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Does a dipole antenna need a minimum frequency of AC to make the electric field detach and propagate?

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No, but to maximize the radiated power at any frequency the dipole (horizontal, center-fed) must be 1/2 wavelength long. This means 5 meters at 30 megahertz, 50 meters at 3 megahertz, 500 meters at 300 kilohertz, 5,000 meters at 30 kilohertz and so on.

Note that the radiation resistance of a center-fed dipole that is shorter than 1/2 wavelength falls off quickly. For example, a 10 meter antenna driven at 300 kilohertz will radiate only a tiny amount of radio frequency power.

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As already remarked in the comments, the optimal frequency is based on the "1/2 wavelength" rule. But to what extent do things get worse at lower frequencies? To make this more quantitative, we have to compute what happens exactly. Computing the antenna input impedance for a linear dipole is indeed the right approach. As can be found in the EM books (like Balanis' "Antenna Theory"), the result for a lossless dipole with not too many further approximations is, for the real and imaginary part:

$$\begin{align} \text{Re}Z=& \frac{Z_0}{2\pi\sin^2 (x/2)} \Big[-\text{Ci}\,x+ {\small \frac{1}{2}} \big(\text{Ci}(2 x)-2\, \text{Ci}\,x+\log {\small\frac{x}{2}}+\gamma\, \big) \cos\,x\ + \\[5pt] & \qquad\qquad\qquad +{\small \frac{1}{2}} \big(\text{Si}(2 x)-2\, \text{Si}\,x \big) \sin\,x+\log\, x+\gamma\,\Big] \\[5pt] \text{Im}Z=& \frac{Z_0}{4\pi\sin^2 (x/2)} \Big[ \left(\text{Ci}(2 x y^2)-2\, \text{Ci}\,x+\text{Ci}(2 x)\right) \sin\,x\,+ \\[5pt] & \qquad\qquad\qquad\quad + 2\,\text{Si}(x)+\big(2\,\text{Si}\,x-\text{Si}(2 x)\big) \cos\,x\,\Big] \end{align}$$

where we define $x=2\pi \ell/\lambda$ and $y=r_w/\ell$ with $\ell$ the length of the dipole and $r_w$ the radius of the rods that it is made of. This result is valid up to and somewhat above the optimal frequency. At half-wave resonance $(\ell=\lambda/2)$ it gives the well-known: $$ Z(x\text{=}\tfrac12) = 73.1296 \dots + i\,42.5445\dots $$ If we go to lower frequencies, i.e. smaller $x$, we can (fortunately) further approximate: $$\begin{align} \text{Re}\,Z_\text{LF} \approx &\ \frac{Z_0\,x^2}{24\pi} \\[5pt] \text{Im}\,Z_\text{LF} \approx &\ \frac{2Z_0(1+\log2y)}{\pi x} \end{align}$$

And there we see the two problems:

  1. The real part, which represents absorbed power, becomes very small, so we would need large current to get any power into the antenna.
  2. The imaginary part becomes very big and therefore the total impedance as well, so we cannot get a big current into the antenna, unless it is tuned with a reactive series impedance with the opposite sign of $\text{Im}\,Z_\text{LF}$. In practice this means putting a coil in series. A quartz crystal would be better, but those are usually specified at the impedance dip, not at the impedance peak (although they definitely have one!) and they cannot handle high power.

Depending on the $Q$ factor of the available tuning element the equations above will tell you how much extra loss you get. Just taking $$R_{\,\text{loss}} = \frac1Q |Z_\text{tune}| = \frac1Q |\text{Im}\,Z_\text{LF}|$$ will give us the ratio of the radiated power and the loss, in other words the efficiency: $$ \eta = \frac{\text{Re}\,Z_\text{LF}}{\text{Re}\,Z_\text{LF}+R_{\,\text{loss}}} = \frac{\text{Re}\,Z_\text{LF}}{\text{Re}\,Z_\text{LF}+|\text{Im}\,Z_\text{LF}|/Q} \approx \frac{Q \ \text{Re}\,Z_\text{LF}}{|\text{Im}\,Z_\text{LF}|} = \frac{Q\ x^3}{48\ |1+\log2y|} $$ so with a realistic value of, say, $Q=100$, taking $\ell$=1/100th of $\lambda$, meaning $x\approx 0.06$, and using a relatively "fat" antenna with the rod diameter 1/10th of the length, so $y=1/20$, would already drop the efficiency to $0.04$%! But with $\ell$ only 1/10th of $\lambda$ we would still have a reasonable $\eta$ around 30%.

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  • $\begingroup$ Jos, want you to add a bit about top load? $\endgroup$ Commented May 6 at 4:46
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    $\begingroup$ That would be two end-point loads for the symmetric case, but the expressions of Balanis are not suited to easily include that... It could best be done by re-derivation of the LF limits for the more general case without looking at the high-frequencies at all. (But for this question about the frequency range, the comparison of frequencies was of course the way to go). $\endgroup$ Commented May 6 at 6:32

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