Calculating mW/m2 from dB

Practically nonexistent.

-10dB does not mean -10mW. If you mean -10dBW (as in "decibels referred to 1 watt"), then -10dBW is 100mW (1W / 10 = 100mW).

But dB can refer to quite a bit of different things. I think the only place where a dB can stand alone as a unit of measure (as opposed to a relative multiplier) is in sound measurement (i.e. "90dB" means "put on earplugs"), but I'm not even sure if that's official, or what the reference sound level is.

Normally in electrical engineering practice, an un-annotated dB is a way of expressing a ratio: a filter with a gain of -10dB cuts the signal by sqrt(10) (or the signal power by a factor of 10), a signal that emerges from a system 20dB down has had it's power cut by a factor of 100, etc.

If you want dB to refer to a physical measurement, then you need to write dBV (decibels referred to 1 volt), or dBW (decibels referred to 1W), or dBm (decibels referred to 1 mW).

If you haven't figured it out by now, your question doesn't make much sense. I suspect that you want to read a power level at an antenna and calculate power density, or perhaps push power into an antenna and calculate power density -- if so, there is a huge amount of dependence on the antenna geometry.

Once you get into the antenna's far field, the power density from the antenna in any particular direction from it will go as distance^2 -- but that's only in far field.

EMF? Do you mean electromagnetic compatibility measurements? Electromagnetic field? What?

If you mean electromagnetic field measurements, my experiences of looking over people's shoulders as they do this stuff tells me that you're not going to learn it from a small document. Even a book will leave you with lots of holes in your knowledge.

I do know that you can send the same piece of equipment to two different electromagnetic testing labs, and get considerably different results back for it's electromagnetic compatibility and susceptibility -- there's at least as much art in this stuff as there is science.

--
www.wescottdesign.com

The official unit for measuring sounds is SPL (sound pressure level)

--
umop apisdn 


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Your measuring Near Field disturbances. Searching for 'near field EMF' I came up with this...

Maybe this will help

Cheers

** Sound levels are measured in "dB SPL" with reference to "0dB SPL" - equal to a sine wave sound with pressure of 20uPa rms (or 28.3uPa peak).

FYI: 0dB SPL is the approximate threshold of hearing ( at 2 to 3 kHz) for young persons.

dBs always represent power ratios while microphones respond to sound pressure linearly - so 10 times the sound pressure produces 10 times the voltage from a microphone and equates to a 20dB increase in the SPL.

90dB SPL is very common, typical loudspeakers will produce it with 1 watt of input at 1 metre. Headphones and earpieces need less than a milliwatt.

.... Phil

dB is a ratio. mW/m^2 is the power density. There's no direct relationship.

After you unscramble your units of measure, this might help. I use it for a field intensity and power density calculation cheat sheet: I forgot which book I stole it from.

More:

--
Jeff Liebermann     jeffl@cruzio.com 
150 Felker St #D    http://www.LearnByDestroying.com 
Santa Cruz CA 95060 http://802.11junk.com 
Skype: JeffLiebermann     AE6KS    831-336-2558

Yes, I can see that my question was insufficient. Sorry.

To take a specific example. let's say I have a wireless access point that will transmit a maximum of 29dBm, or 800mW.

The antenna is omnidirectional with 4dB gain. If we disreagrd connector loss, that is an EIRP of 33dB.

From this, how can I calculate the field density in mW/cm2 at 2 metres?

Robert Wade

When the sensor is positioned in the maximum radiation field, it will be about 3.9uW/cm^2 at 2m. At 2.4GHz, 2m is about 16 wavelengths so it will be in the far field.

Maybe.

How was that calculated? Intuitively, it seems a bit low to me.

So if we move in to 1M the density is only 15.6uW/cm2?

I am just asking because I really don't know. But I have been near WAP's with a meter and the readings seemed much higher. Over 1mW/cm2 as I recall.

Robert Wade

Robert,

Google: "Friis Equation".

This is the classic engineering formula for the power transfer between two antennas. It can be adapted to a variety of situations. I'm sure you'll find it helpful.

It is an estimate without knowing such things as antenna efficiency, what the transmitter does when it has reflected signals interacting, reflections from surrounding objects, the equipment being used, etc. Read Tim's reply and study it deeply, especially that last sentence.

What accuracy do you expect from all this?

Free space, far field, I use the following for *estimates* just as a sanity check:

P = (D^2)*(E^2)/30

where P is in watts, D is in meters, and E is in volts/meter. You can convert V/M using the intrinsic impedance of free space, which is 377. And, you know how to convert square meters to square centimeters, I do hope.

Theoretically, yes, based on your inputs. Are you confusing W/m^2 with W/cm^2, perhaps?

Is to be laboratory-grade measurements? What are you using to measure the field? Is your meter, and its antenna, traceable to NIST? Are you in a controlled environment?

What are you trying to accomplish and how are you going about it?

If you are asking because you really don't know, then you need a professional.

33 dBm

If you like to start with absolute fundamentals, this is how to do it:

If the antenna were perfectly isotropic (meaning it radiates equally well in all directions), then the field density would simply be the total power radiated divided by the area of a 2m radius sphere.

EIRP means that an antenna radiates as much power in its most-favored direction as an isotropic antenna would in all directions -- so take your

2W of effective radiated power and divide it by that surface area, and you have your answer.

--

Tim Wescott 
Wescott Design Services 
http://www.wescottdesign.com

So then the area of a 2M radius sphere is about 50.

If I divide my omni 33dBm EIRP by 50, I get 0.66dBm or 0.00116W/m^2.

This is 0.000116 mW/cm2 or 0.116uW/cm^2

That is less than John's estimate of 3.9uW/cm2.

Can someone please tell me where I have gone wrong?

Thanks,

Robert Wade

I just checked the Wi-Fi modem in my printer. It reads 1-2mW/cm^2 at 2 metres distance.

I get 20uW/cm2 from the phone tower 500M away.

Can someone please set me staight on this?

Robert Wade

33dBm is 2Watts which, when spread over 50 square meters, is 39.9mW/m^2.

One square meter is 10,000 square centimeters. Can you take it from there?

No.

Only because your answer is wrong.

I tried.

If you want to divide power by 50, you don't divide the dB number. dB is a logarithmic scale. To divide by 10 you subtract 10. To divide by

50 I would divide by 100 and multiply by 2 which means subtract 20 and add 3 to get... 16 dBm which is... wait for it... 40 mW.

You were making a mistake using logarithms.

Reads? What are you measuring it with?

When you measure real fields keep in mind that your antenna is *not* isotropic. That is very different from "omni-directional" which means in all directions in the horizontal plane.

--

Rick

Here's the longer version:

You either need to convert your power to regular units, in which case you divide 2W by 50 to get 40mW. Then you can convert that to 16dBm.

Or you can convert the factor of 50 do dB -- 1/50 is a loss of 17dB, and

33dBm - 17dB = 16dBm (Note the oddity: normally you can't "add apples and oranges" - but when you're playing games with decibels you must. Rather than trying to learn a bunch of new rules, it's best to just remember that you're really doing multiplication and division using logarithms, and make sense of it that way).

--
www.wescottdesign.com

Thanks for sticking with it. Just so I am sure, here is another example.

25dBm EIRP = 320mW

At 4 metres field density is 320mW / 201 = 1.59mW/m^2 / 10,000 =

0.000159mW/cm^2 * 1000 = 0.159uW/cm^2

Am I getting closer?

Robert Wade