11.1 Maxwell

The laws of Maxwell relate the electric field to charge, current and the magnetic field:

$$\begin{array}{rcll}\mathrm{\text{rot}}𝐄& =& -\mu \frac{\partial 𝐇}{\mathit{\partial 𝑡}}& \\ \mathrm{\text{rot}}𝐇& =& J+𝜖\frac{\partial 𝐄}{\mathit{\partial 𝑡}}& \text{(11.1)}\\ \mathrm{\text{div}}𝐄& =& \frac{\rho }{𝜖}& \\ \mathrm{\text{div}}𝐇& =& 0& \end{array}$$

Here, $𝐄$ is the electric field, $𝐇$ the magnetic field and $\mathit{𝑟𝑜𝑡}$ and $\mathit{𝑑𝑖𝑣}$ are the well known operators for vector calculus: the rotation and divergence. As the names of these operations already suggest, these operators calculate how much a vector field rotates or changes. We will not go into these operations and equations: we will work towards a — within the context of this book — usable result. The magnetic field is related to currents and voltages through relativity, meaning that the constants $𝜖$ and $\mu$ are related to the speed of light $c$:

$$c=\frac{1}{\sqrt{{\mu }_0{𝜖}_0}}$$

It follows from (11.1) that a change in $𝐄$-field causes a change in $𝐇$-field, and vice versa. From the vector operations, it also follows that the $𝐄$-field caused by a time-varying $𝐇$-field is perpendicular to that $𝐇$-field. The same holds for an $𝐇$-field caused by a time-varying $𝐄$-field: this $E$ is also perpendicular to $𝐇$. It now can be derived that the power density of the $𝐄$ and $𝐇$ fields (called an EM-field) is given by the so-called Poynting vector, which is perpendicular to the $𝐄$ and $𝐇$ fields⁶⁰:

$$𝐒=𝐄\times 𝐇[W∕m^2]$$

(11.2)

11.2 Maxwell and Kirchhoff

The Maxwell equations are an expansion of the voltage and current laws of Kirchhoff. This can also be seen from the equations themselves: for any mesh using the voltage law:

$$\begin{array}{rcll}\sum _{\mathit{𝑚𝑒𝑠h}}\mathrm{\Delta }{V}_{\mathit{𝑚𝑒𝑠h}}& =& 0& \end{array}$$

For the same mesh, a similar equation can be written down in terms of electric field strength $E$. The summation then becomes a contour integral — an integral over a closed contour — and results in:

$$\begin{array}{rcll}{\oint }_C𝐄\mathit{𝑑𝑙}& =& 0& \end{array}$$

The integral form of $\mathrm{\text{rot}}𝐄=-\mu \frac{\partial 𝐇}{\mathit{\partial 𝑡}}$ is given by ${\oint <!--\; FUNCTION\; APPLICATION\; -->}_C𝐄\mathit{𝑑𝑙}=-\frac{\partial {\mathrm{\Phi }}_B}{\mathit{\partial 𝑡}}$. Here, ${\mathrm{\Phi }}_B$ is the total magnetic flux through the surface that is enclosed by the contour. Trying to equate the Maxwell relation equal to the Kirchhoff voltage law relation reveals that:

We can derive something similar for Kirchhoff’s current law. If we take the divergence of Ampère’s law — the Maxwell equation for $\mathrm{\text{rot}}𝐇$ — then we get⁶¹for any 3-dimensional vector:

$$\begin{array}{rcll}\mathrm{\text{div}}(\mathrm{\text{rot}}(𝐇))& \equiv & 0=\mathrm{\text{div}}𝐉+𝜖\mathrm{\text{div}}\frac{\partial 𝐄}{\mathit{\partial 𝑡}}\iff & \\ \mathrm{\text{div}}𝐉& =& -\frac{\mathit{𝑑𝜌}}{\mathit{𝑑𝑡}}& \end{array}$$

This relation states that the change in current (density) in a certain volume is due to the accumulation of charge within that volume. This accumulation happens for every current or voltage change, since charge cannot leave infinitely fast from that volume.Trying to equate the KCL to the Maxwell result above, it follows that:

In all previous chapters in this book, the analyses were based on Kirchhoff’s voltage and current law. This implicitly means that the signal frequencies must be low enough: low enough for the wavelength of an EM-wave $c∕f$ to be much larger than the physical dimensions of the circuit. For an audio amplifier, which has to operate up to 20 kHz, these assumptions are true if the amplifier is much smaller than 15 km, which is usually satisfied. For a GSM in the 1.8 GHz band however, this already becomes a problem. In this case, the entire circuit must be much smaller than 15 cm to be able to use the current and voltage laws. The internal IC’s within the GSM typically are much smaller than this 15cm and then can be designed and analyzed using the KVL and KCL, but as soon as you connect these ICs with something at the outside — a package, matching network or antenna — then the distances increase to such an extent that the laws of Kirchhoff are not applicable anymore.

A basic rule? Well alright: in general, you may use the Kirchhoff laws if the dimensions of the circuit are smaller than $\lambda ∕10$.

Calculating this can be done using the following set of equations [10], in case you’d be interested or if you want to get a numerical value. The impedance of the antenna connector is related to the “internal impedances” as follows:

$$\begin{array}{rcll}{Z}_{\mathit{𝑎𝑛𝑡𝑒𝑛𝑛𝑎}}& =& \frac{1}{\mathit{𝑠𝑖}{n}^2(\frac{\mathit{𝜋𝑙}}{\lambda })}(R_{\mathit{𝑟𝑎𝑑}}+j{X}_a)& \\ R_{\mathit{𝑟𝑎𝑑}}& =& \frac{2P_{\mathit{𝑟𝑎𝑑}}}{I_{\mathit{𝑎𝑛𝑡𝑖𝑛𝑜𝑑𝑒}}^2}=\frac{\eta }{2\pi }{\int \; <!--\; nolimits\; -->}_0^{\pi }\frac{{\left[\mathit{𝑐𝑜𝑠}(\frac{\mathit{𝜋𝑙}}{\lambda }\mathit{𝑐𝑜𝑠}(𝜃))-\mathit{𝑐𝑜𝑠}(\frac{\mathit{𝜋𝑙}}{\lambda })\right]}^2}{\mathit{𝑠𝑖𝑛}(𝜃)}\mathit{𝑑𝜃}& \\ X_a& =& \frac{\eta }{4\pi }\{2S_i\left(\frac{2\mathit{𝜋𝑙}}{\lambda }\right)+\mathit{𝑐𝑜𝑠}\left(\frac{2\mathit{𝜋𝑙}}{\lambda }\right)[2S_i\left(\frac{2\mathit{𝜋𝑙}}{\lambda }\right)-S_i\left(\frac{4\mathit{𝜋𝑙}}{\lambda }\right)]\}-& \\ & & \frac{\eta }{4\pi }\mathit{𝑠𝑖𝑛}\left(\frac{2\mathit{𝜋𝑙}}{\lambda }\right)\{2C_i\left(\frac{2\mathit{𝜋𝑙}}{\lambda }\right)-C_i\left(\frac{4\mathit{𝜋𝑙}}{\lambda }\right)-C_i\left(\frac{4\pi a^2}{\mathit{𝑙𝜆}}\right)\}& \\ & & \mathrm{\text{with}}& \\ S_i(x)& & ={\int \; <!--\; nolimits\; -->}_0^x\frac{\mathit{𝑠𝑖𝑛}(x)}{x}\mathit{𝑑𝑥}{C}_i(x)=-{\int \; <!--\; nolimits\; -->}_x^{\infty }\frac{\mathit{𝑐𝑜𝑠}(x)}{x}\mathit{𝑑𝑥}& \end{array}$$

The constants $a$ and $\eta$ are the wire thickness respectively the free space impedance for an EM-wave $\eta =\sqrt{{\mu }_0∕{𝜖}_0}\approx 120\pi \approx 377\mathrm{\Omega }$ In vacuum or air, the radiation resistance of a half-wave dipole antenna equals $73.14\mathrm{\Omega }$. If the antenna shows no other significant resistive losses due to e.g. Ohmic losses in the antenna, then the radiation resistance is equal to the resistance of the antenna as a whole.

The impedance of an antenna consists of a resistance and a reactance (inductive or capacitive). For a half wave dipole, the antenna thickness does not seem to be of importance and its reactance in vacuum or air is $j42.55\mathrm{\Omega }$. For other antenna lengths the relation is dependent on the thickness of the antenna. For the half wave dipole the reactance is — as you can see from the equation — positive for certain values of $l∕\lambda$, and negative for other values. It might seem scary, a negative reactance, but it isn’t. As you know, the impedances of reactive elements are

$$\begin{array}{rcll}& Z_C=\frac{1}{\mathit{𝑗𝜔𝐶}}=\frac{-j}{\mathit{𝜔𝐶}}& \iff X_C=\frac{-1}{\mathit{𝜔𝐶}}& \\ & Z_L=\mathit{𝑗𝜔𝐿}& \iff X_L=\mathit{𝜔𝐿}& \end{array}$$

thus a positive $X$ corresponds to an inductance and a negative $X$ to a capacitance.

11.6 Other antenna characteristics

We can keep on talking about antennas just about forever, but for this introductory course on electronics, that would not be very useful. However, it is useful to get acquainted with a few concepts concerning the antenna: the most important ones are discussed below.

Directivity and gain The terms directivity and gain of an antenna are often mixed up. The directivity of an antenna gives a measure of how well the antenna is capable of bundling its radiation to a specific direction. It doesn’t matter whether or not it is the transmit or receive antenna, since $\mathit{𝑎𝑛𝑡𝑒𝑛𝑛𝑎}=\mathit{𝑎𝑛𝑡𝑒𝑛𝑛𝑎}$.

• a so-called isotropic antenna transmits or receives just as well in every direction in 3-dimensional space. This isotropic antenna is an ideal antenna, which cannot be realised. This hypothetical isotropic antenna is however used as a reference to relate directivity and antenna gain to.

• a dipole and monopole antenna are symmetric in a (2-dimensional) plane perpendicular to the arm(s). This implies that there is no reason whatsoever that in that plane the radiation in a specific direction is different from that in another direction. As a result, monopole and dipole antennas transmit and receive omnidirectionally perpendicular to the antenna arms.

  Note that the receive and transmit power in the direction of the arms is low to zero, due to the same symmetry. sufficiently far away from the antenna then the radiated or received field adds destructively.

• Antennas with other shapes may radiate (or receive) more power in specific directions.

Numerically, the directivity or gain of an antenna give the ratio of transmitting power in one specific direction, related to the transmitting power of the isotropic antenna:

$$\begin{array}{rcll}G& =& \frac{P_{\mathit{𝑚𝑎𝑥},\mathit{𝑎𝑛𝑡𝑒𝑛𝑛𝑎}}}{P_{\mathit{𝑖𝑠𝑜𝑡𝑟𝑜𝑝𝑖𝑐𝑎𝑛𝑡𝑒𝑛𝑛𝑎}}}& \end{array}$$

where the value of $G$ is usually given in $\mathit{𝑑𝐵𝑖}$: the gain or directivity compared to an isotropic antenna. The gain or directivity of a dipole antenna is $2.15\mathit{𝑑𝐵𝑖}$. Antennae with a high directivity (or gain) radiate and receive mainly a narrow beam. Antennas that receive signals from many directions (almost) equally well have, by definition, a low gain and low directivity.