To List,
Do you remember when you started reading the articles about making
cs, and finding where to buy 0.999 fine wire to make your own?
Well, it probably came in 3 ft lengths. So you cut 5 or 6 inches for
each rod, and put the rest back in the drawer for safekeeping.
Maybe it's still there! If so here's something to consider.
We all started with the same articles, and everyone pretty much
built the same thing, and everyone got fairly similar results.
But if you take a suitable length of wire, and bend it into a "W",
you can get 4 times the wetted area. Do it again, and you have a
funny-looking but very effective anode and cathode.
This does good things. First, it reduces the current density by a
factor of 4.
If you recall my earlier article on the ULV process, the ratio
between the 1.4 mA and 250 uA process was only 5.6. The higher
current produced a lot of mist and black oxide.
The low current process had a higher ion content because less of the
silver ions was wasted making mist and black oxide.
The other good thing is it reduces the initial cell resistance.
If you are using a current regulator, it will reach current limiting
faster, so the process is less sensitive to changes in the
temperature and purity of the distilled water.
It is also less sensitive to bubbles forming on the rods, and less
sensitive to changes in the ion cloud due to movement of the water.
If you are using constant voltage, get thee to Radio Shack and buy a
dvm and some resistors.
You need to know the initial cell resistance, the wetted area, and
the supply voltage.
For a start, set the desired current density to 100 uA per square
inch. To get the actual operating current, use the following
formula:
D = I / A
where
A = wetted area
D = current density
I = actual current
therefore
I = D * A
If you have 4 square inches of wetted area, and want 100 uA per
square inch,
I = 100e-6 * 4
= 400e-6
= 400 uA
(Remember this number. We will use it at the end.)
Now comes the tricky part. What is the initial cell resistance?
It depends on the purity of the water and the wetted area. So yours
will be different from everyone else.
But you can measure it with your new dvm.
One of the resistors you need is a 100k. Put this in series with the
supply and connect it to the rods.
Measure the voltage between the rods, and quickly disconnect the
battery so you don't change the initial cell resistance too much by
depositing silver ions into the solution.
Say you get 0.5V for the cell volage, and you have a 12V supply. I
know these are fictitious numbers, but it is just to illustrate the
calculation process. Add your own number when you get them:)
First, calculate the voltage across the 100k resistor as V2:
V2 = Vsupply - Vcell
= 12 - 0.5
= 11.5V
Now we can calculate the current through the 100k resistor:
I = V2 / 100e3
= 11.5 / 100e3
= 115e-6
= 115 uA
Since it is a series circuit, the same current goes through the
cell. We can now calculate the initial cell resistance:
Rcell = E / I
= 0.5 / 115e-6
= 4347 ohms
Assume the resistance will drop by a factor of 4 when the silver
ions are released. In our case, the final cell resistance will be
approximately 1k.
The last step is to calculate the external resistor neded in series
with the supply to give the desired operating current.
We get the total resistance needed by using the operating current we
calculated above:
R = E / I
= 12 / 400e-6
= 30,000 ohms
Now, the 1k we estimated for the final cell resistance is only 1/30
of this value, so we can simply ignore it.
(See - didn't I tell you putting more silver under water was a good
idea?)
If the cell resistance is an appreciable fraction of the total
resistance needed, simply subtract it from the total to get the
external resistor value. This works up to a ratio of 1/5 or so.
If you cannot arrive at a satisfactory solution, you can increase
the wetted area, or increase the supply voltage, or both.
So what's it like to run at microamp levels? Boring. Nothing
whatsoever happens until the process nears completion.
You will wait and wait, then suddenly notice very tiny gray spikes
on the bottom of the cathode. Bingo! You are starting to make a bit
of oxide, or maybe it's hydrogen bubbles that have captured some
ions. Anyway, it's a very subtle effect compared to the brown mist
we are used to seeing.
If you are monitoring the voltage across the cell, you may see it
has plateaued at some value.
Now comes the reason for equal anode and cathode area.
Reverse the polarity across the cell by swapping the + and - leads
from the power supply. If you monitor the cell voltage, you will see
it drop substantially, then start to rise again to the plateau
value.
You can reverse the polarity several times, then just leave it for a
while longer, perhaps one-half of the total time since starting the
brew.
When you are satisfied, stop the process and look at the result.
You may have the finest pale straw cs you have ever seen. Pour about
one inch in another glass and add three shakes of salt.
You should see a very strong dispersion start immediately, and get
stronger as the salt dissolves. It should end up almost milky white.
This is the strongest stuff I have ever seen.
Now taste it.
I think the initial taste impression is slightly sweet. I and my
guinea pig friend can detect no metallic aftertaste at all.
The pale straw color was once considered an indication of the finest
cs you could get. But few could reproduce it reliably.
But I'll bet you can anytime you want.
Best Regards,
Mike Monett
(the inevitable corrections will be posted later:)
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