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[Mrcelloman99]

I'm building my first antenna for a university capstone project; it's going to be a half wave dipole. I'm concerned about abiding by FCC regulations. They state that at 3 meters, the maximum E-filed strength in the 902-928 MHz band is .5 V/m. I computed that for a half wave dipole, the driving current necessary to achieve a .5 V/m field strength at 3 m broadside is 12.5 mA. Does this mean I cannot let the current going into my transmission line exceed 12.5 mA? Is my computation correct? How do transmitter designers usually deal with this?

Thanks

[R. Fry]

The equation for the power needed by a 1/2-wave, center-fed dipole to radiate a given peak field at a given distance in free space is:

P = (E * D)^2/49.2

where

P = Power in watts
E = Field Intensity in V/m
D = Distance in meters

So after plugging in your numbers and doing the units conversions, we get an answer of about 45.7 nanowatts (0.000 000 045 7... watts).

That power level, and the r-f current it produces across the feedpoint Z of that dipole would be very difficult to measure accurately.

The best approach would be to measure the radiated fields produced by your system using a calibrated field intensity meter (including the use of its calibrated receive antenna), and adjust the transmitter output power or antenna input power so that your FCC limit is not exceeded in the direction of maximum field.

[Mrcelloman99]

I think your computation is off:

with a peak field of .5 V/m at a 3 m distance, the power required is 45.7 mW, not nW.

Also, where did you get that equation? Just curious.

Thanks for the equation and advice!

[R. Fry]

I did make an error in a data entry for my earlier post (sorry). If the 0.5 V/m field limit you first posted is accurate, then your antenna input power would indeed be 45.7 mW, not 45.7 nW.

However from your post I expected that your application had to meet 47 CFR §15.209(a), which limit for your band 902 MHz to 928 MHz is 200 µV/m at 3 meters.

If 200 µV/m at 3 meters is your operational limit, then the power needed at the antenna feedpoint is 7.32 nanowatts -- which is even more difficult to measure than before.

The equation is a re-write of Equation 15 on page 27-7 of Reference Data for Radio Engineers (ITT), to solve for power instead of field.

[Mrcelloman99]

Alright, I'll take that into account.

Thanks again!

[Mrcelloman99]

Wait, I'm seeing conflicting documents. One of the statements of the FCC rules you cited above said 200 uV/m was the peak emission limit of an UNintentional radiator, while the FCC regulations, Part 15, Section 15.245 states that the peak field of an intentional field disturbance sensor (which I'm building) is 500 mV/m or .5 V/m.

[R. Fry]

FCC §15.209 also applies to intentional radiators. Here is a clip from the first part of it:

§15.209 Radiated emission limits; general requirements.

(a) Except as provided elsewhere in this subpart, the emissions from an intentional radiator shall not exceed the field strength levels specified in the following table:
Frequency (MHz) Field strength (microvolts/meter) Measurement distance (meters)
0.009-0.490 2400/F(kHz) 300
0.490-1.705 24000/F(kHz) 30
1.705-30.0 30 30
30-88 100** 3
88-216 150** 3
216-960 200** 3
Above 960 500 3
<clip>

But the limit is higher for devices covered by §15.245.

[Mrcelloman99]

I thought I saw something that said unintentional radiator instead but I can't find it now so I probably misread it.

Regardless though, aren't I fine with the .5 V/m limit under section 15.245?

thank you for bearing with me.

[R. Fry]

Likely so, if your application fits the FCC provisions permitted for such devices.

[Mrcelloman99]

Alright, thanks!