equation: dBm to mW/cm^2

You mean the site that has a page titled "Biologically Alien Electromagnetic Energies and EMF Pollution (Electrosmog)"

For some reason, I don't want to trust that site for technical correctness.

What search terms are you using? Have you combed through the ARRL Handbook?

I'm pretty sure that something like a quarter-wave dipole is only going to be accurate at it's resonant frequency, and its sensitivity is going to be in volts/wavelength, not watts/area. Of course there will be a conversion, though.

If it's just field strength you want, an electrically short antenna driving a correctly (and severely) mismatched load will probably be much more accurate in broadband, but will still need calibration. And I suspect it's not what you want.

Do you have the ARRL antenna book? Have you read it?

--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com

Okay. Right now I'd happily settle for an equation published by enthusiasts of trans-gendered seafood, if it were correct. :)

'dBm mW/cm^2 formula' 'dBm mW/cm^2 equation'

I've seen nomographs and charts using these terms that show the relationship I'm seeking; just not the equation that will reveal how many mW per square centimeter I can expect from a given 1/4 - wave ground plane for a given electrical field with a number of decibels normalized to one milliwatt.

That's a good suggestion. I didn't see anything that looked useful within, however. My copy has a tutorial on Electromagnetic Compatibility but they manage to get through it without revealing this equation.

I did see a couple things in the index that looked hopeful, but they didn't pan out on the actual pages.

I suspect I am looking for the 'aperture' of my 1/4 wave ground plane so that I can convert received power to received power/aperture.

I'm just parrotting what little info I *have* seen so that is probably a misinterpretation.

I see a citation revealing that a 1/2 wave dipole has an aperture measuring 5/16 of a wavelength for example.

I suspect that is true.

I will have to slip out to my local library and paw their copy. Thanks for your help. :)

--Winston

(...)

I didn't have to go that far or even get into the jalopy. :)

From:

Given a transmitting antenna with a gain of 10, a 100 W transmitter will have an Effective Radiated Power of:

ERP = 10*log(100/.001) = 50 dBm

If our receiver is located 100 feet (or 3047.85 cm) away from the transmitter, we will measure a power density of:

Pd =(100,000*10)/(4*PI()*3047.85^2) = 0.0086 mW/cm^2 ^ (Where transmitted power is expressed in mW)

I'm so happy I could eat a tuna!

Thanks. :)

--Winston

Correct, also, only in its direction of maximum radiation/sensitivity is its gain figure correct.

They're the same thing (assuming you mean volts/meter, which is actually not per wavelength, but per lineal meter):

Watts/area = volts/length squared divided by the intrinsic resistance of free space (120*pi ~ 377 ohms).

W=V^2/R again!

The same applies to H field: watts/area = amp(turn)s/length squared times

377.

W=I^2R

SQRT(area) has dimensions of length.

SI units assumed.

The biggie here is that most simple formulas assume plane wave conditions. ie. far-field. Near-field conditions are more complicated. I suspect that the OP is trying to measure near fields.

Most amateur radio publications only concern themselves with far-field conditions.

"Telecommunications Engineering", Duncan and Smith, (Van Nostrand) has appropriate information, ISBN 0-442-30585.

--
"For a successful technology, reality must take precedence 
over public relations, for nature cannot be fooled."
                                       (Richard Feynman)

This'll be a 1/4 wave ground plane so I assume it'll receive equally well or poorly in all directions.

Five wavelengths at 1000 MHz is about 1.5 m. I'm interested in field intensity at that radius and somewhat beyond, so the far-field answer will work for me.

That's reasonable.

Thanks Fred!

--Winston

[snip] [snip]

So mW/cm^2 (power-density) is your absolute Watts divided by the surface area of a sphere with radius equal to the distance of interest (uniform radiator assumed). ...Jim Thompson

--

| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
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I love to cook with wine.     Sometimes I even put it in the food.

Yup.

4 pi r^2

Modern practice is to use W/m^2, V/m, A/m.

That's where the apparent paradox regarding the inverse square law originates. Watts per square meter does fall off as the square of the distance. Volts per meter, which is what most people call "field strength", falls of as the first power.

--
"For a successful technology, reality must take precedence 
over public relations, for nature cannot be fooled."
                                       (Richard Feynman)

Unfortunately no. There's no such thing in reality as the theoretical isotropic radiator. The E-plane polar pattern will be similar to half a half-wave dipole.

End-on, the gain/sensitivity will be zero, with maximum at right angles to the antenna axis. Imagine, if you will, a half bagel, lying cream-cheese-side-down on the ground plane:-)

Five wavelengths is certainly getting there. The trick is to do the math for a uniform isotropic, then adjust for the antenna gain in the direction of interest.

It's easier with receiving antennas to use volts/meter, rather than watts/square meter.

For example, the EMF induced in a half-wave dipole is usually taken to be:

e = E * lambda / pi, where e is the induced emf and E is the field strength in volts per meter.

Using your antenna's stated isotropic gain of 1dB, and the theoretical gain of a half wave dipole (2.15dB), that means that your antenna will be

1.15dB down on E lambda / pi.

This is (open circuit) EMF. For a perfectly matched antenna, the voltage at the receiver terminals will be half that.

To obtain watts per square meter from volts per meter, square, and divide by 377.

Forget centimeters, they're not used in engineering anymore.

--
"For a successful technology, reality must take precedence 
over public relations, for nature cannot be fooled."
                                       (Richard Feynman)

That sounds very reasonable isotropically speaking, though I recall that the website example was for a directional radiator with a gain of 10.

The example implied that my receiver is *always* on-axis with the main lobe on both planes, so I understand that the field strength would measure the same as if the transmitter were 1000 W with an isotropic radiator, all else being equal.

I'm here to learn though, so I *am* willing to listen to reason. :)

--Winston

Ah. I stand corrected. (Orthopedic sneakers).

For my purpose, that 'half bagel' pattern will work just fine. (I assume that the 'reception pattern' for a receiver exhibits the same geometry as does the 'radiation pattern' for a transmitter for a given antenna, yes?)

So, the on-axis 'near field' of an antenna with a gain of 10 is actually more like 50 wavelengths - rather than say 5 wavelengths - away?

I conjecture that I *should* be calculating watts / cm to fall inversely with distance (rather than as the inverse cube...) if my 1 GHz receiver is within say 45 meters of the (gain of 10) antenna? Holy Moley!

My application is EMC / agency compliance related, where W/cm^2 is the lingua franca. So I want to be able to express field intensity using that radix, even if it means doing the involved math to convert from W/m^2 :)

'Sounds as if I need to build a prototype and calibrate it against a leveled 1 GHz source of known wattage (and known radiation pattern).

Excellent! That is another good piece of info.

Thanks for your help, Fred.

I sincerely appreciate it.

--Winston

(...)

Er. make that '(rather than as the inverse square...)'

--Winston

As soon as you have an anisotropic radiator, all "nice" equations are meaningless.

Your roll-off with distance will be somewhere between 1/r and 1/r^2 depending on the lobe pattern, and how much you are off axis.

So measure it. ...Jim Thompson

--

| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    |
| Phoenix, Arizona  85048    Skype: Contacts Only  |             |
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  |
| E-mail Icon at http://www.analog-innovations.com |    1962     |
             
I love to cook with wine.     Sometimes I even put it in the food.

Aye aye, Sir.

--Winston

Some of them are.

I disagree. At any given angle, the radiated watts per steradian are constant, irrespective of distance. What changes with radial distance is the surface subtended by one steradian, which increases as r^2. Hence, along any radial line, power density always follows an inverse square law, and the field intensity, an inverse linear law.

The lobe pattern merely defines the radiated power in a particular direction, ie watts per steradian along a *radial* line.

Agreed.

--
"For a successful technology, reality must take precedence 
over public relations, for nature cannot be fooled."
                                       (Richard Feynman)

No. A lot of people get this wrong, in one way or the other.

Power density (watts per unit area) falls inversely as the square of the distance.

Field intensity (volt per unit length) falls inversely as the distance.

--
"For a successful technology, reality must take precedence 
over public relations, for nature cannot be fooled."
                                       (Richard Feynman)

Huh?

--
"For a successful technology, reality must take precedence 
over public relations, for nature cannot be fooled."
                                       (Richard Feynman)

Yes.

No, near field is dominated by induction, rather than radiation, which is not necessarily a function of antenna gain, From (now quite distant) memory, the inductive field is 3dB down at about 2/3 wavelength. I'll have to check this.

There's no such thing as watts/cm. Power density is watts/area, ie. watts/square meter, or watts/square cm (if you must).

Power density (watts/square meter) falls off as the inverse square of the distance.

Field intensity (volts/meter) falls off as the inverse distance.

What involved math?. For example 10mW/cm^2 is 100W/m^2. You can do that in your head, just multiply by 100^2 ;-)

Any regulatory standards in the last 30 years should be in SI units, anyway. Last thing I had to do in this area (which admittedly was for a European directive in about 1990), levels were quoted in W/m^2.

That's how it's done. That's why we have expensive measuring antennas with NIST traceable calibration.

I'd recommend that you get a copy of Kraus's "Antennas", if you really want to get into this stuff. I guess most university bookstores still have it, or can get it.

--
"For a successful technology, reality must take precedence 
over public relations, for nature cannot be fooled."
                                       (Richard Feynman)

Picky! Picky! Picky! ;-) ...Jim Thompson

--

| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    |
| Phoenix, Arizona  85048    Skype: Contacts Only  |             |
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  |
| E-mail Icon at http://www.analog-innovations.com |    1962     |
             
I love to cook with wine.     Sometimes I even put it in the food.

Yes, I meant W/cm^2 not W/cm. My bad.

Yes, in the far field. I've measured that to be the case. That ain't my question.

1) The old tale about power density falling inversely (not as the inverse square) within the *near field* is true, yes? (I know. 'Just Measure It'.) :) 2) The center lobe of a directional antenna with a gain of say 10 would be expected to have a near field border about 10 x further from it than would an isotropic, all else being corrected, yes? (That sounds perfectly reasonable, but I don't want to assume.)

Thanks for your help! :)

--Winston