E-field probe question
On Sat, 08 Jul 2006 13:08:18 -0400, chuck wrote:
Thanks for the info, Richard.
What I was looking for was information
on the charging part of the circuit. It
is not clear to me how or why the 1.8 pF
should recharge from the E-field in 0.1 uS.
Hi Chuck,
As I said, an error with shooting from the hip. Besides, it is not
the plates of the probe that are charging when compared to the 0.01µF
cap, lead resistors, and meter. Rather, that probe capacitance
(according to the author) is part of the reactance that is in series
with the reactance of the diode. I'm not entirely convinced of the
text's accuracy, but I'm not motivated to read it deeply either. The
diode capacitance is a moving target and even its average is
unsuitable for the analysis presented by the author.
The 0.1 uS number was drawn from a hat for an average over the HF
region when the series resistance is low - guaranteed for at least
half a cycle when there is no stored charge. This returns us to the
interpretation of the clamping circuit, but this interpretation
remains tenuous.
Put differently, what limits the rate at
which energy can be extracted from an
E-field through a capacitance of 1.8 pF?
The source resistance, which in this case is the field. To describe
further embarks us upon the nature of extracting the signal's strength
from the field, and then on to matters of matching load to the source.
Beyond that, it becomes an issue of efficiency.
However, with respect to the proposed probe, matching and efficiency
are not a concern. The point of the 0.01 capacitor (not the plate
structure of the probe) is to store sufficient charge (undrained by
the meter load) so as to effectively become an open circuit to power
demand - for half a cycle at least in the design offered. The point
of a probe is to not disturb the field (not take power, and to return
as much energy as was extracted).
73's
Richard Clark, KB7QHC
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