#### Re: Reforming Electrolytics

Craig Sawyers <c.sawyers@...>

Hi Miroslav

it is a
variety of web
sites, that you always seem to take as a gospel,
Well, can you give me one instance in which the data I have quoted from a
web site has been wrong? You might distrust data from web (you might be
surprised that I do too), but that is not to say it is *necessarily* wrong,
or that I'm daft enough to quote something from a flakey source that is
clearly incorrect.

A book published in 30s has pretty good
chance of being
reviewed and proof read carefully, even if 'self published'.
Ah - but what I found objectionable, Mirolav, was your reply at 00:35UTC
*after* I had quoted the bibliographic citation at 23:28UTC.

'160-200 uS-cm'
I was looking at page 27 "Measurement of rise time and withstand voltage of
Middle and High voltage foils", which uses the figure 160-200uS/cm. And
yes, in my haste to prove the argument, I got the cm on the wrong line (top
instead of bottom - which is as you point out incorrect).

I chose the high voltage specification, because Deeley's book was written at
a time when high voltage capacitors were the lion's share of market.

The lower voltage formulations, which you pick, do indeed have a
significantly higher conductivity - but high voltage films use a solution of
Boric Acid, which is very poor conductor. In fact, at room temperature,
only 57g of boric acid will dissolve in 1 litre of water (that data is from
a web site...), so the bath temperature specified by Panasonic is 70C so
that 80g will dissolve easily. Deeley's book has heaters on the baths too,
and cooling pipes. I'm guessing that you need to heat the bath to start
whith, when the forming electrolyte is cold. Then when it is going, and
tens of kW (see below) are going into tank electrically you need to start
cooling.

So converting the data to SI, conductivity of 200uS/cm goes to 20mS/m

And thus my 2m^2 x 2m tank will have a resistance of 50 ohms (two orders
lower than my original figure).

So I hold that my conclusion still holds - I'll take Panasonic's figure here
for their test sample, which is 20mm x 50mm with a current of 50mA. This
gives a current density of 50A/m^2 (an order of magnitude *less* than my
earlier mail).

Now the voltage drops in a forming tank depend on the geometry, but there is
certainly scope for a very large voltage drop in the electrolyte where the
foil enters the tank and the current is high. The power supply connected to
the tank would have to have a specification something like 500V at 100A, say
(remember that Panasonic specify 1kV at 50mA for 10cm^2). At the foil
itself, of course, the voltage between the foil and the electrolyte is
*low* - the foil is essentially shorted to the eletrolyte. Further down the
tank, where the oxide film has formed, there is very little current flow, so
the foil is in an evironment where it sees the full 50V from the supply - so
the voltage between the foil and the forming electrolyte is the power supply
voltage (say 50V).

Which is precisely the behaviour that Deeley shows in his picture on