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UV nail lamps for EPROM

B

Bud Finley

Jan 1, 1970
0
Can anyone tell me if the 9W UV lamps sold as replacements for
cosmetic nail driers are suitable for EPROM erasing.

IOW are they genuine UV spectrum or some lesser version?

The price is certainly right.

Bud Finely
 
M

Michael A. Terrell

Jan 1, 1970
0
Bud said:
Can anyone tell me if the 9W UV lamps sold as replacements for
cosmetic nail driers are suitable for EPROM erasing.

IOW are they genuine UV spectrum or some lesser version?

The price is certainly right.

Bud Finely


Brand and type? I doubt most of the men on this newsgroup have even
seen what you're talking about, unless it can be used to harden epoxy.


--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida
 
J

Joerg

Jan 1, 1970
0
Michael said:
Brand and type? I doubt most of the men on this newsgroup have even
seen what you're talking about, unless it can be used to harden epoxy.

Snooort!

But we do use nail polish. For fixing enameled wire onto ferrites...

Anyhow, can't this be tried out? Like running it for x minutes, then
checking whether it has erased and once finding the point where it did
work multiply the time by a safety factor y?
 
D

D from BC

Jan 1, 1970
0
Snooort!

But we do use nail polish. For fixing enameled wire onto ferrites...

Anyhow, can't this be tried out? Like running it for x minutes, then
checking whether it has erased and once finding the point where it did
work multiply the time by a safety factor y?

--being goofy---
Glue some EPROM's on your nails and drive over to the local nail spa
for a drying.. :)
Bring a timer...and use one finger at a time..
D from BC
 
J

Joerg

Jan 1, 1970
0
D said:
--being goofy---
Glue some EPROM's on your nails and drive over to the local nail spa
for a drying.. :)
Bring a timer...and use one finger at a time..


Could turn into a nice biz opportunity when some kids with colorful
Mohawk haircuts and leather clothes walk in. "Hey, cool, I want those
27256-2 nails!"
 
M

Michael A. Terrell

Jan 1, 1970
0
Joerg said:
Could turn into a nice biz opportunity when some kids with colorful
Mohawk haircuts and leather clothes walk in. "Hey, cool, I want those
27256-2 nails!"


No, not the 27256. Use the 1702 for the gold! the little idiots do
love their bling. :(

<http://images.google.com/images?um=1&tab=wi&ie=UTF-8&rls=GWYA,GWYA:2006-31,GWYA:en&q=1702+eprom>


--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida
 
D

D from BC

Jan 1, 1970
0
Could turn into a nice biz opportunity when some kids with colorful
Mohawk haircuts and leather clothes walk in. "Hey, cool, I want those
27256-2 nails!"

It'll go with the resistor nipple piercings, power diode tongue
piercings,10Mhz clock crystal necklaces and schematic symbol tattoos..

I'd get the hysteresis symbol tattoo...that one's cool..

D from BC
 
T

The Phantom

Jan 1, 1970
0
Can anyone tell me if the 9W UV lamps sold as replacements for
cosmetic nail driers are suitable for EPROM erasing.

IOW are they genuine UV spectrum or some lesser version?

The price is certainly right.

Bud Finely

Mercury vapor UV lamps come in two varieties; long wave, 365 nm, and short
wave, 254 nm.

Low pressure mercury vapor tubes internally generate mainly short wave UV.
Tubes intended for producing visible light therefore have a phosphor which
glows in the visible when irradiated by short wave UV.

Tubes intended to produce long wave UV have a phosphor which glows in long
wave UV when irradiated by short wave UV.

The short wave UV doesn't penetrate glass, but long wave does (more or
less), so the "glass" of which the tube is made for visible and long wave
use is actual glass.

If the lamp must emit short wave UV, then the envelope is made of quartz,
or a high silica "glass", which does transmit short wave UV, and there is
no phosphor in this case. Such lamps are often designated "germicidal",
and when they're operating, you can usually smell some ozone near the lamp,
although there are so-called "ozone free" lamps.

Short wave UV will harm your eyes if you look at the operating lamp for
very long; a minute or two is enough to give you an object lesson. There
is no sensation of pain while you're looking, and you won't notice the
damage for hours. The short wave UV "sunburns" the conjunctiva and that
night while you're trying to sleep, it feels like your eyes are full of
sand. Every time you move them while they're closed, it hurts. Even
looking at a specular reflection of the lamp can burn your eyes.

Only short wave UV can erase EPROMS (that's why the transparent lid is made
of quartz), and if you look at these lamps:
http://cgi.ebay.com/9-watt-UV-nail-...categoryZ67653QQcmdZViewItem#ebayphotohosting
you will see that the tubes are white looking. That's because there is a
phosphor coating the inside of the tube. This means that this tube
produces long wave UV, which is what the nail dryers use.

On the other hand, these lamps:
http://cgi.ebay.com/9-watt-TUV-UV-l...2QQihZ016QQcategoryZ67653QQrdZ1QQcmdZViewItem
are transparent. There's no internal phosphor coating. These are short
wave UV lamps. But the socketing looks the same as the long wave tubes,
and I'll bet you could put one of the short wave tubes in this nail dryer:
http://cgi.ebay.com/9-watt-GEL-UV-l...4QQihZ016QQcategoryZ67653QQrdZ1QQcmdZViewItem
and have yourself a cheap EPROM eraser. Just rig a shield so you won't be
exposed to the UV while it's turned on.
 
M

Mike Monett

Jan 1, 1970
0
[... very good info on UV]

Thanks for posting this information, Phantom.

I have a question. I notice that silver chloride turns black when
exposed to the light from ordinary office flourescent lights.

This reaction occurs when a UV photon forces a chlorine ion to give
up an electron, which then converts a silver ion to a metal atom.
The metal absorbs visible light and appears black. The reaction is
quite strong with only two overhead lights. Here is a description:

2AgCl + 2UV --> Ag(s) + Cl2(g)

The same reaction occurs outdoors in sunlight. Since the short wave
UV cannot penetrate ordinary glass, I assume the UV in this reaction
is long wave UV.

However, manufacturers of flourescent lights, such as GE, insist
that no UV escapes from their product. But obviously a great deal
does escape.

Do you have any idea how the UV gets through the phosphor coating?

Regards,

Mike Monett
 
R

Roger Hamlett

Jan 1, 1970
0
Mike Monett said:
[... very good info on UV]

Thanks for posting this information, Phantom.

I have a question. I notice that silver chloride turns black when
exposed to the light from ordinary office flourescent lights.

This reaction occurs when a UV photon forces a chlorine ion to give
up an electron, which then converts a silver ion to a metal atom.
The metal absorbs visible light and appears black. The reaction is
quite strong with only two overhead lights. Here is a description:

2AgCl + 2UV --> Ag(s) + Cl2(g)

The same reaction occurs outdoors in sunlight. Since the short wave
UV cannot penetrate ordinary glass, I assume the UV in this reaction
is long wave UV.

However, manufacturers of flourescent lights, such as GE, insist
that no UV escapes from their product. But obviously a great deal
does escape.

Do you have any idea how the UV gets through the phosphor coating?

Regards,
GE, publish the spectrum of their lamps, and show a small peak at
404.656nm, and a second somewhat larger one at 435.833nm, together with a
general very low level of radiation beyond this. The 'black light' type
lamps, typically produce perhaps 10* the peak intensity of either of these
lines, at the shorter wavelength of about 370nm. GE, say that their lights
should be used with 'CovRguard' fittings, or with UV sleeves, to
completely eliminate UV (I don't know who said the bulbs themselves
produce 'no' UV, if you talk to their commercial division, and specify
that there must be no radiation above 400nm, in the illuminated area, they
specify sleeves to be used...). They say that the total UV (beyond 400nm),
is around 1/500th the level from noonday Sun, but not that their is
'none'.
You'd need to work out whether it might be the two visible lines at the
top of the spectrum, that are producing the effect, or 'UV'. I'd suspect
it might be the visible lines, rater than 'UV'.

Best Wishes
 
M

Mike Monett

Jan 1, 1970
0
Roger Hamlett said:
GE, publish the spectrum of their lamps, and show a small peak at
404.656nm, and a second somewhat larger one at 435.833nm, together
with a general very low level of radiation beyond this. The 'black
light' type lamps, typically produce perhaps 10* the peak
intensity of either of these lines, at the shorter wavelength of
about 370nm. GE, say that their lights should be used with
'CovRguard' fittings, or with UV sleeves, to completely eliminate
UV (I don't know who said the bulbs themselves produce 'no' UV, if
you talk to their commercial division, and specify that there must
be no radiation above 400nm, in the illuminated area, they specify
sleeves to be used. ). They say that the total UV (beyond 400nm),
is around 1/500th the level from noonday Sun, but not that their
is 'none'.
You'd need to work out whether it might be the two visible lines
at the top of the spectrum, that are producing the effect, or
'UV'. I'd suspect it might be the visible lines, rather than 'UV'.
Best Wishes

Hi Roger,

Thanks for your very interesting post. Can you give the url for the
info on the GE spectrum?

Untill now, my sources for the reaction of light on silver chloride
claimed that UV light was needed to knock an electron loose from the
chlorine.

However, I just discovered the following entry that shows the
reaction is stronger at UV, but it still occurs even with red light:

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
After hearing about William Herschel's discovery of infrared light
in 1800, Johann Ritter decided to see if he could detect light
beyond the other end of the spectrum - past violet. In 1801, he was
experimenting with silver chloride, which turned black when exposed
to light. He had heard that blue light caused a greater reaction in
silver chloride than red light did and decided to conduct an
experiment to see if this was indeed true. Ritter directed sunlight
through a glass prism to create a spectrum and then placed silver
chloride in each color. He found that the silver chloride
increasingly darkened from the red to the violet part of the
spectrum as predicted. Ritter then decided to place silver chloride
in the area just beyond the violet end of the spectrum in a region
where no sunlight was visible, and was amazed to see an even more
intense reaction there. This experiment showed for the first time
that an invisible form of light existed beyond the violet end of the
visible spectrum. This is now know as the ultraviolet part of the
electromagnetic spectrum.

http://coolcosmos.ipac.caltech.edu//cosmic_classroom/classroom_activities
/ritter_example.html

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
I am very pleased to find this out. I was about to invest some time
on an instrumentation project that would have not worked the way I
planned:)

Regards,

Mike Monett
 
R

Rich Grise

Jan 1, 1970
0
Could turn into a nice biz opportunity when some kids with colorful
Mohawk haircuts and leather clothes walk in. "Hey, cool, I want those
27256-2 nails!"

I once made a tie-tac out of an 8751 chip.

Nobody noticed. )-;

Cheers!
Rich
 
R

Rich Grise

Jan 1, 1970
0
Could turn into a nice biz opportunity when some kids with colorful
Mohawk haircuts and leather clothes walk in. "Hey, cool, I want those
27256-2 nails!"

It'll go with the resistor nipple piercings, power diode tongue
piercings,10Mhz clock crystal necklaces and schematic symbol tattoos..

I'd get the hysteresis symbol tattoo...that one's cool..[/QUOTE]

I'd like to make some earrings (or any other pierced area rings) from
machine shop chips and sell them in Hollyweird or Laguna Beach or
something. ;-)
http://www.neodruid.net/images/MetalChips.jpg

Cheers!
Rich
 
R

Roger Hamlett

Jan 1, 1970
0
Mike Monett said:
Hi Roger,

Thanks for your very interesting post. Can you give the url for the
info on the GE spectrum?
Paper version sent to me at work, when I raised this question with regards
to a commercial application which would be affected by light in the very
near UV.
I'd think they would send another copy, or may have it somewhere on their
website as well.
Just looked. They have a much lower resolution copy, showing the same
peaks, but so poorly resolved, that it is hard to tell the exact
frequencies. The spectra are at:
http://www.gelighting.com/na/busine...ces/learn_about_light/distribution_curves.htm
The curve I had, was for the 'WWX' lamp, with the paper copy ending at
370nm, where some significant output is still shown. The computer version
seems to lose the bottom percent or so.
Untill now, my sources for the reaction of light on silver chloride
claimed that UV light was needed to knock an electron loose from the
chlorine.

However, I just discovered the following entry that shows the
reaction is stronger at UV, but it still occurs even with red light:

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
After hearing about William Herschel's discovery of infrared light
in 1800, Johann Ritter decided to see if he could detect light
beyond the other end of the spectrum - past violet. In 1801, he was
experimenting with silver chloride, which turned black when exposed
to light. He had heard that blue light caused a greater reaction in
silver chloride than red light did and decided to conduct an
experiment to see if this was indeed true. Ritter directed sunlight
through a glass prism to create a spectrum and then placed silver
chloride in each color. He found that the silver chloride
increasingly darkened from the red to the violet part of the
spectrum as predicted. Ritter then decided to place silver chloride
in the area just beyond the violet end of the spectrum in a region
where no sunlight was visible, and was amazed to see an even more
intense reaction there. This experiment showed for the first time
that an invisible form of light existed beyond the violet end of the
visible spectrum. This is now know as the ultraviolet part of the
electromagnetic spectrum.


http://coolcosmos.ipac.caltech.edu//cosmic_classroom/classroom_activities
/ritter_example.html

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
I am very pleased to find this out. I was about to invest some time
on an instrumentation project that would have not worked the way I
planned:)

Regards,

Mike Monett
So it is responding to the visible, but more strongly to UV. A
photo-chemist would probably be able to tell you the minimum energy photon
required to trigger it.

Best Wishes
 
M

Mike Monett

Jan 1, 1970
0
Paper version sent to me at work, when I raised this question with
regards to a commercial application which would be affected by
light in the very near UV. I'd think they would send another copy,
or may have it somewhere on their website as well.
Just looked. They have a much lower resolution copy, showing the
same peaks, but so poorly resolved, that it is hard to tell the
exact frequencies. The spectra are at:

The curve I had, was for the 'WWX' lamp, with the paper copy ending
at 370nm, where some significant output is still shown. The computer
version seems to lose the bottom percent or so.

Great - thanks. The graphs show very little output above 375nm, and
the FAQ states:

"UV exposure from sitting indoors under fluorescent lights at
typical office light levels for an eight hour workday is equivalent
to just over a minute of exposure to the sun in Washington, D.C. on
a clear day in July."

So the UV output is negligible for any practical applications.

[...]
So it is responding to the visible, but more strongly to UV. A
photo-chemist would probably be able to tell you the minimum
energy photon required to trigger it.
Best Wishes

Thanks for your help. I'll try infrared and see if that works.

Regards,

Mike Monett
 
J

Joerg

Jan 1, 1970
0
Rich said:
I once made a tie-tac out of an 8751 chip.

Nobody noticed. )-;

That's because you live in the Silicon Valley ;-)
 
R

Rich Grise

Jan 1, 1970
0
That's because you live in the Silicon Valley ;-)

No, actually John L is closer to Silicon Valley than I am - at the time, I
was living and working in Orange County, a little quasi-libertarian
bastion between LA and San Clemente. ;-)

Cheers!
Rich
 
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