Re: Water probe: improvements

["Tesla list" <tesla@xxxxxxxxxx>]

Original poster: "Bob (R.A.) Jones" <a1accounting@xxxxxxxxxxxxx>

Hi Marco



> Original poster: "Denicolai, Marco" <Marco.Denicolai@xxxxxxxxxxx>
>
> Hi Bob,
>
>  > I just reread your post and realized you had some
>  > compensation for input C of 100pf that's huge figure to
>  > compensate for and the amp may have run out hf gain hence the
>  > wrong edge height.
>
> The original capacitance of the lower voltage arm of the divider is
> about 500 pF. The capacitance of the schottky diodes I added (one
> between input and V+, the other between input and V-) is 50 pF. So two
> of them makes 100 pF.

Oh I did not realize the out put impedance was so high without the diodes.
So removing them will only have a 20% change in your frquency response.
500pF does seem very high even given the high dielectric constant of water.
The out put resistance is what?. At what frequency is the outputs C's
impedance equal to the output resistance.
This is the corner frequency of the hf drop off of your probe or lag.
Lets say that's 0.056MHz so at 0.56MHz it will be 20dB down at 5.6MHz 40dB
down

> I am really not confident the amplifier would last a second without them
> if the TC is turned on.

They usually have internal protection diodes. If the output R of your probe
is high you may not exceed the input current spec.
More problematic is after such an event the opamp/probe output C my be very
slow to recover so if your relying on that for clipping spikes during normal
operation the spikes could get turned in to pulses with long tails.  An
effect I belive in Terry's opto current probe.
>
> The opamp gain (with my simulations) should go to 0 dB at -100 degrees
> at 56 MHz
which means its 20dB at 5.6MHz and 40dB at 0.56Mhz. assuming a first order
roll off i.e. a lag
so you would just have enough gain if your probe start to roll off at
0.056MHz .

Bob
.