At 05:23 PM 6/27/2004 -0500, you wrote:
>Evening Ode,
>
>You are trying hard to confuse me.  <grin>
>
> >>I use about 32 volts out of a 24 volt transformer to start, then, of
>>course, the voltage goes down from there as the current gets adjusted to
>>stay constant.
>
>     Are you talking about the load voltage or the power source voltage?
###That is source voltage..actually 81 volts peak to peak AC rectified and
regulated to 32
>
>
>>  Strangely, the voltage levels off at around 4vdc and conductivity keeps
>>rising at a more or less linear rate while current stays rock steady.

>
>In the simple circuit, only a few things that are readily apparent ( to 
>most of us ) enter in.
>One important aspect is the percent regulation of the power supply.   Few 
>are 100 %
 ## This one is pretty close. As soon as conductivity becomes high enough
to pull the desired current,  it locks in and stays there for quite a
while, then changes up by +1 hundreth of a milliamp and stays there till I
get tired of watching. [many many hours up to 78 uS]
>
>In the simple circuit, the ones of us that understand ohms law can see it 
>working, understand the changes I mentioned above, and not doubt Mr. Ohms.
>
>>  According to ohms law, that ain't possible.  I'm trying to figure out
>>what's going on.
>     I think it is possible, and I am tying to understand what is happening 
>to make it appear that ohms law does not work.
##  Part of it may be particle formation.  The voltage line appears to
start to flatten a little around 13 to 15 uS...about where a TE becomes
visible [using constant stirring]
 Beyond 30 uS it get pretty darned flat and the conductivty rise line
starts to flatten out some too.

 ##I was thinking about using a voltmeter like a conductivity meter by
drawing a correlation between voltage drop and 'PPM'. It turns out that
such a thing might be somewhat useful over a fairly short range between 10
and 20 uS but goes off below and beyond that range.  Oh well...
>
>>  Strangely, the voltage levels off at around 4vdc and conductivity keeps
>>rising at a more or less linear rate while current stays rock steady.
>
>     4 VDC and a current factor, compared to   ( 32 volts out of a 24 volt 
>transformer ) is working with a different load.
>
>      By "load" I mean  resistance or wattage factor. ( this wattage factor 
>is something I mentioned a few times in the past.  Seems only one person 
>commented and agreed with me )
##  Load does drop the source voltage.  That's why transformers have max
current ratings. [well, one reason]
 I'm using a 100 ma transformer. If I recall, the draw is never over 7 ma.
 I tried it again with a 400 ma transformer putting out 15.7 vdc at no load.
 The voltage drop line got really flat at around 32 uS. 





>
>      A fixed current ( constant current ) will produce less voltage drop 
>across a lower value of resistance ( load ).  The finished mass of CS has a 
>much lower resistance at the end of the process.
>
>      As you stated, the conductivity increases  ( = the resistance 
>decreases ) so the  series circuit has changed.   Part of the voltage drop 
>is being dropped some place else in the circuit.
## Part of it might be due to the characteristics of transistors.
Semiconductors in general tend to leak a little and might be shunting off
wattage after some point is reached.
  I really don't now that much about electronics...nothing like the
engineer that popped that circuit out of his head like yesterdays news.
[But he doesn't think he knows anything at all about electrolysis like I
sometimes 'think' I do...emphasis on the 'anything at all' part]

 Part of it may be particle formation. Conductivity drops over time as
particles form..
 At those higher uS ranges, ions should be forming crystals faster and
faster that far past the saturation point which appears to be somewhere
around 13 to 14 uS. [I see a slight drop in the drop rate there that gets
more pronounced at around 20 uS going nearly flat at 30 uS]
 My laser doesn't give me numbers but it 'looks' that way.
..then, there's ever increasing plateout rate and oxide deposit formation
max rate at a given current.  That would suck up wattage without increasing
water conductivity...right?
 Removing the deposits doesn't change the voltage line but they would still
be forming at the same rate with plateout increasing with conductivity rise.


>     The "Constant Current Circuit" is working in reverse of the CS 
>conductivity.
>As the CS mass changes in conductivity and resistance, the constant current 
>circuit equals these changes in reverse.
>
>An easy, non technical way to view the constant current circuit is like a 
>manually controlled potentiometer.
##  Yup
>
>I have some very old messages where we touched on this subject.  I stated 
>some facts and opinions about this same thing.
>
>I got ZERO comments or disagreements.  I felt that ever one considered it a 
>NO ISSUE or thought I was so far off the deep end, they did not want to 
>discuss it.

## Ya gotta know 'something' before questions can even be asked or it's
like diving into an empty swimming pool. All you find out is that gravity
works.
  I have maybe a puddle in there and only have an 'idea' where the deep end
is from wading in the puddle. [Never mind what may be lurking there]
 I get lots of ideas, an engineer pops out the picture frame with napkin
squiggles and I cut and try the scenerio till it looks something like the
idea. [Talking to the engineer feels something like freefall..but I do
sometimes manage to wiggle enough to alter the rocks glide path.]
 That's how it works for this nearly ignorant one whos major talent is
using gutter parts with plumbing parts and masons trowels to build bicycles
that roll somehow. [Then, wonder how. :-) ]
 Junk yard wars.  Sometimes the crazy idea works better than expected,
sometimes not.
This one [voltage drop = conductivity gain] apparently broke down just
across the finish line.

 My guess is that that's why reasonably good conductivity meters cost $50
and up.
 And that's not EVEN going into what conductivity means to PPM...or doesn't.

Ode
>
>Maybe I still am.   I look forward to corrections in my analysis above.
>
>Wayne
>
>
>
>
>
>
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