Calculating mW/m2 from dB

[Robert Wade]

What is the mathematical relationship between dB and mW/m2?

Can someone please give an example, such as:

-10dB = 10mW = X mW/m2

How is distance from the source be factored in?

Is anyone aware of a free online document that provides instruction on
taking ambient EMF measurements?

Robert Wade

[tim]

On Sun, 28 Sep 2014 06:44:32 +1000, Robert Wade wrote:

> What is the mathematical relationship between dB and mW/m2?

Practically nonexistent.

> Can someone please give an example, such as:
>
> -10dB = 10mW = X mW/m2

-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).

> How is distance from the source be factored in?

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.

> Is anyone aware of a free online document that provides instruction on
> taking ambient EMF measurements?

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

[Jasen Betts]

On 2014-09-27, tim <t...@seemywebsite.com> wrote:
> On Sun, 28 Sep 2014 06:44:32 +1000, Robert Wade wrote:
>
>> What is the mathematical relationship between dB and mW/m2?
>
> Practically nonexistent.
>
>> Can someone please give an example, such as:
>>
>> -10dB = 10mW = X mW/m2
>
> -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.

The official unit for measuring sounds is SPL (sound pressure level)
the reference is (20µPA)

--
umop apisdn


--- news://freenews.netfront.net/ - complaints: ne...@netfront.net ---

[Martin Riddle]

On Sun, 28 Sep 2014 06:44:32 +1000, Robert Wade <rw...@santos.com>
wrote:

Your measuring Near Field disturbances.
Searching for 'near field EMF' I came up with this...
<http://www.edn.com/design/analog/4405857/3/Electromagnetic-Near-Field-and-the-Far-Field--Chapter-two>

Maybe this will help
<http://www.keysight.com/upload/cmc_upload/All/NSI-near-far.pdf?&cc=US&lc=eng>

Cheers

[Phil Allison]

tim wrote:

>
> 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.

** 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

[Jeff Liebermann]

On Sun, 28 Sep 2014 06:44:32 +1000, Robert Wade <rw...@santos.com>
wrote:

>What is the mathematical relationship between dB and mW/m2?

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

>Is anyone aware of a free online document that provides instruction on
>taking ambient EMF measurements?

After you unscramble your units of measure, this might help. I use it
for a field intensity and power density calculation cheat sheet:
<http://802.11junk.com/jeffl/coverage/VZW-water-plant/Field%20Intensity%20and%20Power%20Density.pdf>
I forgot which book I stole it from.

More:
<http://www.giangrandi.ch/electronics/anttool/field.html>
<http://en.wikipedia.org/wiki/Field_strength_in_free_space>


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

[Robert Wade]

On Sat, 27 Sep 2014 17:46:09 -0500, tim <t...@seemywebsite.com> wrote:

>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.
>

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

>

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

[John S]

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.

[Robert Wade]

On Sun, 28 Sep 2014 06:04:46 -0500, John S <Sop...@invalid.org>
wrote:

>> 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

[mpm]

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.

[John S]

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.

[John S]

On 9/28/2014 7:38 AM, Robert Wade wrote:

> On Sun, 28 Sep 2014 06:04:46 -0500, John S <Sop...@invalid.org>
> wrote:
>
>
>>> 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?

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

> 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

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.

[Tom Miller]

"Robert Wade" <rw...@santos.com> wrote in message
news:2enf2a5f13ms6e7bg...@4ax.com...

33 dBm

[Tim Wescott]

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

[Robert Wade]

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

[Robert Wade]

On Mon, 29 Sep 2014 06:56:37 +1000, Robert Wade <rw...@santos.com>
wrote:

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

[John S]

On 9/28/2014 3:56 PM, Robert Wade wrote:
> On Sun, 28 Sep 2014 14:09:27 -0500, Tim Wescott
> <seemyw...@myfooter.really> wrote:
>
>>> On Sat, 27 Sep 2014 17:46:09 -0500, tim <t...@seemywebsite.com> wrote:
>>>
>
>> 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.
>>
>
> 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.

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?

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

No.

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

Only because your answer is wrong.

> Can someone please tell me where I have gone wrong?

I tried.

> Thanks,
>
> Robert Wade
>

[rickman]

On 9/28/2014 5:11 PM, Robert Wade wrote:
> On Mon, 29 Sep 2014 06:56:37 +1000, Robert Wade <rw...@santos.com>
> wrote:
>
>> On Sun, 28 Sep 2014 14:09:27 -0500, Tim Wescott
>> <seemyw...@myfooter.really> wrote:
>>
>>>> On Sat, 27 Sep 2014 17:46:09 -0500, tim <t...@seemywebsite.com> wrote:
>>>>
>>
>>> 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.
>>>
>>
>> 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.

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.

>> 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.

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

[tim]

On Sun, 28 Sep 2014 21:46:24 -0400, rickman wrote:

> On 9/28/2014 5:11 PM, Robert Wade wrote:
>> On Mon, 29 Sep 2014 06:56:37 +1000, Robert Wade <rw...@santos.com>
>> wrote:
>>
>>> On Sun, 28 Sep 2014 14:09:27 -0500, Tim Wescott
>>> <seemyw...@myfooter.really> wrote:
>>>
>>>>> On Sat, 27 Sep 2014 17:46:09 -0500, tim <t...@seemywebsite.com>
>>>>> wrote:
>>>>>
>>>>>
>>>> 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.
>>>>
>>>>
>>> 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.
>
> 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.

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

[Robert Wade]

On Sun, 28 Sep 2014 16:38:19 -0500, John S <Sop...@invalid.org>
wrote:

>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?
>

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

[Robert Wade]

Dumb question perhaps, but why are calculated field densities so far
removed from what my $200 RF meter indicates?

I am using the ED85EXS shown on this webpage:
http://www.cornetmicro.com/

It supposedly reads down to a few microwatts/m^2

Not top-of-the-range, but I had assumed it would be accurate enough
for informal readings.

Robert Wade

[John S]

Yes.

[John S]

Because the calculated field strengths mentioned in this thread make
certain assumptions. Such as, free space, no reflections, no earth, no
water bag in the way, etc.

The meter is to be used to measure the field strength where it is
placed, not to indicate the power level of a non-perfect source.

[Phil Allison]

John S wrote:

>
>
> Because the calculated field strengths mentioned in this thread make
>
> certain assumptions. Such as, free space, no reflections, no earth, no
>
> water bag in the way, etc.

** The main assumption being made is that the signal source is not discontinuous.

Analogue signals like broadcast AM and FM have average carrier power levels that remain steady - but they are just about the only common ones that do.




... Phil



... Phil

[John S]

Correct. But, drive along the highway and you might get fading, not just
from distance, but from phase cancellation. This world is not a perfect
place to do his sort of measurement. A special anechoic room for the
measurements is advised.

I know you know this, Phil. This is really for the guy that doesn't get it.

[Robert Wade]

On Mon, 29 Sep 2014 11:26:16 -0500, John S <Sop...@invalid.org>
wrote:

Of course the confusing thing for "the guy that doesn't get it" is
that the meter reads in mW/cm2.

Robert Wade

[Phil Allison]

John S wrote:

Phil Allison wrote:
>
>
> > ** The main assumption being made is that the signal source is not
>
> > discontinuous.
>
> >
>
> > Analogue signals like broadcast AM and FM have average carrier power
>
> > levels that remain steady - but they are just about the only common
>
> > ones that do.
>
> >
>
>

> Correct. But, drive along the highway and you might get fading, not just
>
> from distance, but from phase cancellation.

** I suspect you missed my meaning.

When the power level of a data transmitter is quoted in so many watts - is that the peak level or the average power output during operation ?

For example, SSB transmissions are quoted in "peak envelope power" cos there is no carrier.

Are not most data transmissions discontinuous ( eg GSM mobile phones) with the carrier being modulated off constantly ? That would create a disconnect between the quoted power in watts and the field strength picked up by an average responding instrument.

BTW: info on the signal format and modulation method used in Wi-Fi are not things that jump up at you with a Google search and my 50Mhz scope is not much help either.


... Phil

[upsid...@downunder.com]

On Mon, 29 Sep 2014 13:54:40 +1000, Robert Wade <rw...@santos.com>
wrote:

>

You haven't specified the frequency in use.

Any calculations using the inverse square relation assume far field,
things in the near field can change quite dramatically.

Then there is the question, where is the limit between near and far
fields. some put it about one wavelength, others about 1/2pi
wavelengths or something in between.

For accurate near field measurements, you would have to make both
magnetic and electric field measurements, since you can't assume the
free space impedance to be 120pi = 377 ohms.

[John S]

I did miss your meaning. And you're right about the rest, as usual. I
don't understand all those communications methods so this is where I
stop. Thanks.

[John S]

Where does it say that?
<http://s160930441.onlinehome.us/resources/ED85exs1.pdf>
I see mentioned:
Max, Hold, and mw/m2, v/m, dBm unit display
High sensitivity : �65dBm to 5dBm (8.2mV/m to 26.2V/m)
Peak power density measurement : 0.18uW/m^2 to 1.8W/m^2

Even the little picture of the display shows uW/m^2

No mention at all of cm or cm^2.

In any case, the conversion is 10,000 cm^2 per m^2.

[Robert Wade]

On Tue, 30 Sep 2014 06:49:26 -0500, John S <Sop...@invalid.org>
wrote:

>Where does it say that?
><http://s160930441.onlinehome.us/resources/ED85exs1.pdf>
>I see mentioned:
>Max, Hold, and mw/m2, v/m, dBm unit display

>High sensitivity : �65dBm to 5dBm (8.2mV/m to 26.2V/m)

>Peak power density measurement : 0.18uW/m^2 to 1.8W/m^2
>
>Even the little picture of the display shows uW/m^2
>
>No mention at all of cm or cm^2.
>
>In any case, the conversion is 10,000 cm^2 per m^2.
>

Thanks for pointing that out. My mistake.

Robert Wade