Re: NEC-LIST: induced voltages

At 10:36 23.06.98 -0400, you wrote:
>Grant,
>
>Your question is not so dumb! I think a question is never dumb, per
>se. This is an interesting problem, and I gave it some thought. My
>conclusions are as follows:
>
>After considering your "wire" as a half-wave dipole, since your
>resistor is in the center, I took another approach, since I wanted to
>be independent of wavelength or frequency. I considered the wire as an
>effective receiving area multiplied by the power density impinging on
>the wire. In order to obtain the voltage, I used the radiation
>resistance of a Hertz dipole. This resulted in the formula:
>
> [snip!]
>
>Result : the voltage can never equal the field strength !
>
>Comment: This assumes that you have a matched situation. If you load
>with different impedances, you must consider this in the formula.
>
>Max Schmitt

Hi Max,

Your results agree precisely with a similar closed solution of antenna
theory, using equivalent circuits, giving the opportunity of showing
the influence of a load.

The short dipole can can be represented by the following circuit:

                    C
        ---------I I--------o
        I
        O E * l_eff
        I
        -------------------o

The voltage of the source depends on the electric fieldstrength E and the
effective length l_eff of the dipole, where l_eff for short dipoles can be
derived from the mechanical dipole length L by:

                Lambda 1 - cos(pi * L / Lambda)
  l_eff = ------------ --------------------------------
                     pi sin(pi * L / Lambda)

(with L < 0.9 Lambda)

With L < .4 Lambda (triangular current distribution along the dipole)
this yields to the extremely simple coherence:

            l_eff = L / 2

However, if there is a load Z connected to the antenna, the voltage V
measured at the antenna port will be remarkable smaller:

 V = E * l_eff * Z / (1/(j w C) + Z)

with C for a thin short dipole approximately

 C = 2.5 pF * L

Even if you use a high ohmic input ampifier, lets say a FET, you will
find a much lower voltage because of the amplifier imput capacitance
of typically 2 ... 5 pF, which becomes much worse with a cable between
antenna port and amplifier!

>From the ratio E/V you can now calculate the antenna factor.

Regards and vy 73
Joachim Brose
DL6MES
Received on Mon Jun 29 1998 - 09:14:08 EDT