Converting Watt Ratings to Watts per Square Metre

In order to answer that question at least somewhat accurately, one would need to know the distance at which the measurement was being taken, and the antenna's gain pattern.

For a crude approximation, assume that the antenna's pattern is isotropic (equal power in every direction) - physically impossible but it's a useful fiction.

Pick your measuring distance, and compute the surface area of a sphere of that size (4 pi r^2). Divide the surface area in square metres into the power (200 mW) and you have your crude answer. For a measurement distance of 1 meter it would be 200/4pi or 50/pi or about

16 milliwatts per square metre (unless I'm doing my math wrong, which is always possible).

The theoretical gain of a simple half-wave dipole is 2.15 dBi - so, if that's the sort of antenna your router has, the maximum power per square metre (out along the donut-shaped axis of maximum power) will be about 60% greater than that.

** That number could refer to the DC input power to the final stage.

Supply a link to the actual specs - FFS.

Something like this.

My confusion stems from how the output power is measured. By "mW's" do they mean mW/m2 or something else?

In other words, "Watts" can apparently represent either power consumption or signal density.

It does not appear related to power consumption in this case. So what exactly does it mean?

Martin King

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Something else. 

They mean that the total power rated into space is 200 milliwatts.

Of course, as a practical matter, the antenna probably has gain (or loss). So, in logarithmic form using the popular units of milliwatts and microvolt s per meter we have:

10log(mW) = 20log(uV/m) + 10log(lambda^2g)-126.76

where lambda is the wavelength in meters, and g is the directive gain of th e antenna.

The left side of the equation is the expression for dBm. The first term on the right side is the expression for dBu.

Therefore, dBm = dBu + 10log(lambda^2g)-126.76.

BTW, I thought (at least in the United States), that "wireless internet rou ters" had maximum output powers specified in terms of microvolts per meter, measured at a specified distance, with a specified probe (usually isotropi c). That would at least eliminate the affects of antenna gain/loss. Shoul d be specified in detail in FCC Part-15 rules for low-power, unlicensed dev ices.

Probably time to refresh the memory on that... 2.4/5.8 GHz is not a band I usually work with professionally, except for the personal consumer grade g izmos that are pretty much plug and play. And that said, if I encountered a situation where I was concerned that I needed "more power" to make the tr ip, I might focus more of my attention on receiver sensitivity and line-of- sight obstructions, best mounting location, (etc..) before I got too hung u p over the mW's on the device packaging.

** Thta is not a spec of any kind.

That is purest marketing bullshit.

My god you are a gullible jerk.

.... Phil

No, they mean "milliwatts". Power, or energy-per-time.

Nope. Different dimensions.

200 milliwatts is a power (amount of energy per time).

Signal density is the amount of power being delivered into a specified area, at a specific distance from the transmitter. It's equal to the power, divided by the total area present at the specified distance from the transmitter.

In this case, it may very well be the transmitting system's output power - the amount of radio frequency energy which leaves the antenna and travels out into space towards the receivers.

This will be less than the system's total power consumption, by quite a lot. Most of the power being drawn by the router/access point is being used to run the digital electronics. Only a small fraction of it (probably a watt or so) goes into the radio transmitter stage. Some of that is lost in the transmitter itself (they aren't 100% efficient, and the remainder turns into heat), and some of it is lost in the wiring and in the antenna itself (again, it turns into heat). The remainder is what goes outwards towards your PCs in the form of microwaves.

In the case of a router, the manufacturer may have quoted the 200 milliwatts as being the maximum power which the transmitter itself creates... they may have omitted mentioning the losses in the wiring and antenna.

200 milliwatts (or 23 dBm, meaning "23 decibels greater than 1 milliwatt") is at the upper end of the range for typical consumer WiFi devices. A lot of the less expensive ones (for in-home use) are in the 10-15 dBi range (very roughly 10 - 50 milliwatts).

You need to know its antenna gain over a theoretical isotropic radiator, its radiation pattern, and the distance.

A theoretical isotropic radiator radiates into a sphere, hence at a radius r, the power density will be (output power in watts/(4*pi*r^2)) watts per square meter.

I would specify output power as effective isotropic radiated power (EIRP). The manufacturer of your router might have had other ideas, to make the figures look good.

The actual power density will be antenna gain times that, or the power in dBW plus the antenna gain in dB (which will give dBW/m^2), in the direction of maximum radiation.

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"Design is the reverse of analysis" 
                   (R.D. Middlebrook)

WTF is a dBu?

Where do you get an isotropic probe from?

Lambda don't enter into it, unless you are calculating the EMF in an antenna.

The OP wanted to know the power density:

Assuming that the stated 200mW is effective radiated power, which in not an unreasonable assumption, and further assuming that it was measured in the direction of maximum radiation (also reasonable), then:

Maximum power density = ERP/(4*pi*r^2) So, at 1 meter radius, the maximum power density is 200*10^-3/(4*pi) =

15.91 milliwatts per square meter, in the direction of maximum radiation. (QED)

That equates to a field strength of 2.45 volts per meter, at 1 meter radius.

--
"Design is the reverse of analysis" 
                   (R.D. Middlebrook)