A simple E-field probe for measuring
near-field radiation (to verify
compliance with FCC limits) is described at:

http://www.kn4s.com/rf.html
RF Safety

(about halfway through the document)

My question is this:

Looking only at the two plates separated
by 10 cm, what is the impedance the
plates would present to a measuring
device in the hf spectrum? Is there a
simple way to calculate (estimate) the
impedance? Is the charging circuit time
constant a simple function of the
resistivity of air?

Ordinarily, a 10 megohm instrument would
load such a probe excessively because
the probe's impedance would be much
higher than 10 megohms.

Many thanks.

Chuck
NT3G

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On Sat, 08 Jul 2006 10:38:18 -0400, chuck wrote:

My question is this:

Looking only at the two plates separated
by 10 cm, what is the impedance the
plates would present to a measuring
device in the hf spectrum?

Hi Chuck,

From their own page:
"the capacitance of its plates (1.8 pF)"

There are enough dimensional clues to compute this yourself, however.

Is there a
simple way to calculate (estimate) the
impedance?

Xc = 1/(2 · pi · f · C)

Is the charging circuit time
constant a simple function of the
resistivity of air?

No.

Ordinarily, a 10 megohm instrument would
load such a probe excessively because
the probe's impedance would be much
higher than 10 megohms.

The probe's resistance is what that is important, not its impedance
(which is to say the reactance is immaterial except as a scaling
factor as described). In that regard, the resistance is nearly
infinite for all practical purposes. A 10 Meg Ohm meter would take 10
ms to discharge in comparison to 0.1 µs to recharge. The circuit is a
clamping circuit (or peak detector) that automatically disconnects any
load from the probe. The actual load is the sag in the charge to the
capacitor. The ratio of charge/discharge insures that the volt meter
is quite transparent to the probe - unless you are after an accuracy
that borders on delusion.

73's
Richard Clark, KB7QHC

On Sat, 08 Jul 2006 09:09:02 -0700, Richard Clark
wrote:

A 10 Meg Ohm meter would take 10
ms to discharge in comparison to 0.1 µs to recharge.

This was rather a hip-shot guesstimate where I put the cap on the
wrong side of the isolation line's resistance. Anyway, there's still
a significant charge/discharge ratio that isolates the meter.

On the other hand, having called it a clamping circuit was a stretch
too. It is in fact a rather sloppy design in that regard (it is self
discharging) but this is not to say it lacks its intended utility.

73's
Richard Clark, KB7QHC

Richard Clark wrote:
On Sat, 08 Jul 2006 09:09:02 -0700, Richard Clark
wrote:

A 10 Meg Ohm meter would take 10
ms to discharge in comparison to 0.1 µs to recharge.

This was rather a hip-shot guesstimate where I put the cap on the
wrong side of the isolation line's resistance. Anyway, there's still
a significant charge/discharge ratio that isolates the meter.

On the other hand, having called it a clamping circuit was a stretch
too. It is in fact a rather sloppy design in that regard (it is self
discharging) but this is not to say it lacks its intended utility.

73's
Richard Clark, KB7QHC

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.


The 0.1 uS time constant suggests a 5.6
megohm resistance in the charging
circuit. From where does that resistance
arise?

Put differently, what limits the rate at
which energy can be extracted from an
E-field through a capacitance of 1.8 pF?



Thanks.

Chuck


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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

Thanks for the additional comments,
Richard. The probe seems remarkable in
being able to utilize a 10 megohm DMM to
measure an RF field given that a DMM
would not work at all to measure a
static electric field.

Static field measurements are almost
always done with extremely high input
resistance electrometers, fieldmeters,
and field mills. All of these usually
incorporate voltage follower op amps or
something similar to keep the input
resistance above 10 exp(13) ohms.
Hooking up a DMM would quickly discharge
any electrodes in the field.

Thanks again.

Chuck

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