[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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