothering me is that very few flashlight makers have progressed
to designing their lights to use rechargeable cells. If more AA cell
sized flashlights would be designed to use Ni-MH cells, the world would
have a lot less batteries to trash, recycle, etc.

- Jeff

I've found cheap, off-brand batteries simply don't last very long. A
bought a bunch of el-cheapos from Big-Lots and they lasted about 1/3
as long as name brand ones.

NiMH batteries don't handle the current draw of a incandescent
flashlight very well. The do seem to be a good match for LED
flashlights though.