Mobile phone threat?

Lots of answers here touching on issues of interest, but some wrong answers and no one mentioned the real issue. It is the FCC, not the FAA that forbids aerial cell phone use. The reason is that at altitude and speed, the cell phone reaches many towers simultaneously, confusing the system. Some European airlines are allowing cell phone use, and the way they do it is to provide a cell "tower" in the plane, which then communicates via satellite to the ground. I heartily agree with those who don't like this on the basis of having to listen to loud personal conversations, but that is not a technical issue.

Although some FAA types may claim that cell phones might cause interference with aircraft systems, this is simply not the case. The low power transmitters could only cause interference to an in-band (same frequency) aircraft communications system, and those don't exist. Any non-antenna connected aircraft system critical to the continued safe flight and landing of the aircraft (verbatim FAA terminology) will be qualified to levels of electric field intensity representative of tens to hundreds and even thousands of Watts of transmit power, depending on frequency. At cell phone frequencies, the qualification test performed uses a 200 Watt microwave amplifier connected to a horn antenna with much higher gain than the omnidirectional stub attached to your cell phone sub-Watt transmitter.

In conclusion, the issue with cell phones on aircraft is problems with the system of cell towers, not the host aircraft.

emc_c

This is a very long post, because the post to which this message replies makes an error that is very commonplace and needs dispelling, at least amongst the engineering community. If you're not an engineer, this post will bore you and or lose you. Fair warning.

Strictly speaking, wiring is not interfered with. Wiring is a passive conductor, and it cannot be "upset." What can be upset is electronic circuits at the ends of those wires. At frequencies where cables are a half-wavelength long or less, the longer the cable, the more efficient it is at collecting radiated rf energy. The relationship of wavelength to frequency is given by either of the two following equations, which are equivalent; usage is by convenience:

wavelength = 300/frequency in MHz;

wavelength = 0.3/frequency in GHz.

So the cables pick up more than the short wire runs in an equipment enclosure. Further, although sensitive critical circuits will have their signals run on shielded cables, a flexible braided cable shield is nowhere near as effective as the shielding provided by the solid metal equipment enclosures at the end of the cables.

The above analysis is true below 400 MHz, with 400 MHz having a wavelength of 75 cm, or a half-wavelength of 37.5 cm, just over a foot. But cell phones today operate around 1.875 GHz, with a half-wavelength of 8 cm, or about three inches. So the cables aren't a much better pickup at these frequencies than the wire runs in an equipment enclosure, except for the difference in shielding. Further, once a cable or wire is a half-wavelength long, the pickup efficiency of that wire decreases with the square of increasing frequency.

Bottom line: you never pickup more than when the wire is one half-wavelength long. Some very simple relations can be used to predict the coupling to a wire from a cell phone.

The first relation yields the electric field intensity from a source of known power and antenna gain. The second relation predicts the current induced on the shield of a cable due to that electric field immersion. A last relationship is used to get from the induced common mode shield current to a common mode potential induced in the circuits protected by that shield.

The electric field intensity E (V/m) at a distance r (meters) from an rf power source P (Watts) and some numeric gain G is:

E = sqrt(30*P*G)/r

In order to illuminate 8 cm of wiring, we have to be roughly 8 cm back from the wire, assuming the cell phone has an omnidirectional pattern. That gives us both r = 0.08 m, and G = 3.28 (very worst case estimate for a quarter wave stub over an infinite ground plane). Further assume cell phone power output is 1 Watt, which is also very conservative. That yields an estimated field intensity

E = sqrt(30*1*3.28)/0.08 = 124 V/m

That's a pretty big number, but the second relationship is that the current induced on a cable shield, as mounted in a typical aircraft , is 1.5 mA per Volt per meter of illuminating field intensity. That works out to 186 mA of current, and that current is riding on the shield of the cable.

Current riding on the outside of a perfect cable would stay on the outside, but real cable shields aren't perfect. Because of the braid, current on the outside can "porpoise" into the inside and couple to the protected wiring. High quality cable shields provide an inner foil liner to protect against that coupling: the foil doesn't have a braid and doesn't allow that coupling mode. One might ask why not just use the foil shield, but it doesn't work well at lower frequencies, plus it is fragile all by itself.

Anyway, all shields have a transfer impedance, which is the ratio of the voltage induced in the protected inner conductors relative to the current on the shield itself. A crummy cable shield at cell phone frequencies might have a transfer impedance of 1 Ohm per meter, but we are only interested in the first 8 cm of that meter, so that we are looking at 80 milliohms of transfer impedance. Multiplying 80 milliohms by 186 mA yields 15 millivolts.

Now that is a common mode signal. Depending on wiring configuration, and common mode rejection, or the lack thereof, some fraction of that 15 mV will be interpreted as a real signal by a circuit at the end of the illuminated cable. 15 mV will upset no digital databus. It could interfere with a low-level analog system, but safety critical signals at these levels are not run on long cables through the aircraft. The cables that carry important signals long distance are either digital buses or discretes.

If you think about it, it would be lousy system engineering to have as complex and valuable a system as an aircraft relying on piping safety critical data at levels in the millivolts down long runs of cables. For lots of reasons, such as crosstalk from the 400 cycle power feeders that run all through aircraft.

People who claim that commercial airliners built by huge concerns such as Boeing and Airbus could be brought down by millivolt interference levels ought to stop and think about what they are really saying. The post to which this is a reply did not make that claim, just that the cell phones could cause interference through wire coupling. But others have taken that same position and made quite startling claims about the safety or lack thereof of flight relative to high intensity radiated fields (HIRF).