[Lowfer] Part 15 power limit for Hifers
Garry and Linda Hess
k3siw at sbcglobal.net
Sat Aug 6 16:59:02 EDT 2011
John, thanks for the comment and update on the field strength number.
I didn't mean to imply that a half-wave dipole would be used for field
strength measurements. I simply used an equation for the power received
by a half-wave dipole at distance D and frequency F in the presence of
field strength E. Then solving the standard free-space path loss
equation for the transmit power to a half-wave transmit antenna that
produces E gave 1.86 mW with E=10,000 uV/m. Using 15,848 uV/m instead
produces 4.67 mW as you mentioned, since that E represents 4 dB greater
power.
In case anyone is wondering about the direct formula you mentioned, it
can be derived as follows: The power flux density (i.e., power per unit
area) for an isotropic radiator of Pi watts at distance R meters is
S=Pi/(4*pi*R*R), where the denominator is the area of a sphere of radius
R. The maximum power gain of a linear half-wave dipole relative to
isotropic is 2.15 dB, or a factor of 1.64. Using dipole power Pd instead
of Pi means S=Pd*1.64/(4*pi*R*R). But the power flux density is also
given by the electric field strength squared divided by the intrinsic
impedance of free space, about 377 ohms, or S=E*E/377. Equating the two
expressions for S and solving for E leads to
E=sqrt[(Pd*1.64*377)/4*pi]/R=sqrt(49.2*Pd)/R.
My hifer is currently putting about 3.7 dBm to the dipole feed so I
should be legal by a dB.
Thanks again for the comment,
Garry, K3SIW, EN52ta, Elgin, IL
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