Ed Bailen wrote in message . ..
My wife recently had a defibrillator implanted. The folks that
manufactured it kindly sent her an email discussing the varoius bands
and field strengths it had been tested against E field and H field).
She is concerned about exposure to RF both from our mobile VHF and HF
operations, and from the un-shielded multi-gigahertz processors in her
lab. We would like to come up with an absolute (not relative) field
strength meter that we can use to evaluate field strengths (mainly E
field) at various frequencies. I could not find anything on Google.

I can cobble something together wirh an antenna, a tuned front end, a
detector, and a peak sampler, but how to calibrate it?

Any suggestions?

TNX Ed Bailen - N5KZW

Hi Ed,
I dont think you can, not without accesss to the proper gear in a RF
shielded room (anechoic chamber?). But, and this is an educated guess,
the defibrillator would be constructed to withstand fairy massive RF
fields - ie, if you lived at the base of a 100Kw TV/ tranmitter. The
only practical thing I can suggest is, as most of these devices can be
monitered via an external pod to check operation, get your heart
specialist to have his monitoring gear on it while you fire up HF/VHF
whatever gear - you might have to do a bit of smooth talking, but what
other option is there......

I have a friend who is in a similar position, he was a TV service tech
and was told not to work on the back of TV' s because of the radiation
from the line output stage - sounds a bit like bull****, but then no
manufacturer is EVER going to say its safe to do this as it would
expose them to litigation if it didnt - sort of like dont ask the
government for pernmission to do anything as they will always cover
their backs and say no.

73 de VK3BFA Andrew.

PS - hope your wife is ok, at least she knows if anything happens it
will kick in and do a "cold restart"

I don't know the threshold specs involved, but you can crudely estimate
field strength if you know transmitter power. The "impedance" of free space
/ air is 377 ohms. Thus the ratio of RF field voltage and current will be
fixed and related to the power density - unless the field is distorted.

As another crude approximation, assume that (unless you know a radiation
pattern) that the unwanted signal's power is equally distributed over a
sphere or maybe a hemisphere.

The above will allow you to compute a crude approximation of field strength.

The FCC requires hams to make these calculations with transmitters above
certain power thresholds.


How close does this come to the stated "safe" field strengths? I certainly
wouldn't want to come as close as 20 dB (10 times voltage) to the stated
thresholds.


TV/FM broadcasting stations have very little field strength directly under
their tall towers. All broadcasting stations are not allowed to exceed
certain field strengths on the ground outside their fenced compounds. The
TV/FM stations in a hilltop residential area in Portland, OR, had to greatly
increase their tower heights when this rule came into effect.

de w3otc

The "impedance" of free space/ air is 377 ohms. Thus the ratio of RF field
voltage and current will be
fixed and related to the power density - unless the field is distorted.

As another crude approximation, assume that (unless you know a radiation
pattern) that the unwanted signal's power is equally distributed over a
sphere or maybe a hemisphere.

R. J. & Ed-

The calculations R. J. refers to apply to a field at a distance - called "far
field". Up close, radiation patterns are complex and not easily predicted.

For the sake of discussion, lets carry this free space impedance topic a bit
further. If you know transmitted power and the type of antenna, you can
approximate the effective radiated power. For example, a half wave dipole and
a quarter wave ground plane have about the same gain, 2.1 dB. If you are
concerned about worst-case, you should assume a handheld radio with a
rubber-duckie antenna, has the same gain as a full size ground plane. (You
could also add some gain due to reflecting surfaces, just to be safe.)

Suppose your handheld radio has five watts output. For a gain of 2.1 dB, the
effective isotropic radiated power is about 8 watts. This power illuminates
the inner surface area of a sphere equal to (4/3) Pi R-squared, where R is the
distance.

At some distance the field strength will equal an amount that will interfere
with the defibrillator. Suppose that critical field strength is one Volt per
Meter. One Volt squared, divided by 377 equals 2.7 milliWatts per square
Meter.

Working this backwards with the sphere, 4/3 Pi R-squared divided by 8 watts
equals the power (2.7 mW) per square meter at distance R. You can calculate
that the minimum distance R equals 0.072 Meters, which is less than 3 inches.
(Please don't hold any errors against me. Check it yourself!)

Since this distance is well within the near-field region, there is an
uncertainty that depends on whether there is electric coupling or magnetic
coupling. To be absolutely sure there is no interference, you could back off a
couple of wavelengths, say fifteen feet on two meters. (Others may say ten
times 3 inches would be enough.)

How does this one Volt per Meter assumption compare with the defibrillator
specifications? If the spec is higher, then there seems little to worry about.

73, Fred, K4DII

(Fred McKenzie) wrote in message ...
....

Working this backwards with the sphere, 4/3 Pi R-squared divided by 8 watts
equals the power (2.7 mW) per square meter at distance R. You can calculate
that the minimum distance R equals 0.072 Meters, which is less than 3 inches.
(Please don't hold any errors against me. Check it yourself!)

Since Fred put that caveat in there, this is obviously nothing against
Fred, but...

We started with 8 watts assumed to be radiated (to find the max
farfield value). Area of a sphere is 4*pi*r^2. (Volume is
(4/3)*pi*r^3.) Seems like we want to say 8 watts/(4*pi*r^2) =
..0027w/m^2, and that yields a radius of about 15.4 meters, which would
actually be a good approximation of far field for 2M wavelength.
Units analysis is always a valuable check; also useful is a quick
mental-arithmetic check: a 1 meter radius sphere has a surface area
of roughly 12 square meters, so 8 watts at 1 meter will be about 2/3
watts per square meter.

OTOH, one would hope that the defib would be considerably less
sensitive than that, else cell phones pose a significant problem.

Cheers,
Tom

Seems like we want to say 8 watts/(4*pi*r^2) =
..0027w/m^2, and that yields a radius of about 15.4 meters, which would
actually be a good approximation of far field for 2M wavelength.


Tom-

You caught me!

I don't have a calculator handy, but it seems to me that the correct formula
yields an even lower power when spread over three times the area.

73, Fred, K4DII

Tom-

You caught me!

I don't have a calculator handy, but it seems to me that the correct formula
yields an even lower power when spread over three times the area.



Caught again!

I took another look at my earlier calculations and found that not only did I
have the wrong formula for the area of a sphere, I had my equation upside down!
I ended up with something like watts per square meter equals square meters per
watt. Yes, the power is spread over a greater area, but that did not
compensate for the really stupid mistake.

So, if one volt per meter equals 2.7 milliWatts per square Meter, that field
strength would be generated by a five watt handheld radio at a distance of 15.4
Meters, which is about 50 Feet. This is sounding serious unless the
defibrillator has a greater tolerance. I certainly hope it does, since we are
often irradiated by passing mobile radios running 50 watts or so, at closer
distances than 50 feet.

Lets see - 50 watts is ten dB greater, so the distance would be about 3 times
as great as 5 watts, or about 150 feet for one volt per meter. Or, maybe I
should just keep my mouth shut to keep from screwing up again!

73, Fred, K4DII

One of the tech support people at Guidant actually sent us some
information regardint the testing limits on my wife's device. In the
area of interest to hams, it thas been tested at a field strength of
140 V/m RMS over a frequency of 500 KHz to 6 GHz..

Thanks for the discussion and information,
Ed Bailen - N5KZW

On 04 Jun 2004 00:17:10 GMT, (Fred McKenzie) wrote:

Tom-

You caught me!

I don't have a calculator handy, but it seems to me that the correct formula
yields an even lower power when spread over three times the area.



Caught again!

I took another look at my earlier calculations and found that not only did I
have the wrong formula for the area of a sphere, I had my equation upside down!
I ended up with something like watts per square meter equals square meters per
watt. Yes, the power is spread over a greater area, but that did not
compensate for the really stupid mistake.

So, if one volt per meter equals 2.7 milliWatts per square Meter, that field
strength would be generated by a five watt handheld radio at a distance of 15.4
Meters, which is about 50 Feet. This is sounding serious unless the
defibrillator has a greater tolerance. I certainly hope it does, since we are
often irradiated by passing mobile radios running 50 watts or so, at closer
distances than 50 feet.

Lets see - 50 watts is ten dB greater, so the distance would be about 3 times
as great as 5 watts, or about 150 feet for one volt per meter. Or, maybe I
should just keep my mouth shut to keep from screwing up again!

73, Fred, K4DII

One of the tech support people at Guidant actually sent us some
information regardint the testing limits on my wife's device. In the
area of interest to hams, it thas been tested at a field strength of
140 V/m RMS over a frequency of 500 KHz to 6 GHz..

Thanks for the discussion and information,
Ed Bailen - N5KZW

On 04 Jun 2004 00:17:10 GMT, (Fred McKenzie) wrote:

Tom-

You caught me!

I don't have a calculator handy, but it seems to me that the correct formula
yields an even lower power when spread over three times the area.



Caught again!

I took another look at my earlier calculations and found that not only did I
have the wrong formula for the area of a sphere, I had my equation upside down!
I ended up with something like watts per square meter equals square meters per
watt. Yes, the power is spread over a greater area, but that did not
compensate for the really stupid mistake.

So, if one volt per meter equals 2.7 milliWatts per square Meter, that field
strength would be generated by a five watt handheld radio at a distance of 15.4
Meters, which is about 50 Feet. This is sounding serious unless the
defibrillator has a greater tolerance. I certainly hope it does, since we are
often irradiated by passing mobile radios running 50 watts or so, at closer
distances than 50 feet.

Lets see - 50 watts is ten dB greater, so the distance would be about 3 times
as great as 5 watts, or about 150 feet for one volt per meter. Or, maybe I
should just keep my mouth shut to keep from screwing up again!

73, Fred, K4DII

I'm not going to check you gentlemen's math -- it sounds like you've
gone through that pretty thoroughly. But I'll point out that you're
calculating the field strength from an isotropic source in free space.
Let me remind the readers that this is the absolute lowest possible gain
an efficient antenna can have. A dipole in free space produces a field
strength about 2 dB greater than this in some direction; an isotropic
source over ground produces a field strength 3 dB greater. So this rough
estimate of field strength is always lower than the field that a real
antenna can produce in some direction.

_RF Exposure and You_ (Hare, ARRL) has some excellent information on how
to approximate field strength of real antennas. And of course modeling
programs do the same thing. But under no circumstances would I depend on
either one, or even better-than-average amateur measurements, to make
any determination that might impact human life or safety.

Roy Lewallen, W7EL

Fred McKenzie wrote:

Tom-

You caught me!

I don't have a calculator handy, but it seems to me that the correct formula
yields an even lower power when spread over three times the area.



Caught again!

I took another look at my earlier calculations and found that not only did I
have the wrong formula for the area of a sphere, I had my equation upside down!
I ended up with something like watts per square meter equals square meters per
watt. Yes, the power is spread over a greater area, but that did not
compensate for the really stupid mistake.

So, if one volt per meter equals 2.7 milliWatts per square Meter, that field
strength would be generated by a five watt handheld radio at a distance of 15.4
Meters, which is about 50 Feet. This is sounding serious unless the
defibrillator has a greater tolerance. I certainly hope it does, since we are
often irradiated by passing mobile radios running 50 watts or so, at closer
distances than 50 feet.

Lets see - 50 watts is ten dB greater, so the distance would be about 3 times
as great as 5 watts, or about 150 feet for one volt per meter. Or, maybe I
should just keep my mouth shut to keep from screwing up again!

73, Fred, K4DII

I'm not going to check you gentlemen's math -- it sounds like you've
gone through that pretty thoroughly. But I'll point out that you're
calculating the field strength from an isotropic source in free space.
Let me remind the readers that this is the absolute lowest possible gain
an efficient antenna can have. A dipole in free space produces a field
strength about 2 dB greater than this in some direction; an isotropic
source over ground produces a field strength 3 dB greater. So this rough
estimate of field strength is always lower than the field that a real
antenna can produce in some direction.

_RF Exposure and You_ (Hare, ARRL) has some excellent information on how
to approximate field strength of real antennas. And of course modeling
programs do the same thing. But under no circumstances would I depend on
either one, or even better-than-average amateur measurements, to make
any determination that might impact human life or safety.

Roy Lewallen, W7EL

Fred McKenzie wrote:

Tom-

You caught me!

I don't have a calculator handy, but it seems to me that the correct formula
yields an even lower power when spread over three times the area.



Caught again!

I took another look at my earlier calculations and found that not only did I
have the wrong formula for the area of a sphere, I had my equation upside down!
I ended up with something like watts per square meter equals square meters per
watt. Yes, the power is spread over a greater area, but that did not
compensate for the really stupid mistake.

So, if one volt per meter equals 2.7 milliWatts per square Meter, that field
strength would be generated by a five watt handheld radio at a distance of 15.4
Meters, which is about 50 Feet. This is sounding serious unless the
defibrillator has a greater tolerance. I certainly hope it does, since we are
often irradiated by passing mobile radios running 50 watts or so, at closer
distances than 50 feet.

Lets see - 50 watts is ten dB greater, so the distance would be about 3 times
as great as 5 watts, or about 150 feet for one volt per meter. Or, maybe I
should just keep my mouth shut to keep from screwing up again!

73, Fred, K4DII