RE: Water probe: improvements

["Tesla list" <tesla@xxxxxxxxxx>]

Original poster: "Denicolai, Marco" <Marco.Denicolai@xxxxxxxxxxx>

Hi Bob,

> What's the hash on the edge?  It looks like you don't have
> something correctly terminated or you have current loops on
> the grounds.

Having to stimulate with a clean, sharp pulse means short connections.
The need the leave the nearby of the probe free from any metallic object
means just the opposite. Having all the stuff at least 4 meters from the
probe means also 4 meters long connections (and groundings too). I
suspect the wall grounding, which is not the same grounding than the
pulse generator (floor) or scope (mains). I have to work on that.

> Is your scope a diff input or grounded/earthed reference.

Ground hearted. Not a problem to have it floating with an insulation
trasformer, but that usually doesn't help. Would a differential probe be
better?

> Can you expand on what you mean by dielectric polarization?

The water is a so called "polar" dielectric. Its polarized molecules
take time to organize according to the changing E field. I haven't yet
found a thorough source for this, but dielectric absorption seems to be
related with polarization. Dielectric absorption is responsible for the
memory effect of certain capacitors (gaining back voltage after having
been shortcircuited) and can be modeled as an RC series in parallel with
the ideal capacitor.
Now, this is just what I am seeing in those waveforms. It is not a decay
any more, is more an extra bump following a sharp falling edge.

> The last trace looks like the C division is lower than the  R
> division. i.e.
> if you have a C divider and R divider in parallel and the C
> divider has a lower ration than R divider, then the rising
> edge will rise quickly to a particular level determine by C
> division ratio followed by a slower rise to R division ratio.
> You will be able to compensate for this in the video amp with
> the correct lead lag terms (hf boost) I think. I would need
> to check that analytically to be certain.

I guess you mean what I call "overcompensation", that makes the falling
edge peak to the negative side of the voltage scale. Well, if I
overcompensate that (I did it to see how it goes), the bump stays there
but the horizontal part after the bump starts bending to the negative
side. Encreasing the overcompensation (i.e. reducing even further the
resistance on the divider lower voltage arm) the horizontal part bends
very well down but the bump is still there (!).
Consider also that the two traces have same setup, only different time
scale for a better insight look.

Best Regards