NEC-LIST: Modeling short dipoles.

Good day;

        I didn't read most of the correspondence on this topic, but I
did read the last item from Charley Rhodes. It appears to be time
to bring a lot of people back to basics.

        The gain of a short dipole in free space is a textbook item.
Its directivity is 1.5, which translates to 1.761 dB. The radiation
resistance is

        Rr = 20(pi L f/300)^2 (unloaded) (1)
             80(pi L f/300)^2 (end loaded) (2)

where L is length in m, f is frequency in MHz, and 300 is light speed
in Mm/s. The wire loss resistance for the unloaded case can be found
by assuming

        I(z) = Iin(1-|z|/(L/2)) (3)

which is a linear approximaation to the tail-end of a sine
distribution, good enough for an electrically small dipole. The
incremental power loss is

        dP = (I(z))^2 Rs dz/(2 pi a) (4)

where Rs is the surface resistance per square and 'a' is the radius of
the wire.

Integrating the power loss and dividing by (Iin)^2 gives the input
equivalent loss resistance,

        Rw = Rs L/(6 pi a) (5)

For the end-loaded case,

        Rw = Rs L/2 pi a) (6)

Efficiency is

        Eff = Rf/(Rf+Rw) (7)

and gain is

        G = Eff D = 1.5 Eff (8)

The short dipole is not so inefficient or low-gain as you might
suppose. For example, at 300 MHz, L = 0.055 m, and a = 0.0008139
for #14 wire, the efficiency is 97.36 % from the above formulas and
97.42 % from the NEC example given below. However, the reactance is
nearly 1600 ohms. A series coil to resonate this antenna, with a Q of
200, would have a loss resistance of 8 ohms, swamping the Rr of about
0.6 ohms. Putting a 50 ohm resistor in series with the antenna has
only a small effect on the current because of this high reactance.
The input power is calculated by NEC as Pin = 0.5VI* and this is
used in calculating the gains, so if I doesn't change, neither does
the gain. Dividing the wire radius by 10 gives Eff = 0.7866 from the
above, and NEC gives 0.7846, with an input reactance of 3.3 kohms.
The resistance values are less than those given by the equations
above, with the thinner wire being closer.

        The input file given below produces the input impedance,
radiated and loss powers based on Pr = Pin-Ploss, the gains for theta
in 2 deg. steps and phi = 0 and 90 deg., the average gain, and the
total radiated power calculated by integrating the E-field-based power
densities at the angles specified over the sphere. Ploss is also an
integration over the wire. If the field is sampled sufficiently, and
the input impedance is correct, the two values of Pr will agree, and
the reported efficiency will equal the average power gain. NEC does
have problems with small, complex structures. It is possible to get
negative input resistance. It can also happen that Pin-Ploss is
negative. Pin is dependent on the accuracy of one number, the input
current. Ploss and Pr from E-field values are more accurate because
they only depend on the shape of the overall current distribution,
which is more likely to be accurate than any single value.

Cheers!
Douglas B. Miron

CM SMALL dipole
CE
GW 1 33 0 0 0 0 0 0.055 0.0008139
GE,0
EX,0,1,17,0,1.0E3,0.0,1.0
FR,0,1,0,0,300,0.1
LD,5,0,0,0,5.8E7,1
PT,-1
RP,0,46,2,1004,0,0,2,90
EN
Received on Wed Nov 26 1997 - 09:42:00 EST