On Dec 19, 2005, at 8:48 PM, Michael Foster wrote:
Robin wrote:
If the electrodes do indeed form diodes, and the glow occurs
during reverse bias, then that is when a high voltage falls across
a very thin chemical layer. The electron leakage current could be
sufficiently accelerated to produce energetic electrons capable of
exciting high energy (i.e. blue) transitions within the atoms.
I may be the only Vort having extensive practical experience with
this phenomenon.
I have some practical experience with this.
Yes, the electrodes really form diodes. This is
hundred-year-old stuff. If you use lead or stainless steel for one
of the electrodes it makes a serviceable rectifier, although the
voltage drop is about five volts, as opposed to a silicon rectifier
at about .6 volt.
Yes, but note this means steel and lead do not form the rectifying
layer.
I don't think its accurate to refer to this as electroluminescence.
It's more like electro-scintillation. If you look at the electrode
under about 40X magnification, it resembles a swarm of fireflies.
Most of the light given off is apparently in the UV. If you add a
fluorescent dye to the electrolyte you get a much brighter display.
And of course, you can choose the color you want. Fluorescein or
rhodamine 6G work nicely.
A sign that UV is present in addition to the blue-green.
Given the appearance of the electrodes under magnification, I'm not
sure if Robin's hypothesis would explain the phenomenon. Wouldn't
electron leakage current produce a more uniform light intensity?
Not if the glow itself exist in the electrolyte and thus is subject
to and possibly even creates microscopic turbulence.
Alternate explanations might be simple arcing on a microscopic scale,
or maybe oxygen bubble formation and subsequent collapse, thereby
producing sonoluminescence. Incidentally, you can still form the
semiconductor layer at lower voltage, around 15V, but no light is
given
off. I'm not sure what the voltage threshold is for the glow.
At the low voltage used by Nyle Steiner in the article:
<http://home.earthlink.net/~lenyr/borax.htm> the breakdown
threshold, when AC was applied, varied with the peak to peak
operating voltage.
I quote from the article: "The reverse breakdown voltage of these
rectifiers tends to adjust itself to a point slightly below the peak
of the applied voltage. When the ac voltage is increased, this
breakdown value increases accordingly. When the ac voltage is
decreased, this breakdown value decreases accordingly. This can be
seen in the pictures above as downward dips in left portion of the
curve from the curve tracer and at the bottom of the half wave
rectified waveform."
I used much higher voltages though, up to over 800 V, so I was able
to see the breakdown threshold find a limit in some of the cases I
tested. The limit was high, over 200 V, and depended on the cell
conductivity. Such a dependence may only indicate that the actual
breakdown voltage across the thin layer is constant because cell as a
whole is a voltage divider. Still, electrolyte composition and
concentration greatly affects the threshold too, more than just in
proportion to cell resistance. Also, previously used and conditioned
electrodes formed the glow very fast, as opposed to fresh metal that
required conditioning time of up to 15 min.
Here's a hypothesis. Suppose the critical criteria for the effect is
formation and sustaining of an insulating layer that prevents anion
exchange across its boundary. The anode layer only permits electron
flow across itself, either by tunneling, or by direct conduction.
The main candidate remaining for the light producing reaction I think
might then be:
OH- -> OH + e-
where the anode takes the stripped electron. The anode can similarly
strip electrons from other negative radicals. The anode repels
positive ions, so the concentration of these should be low in the
very close vicinity of the anode, though the shielding effect, ion
encapsulating effect, of neutral H2O is very strong, so that non-
conducting gap would be small. The H3O+ and other cations need only
be separated from the anode by a sufficient distance to prevent
tunneling of charge. This eliminates cations in the vicinity. Since
both cations and anions are for the most part eliminated from the
interface layer, it can not conduct. When the polarity changes, the
protons from the H3O+ are able to tunnel their way through the H2O
interface in the usual manner. The diode effect then essentially
comes from the difference in mobility of protons vs OH- through the
anode interface layer. The boundary can be broken when the
anodeinterface potential drop is sufficient to force OH- radicals
through the (probably two molecule thick) anode interface boundary.
