## Electromagnetic Radiation |

Radiated coupling results when electromagnetic energy is emitted from a source, propagates to the far-field, and induces voltages and currents in another circuit. Unlike common impedance coupling, no conducted path is required. Unlike electric and magnetic field coupling, the victim circuit is not in the electromagnetic near field of the source. Radiated coupling is the only possible coupling mechanism when the source and victim circuits (including all connected conductors) are separated by many wavelengths. Of the four possible coupling mechanisms, Today, we take wireless communication for granted. It is no longer viewed as magical, but the theory is still complex and the equipment used to send and receive signals is still among the most sophisticated of our time. This leads many engineers to believe that electromagnetic radiation is difficult to create and difficult to detect. However, virtually all circuits radiate and most pick up detectable amounts of ambient electromagnetic fields. It is not necessary to attach an antenna to a circuit to make it radiate, the structure and location of most high frequency circuits allows them to act as their own antenna or to couple to nearby objects that act as efficient antennas. The more difficult challenge for the designers of most electronic products is to design circuits that don't produce too much electromagnetic radiation. In order to understand how and why circuits exhibit unintentional electromagnetic emissions, it is helpful to review a few general concepts related to electromagnetic radiation and antenna theory. ## Fields produced by a time-varying currentConsider the short current filament illustrated in Figure 1.
A current with amplitude,
The magnetic vector potential due to this current can be expressed as, , (1) where the term represents the delay between the time the
current changes at the origin and the time the change can be detected at a
point a distance . (2) In free space, the magnetic field intensity can be calculated from the magnetic vector potential as, . (3a) Note that the magnetic field points in the
direction everywhere. It circulates around the
axis of the current and has its maximum amplitude in the plane perpendicular to
the axis of the current (sin θ = 1). There is no magnetic field on the Applying Faraday's Law in point form, we can calculate the electric field as, . (3b) The electric field is perpendicular to the magnetic field at every point in space and has both a component and an component. Although these expressions are fairly complicated, we can appreciate the more important aspects of
these field distributions by considering two separate cases:
. (4) If we are close to the source relative to a wavelength, then
When we are far from the source, (5) and . (6) Note that in the far
field, are perpendicular to each other
and to the direction of propagation, Hr.
The fields are in phase with each other and the ratio of their amplitudes is,(7) at all points in space. These are the characteristics of an electromagnetic plane wave. Far from the source, where the spherical wave front is large relative to the size of the observer, the radiated field is essentially a uniform plane wave.
To answer the question above, we note that in the far field of a radiation source, the field strength is inversely proportional to the distance. Therefore, increasing the distance by a factor of 3.3 would reduce the field strength by a factor of 3.3. This is approximately a 10-dB reduction in the field strength, so the correct response would be 30 dB(μV/m). ## Fields Produced by a Small Current LoopConsider the small loop of current illustrated in Figure 2.
This current loop can be modeled as 4 current filaments oriented to form a
square. We'll let the current amplitude be ω as in the previous example. Using the
principle of superposition, we can add the electric fields from each current
filament to calculate the fields due to the loop. This is a straightforward (though
somewhat tedious) process, which is described in many antenna texts.
However, for our purposes, the result is more interesting than the derivation,
so only the results are presented here.
In free space, the electric field intensity produced by a small loop of current is given by the expression, (8) where Applying Ampere's Law in point form, we can calculate the magnetic field as, . (9) The magnetic field from the loop looks a lot like
the electric field from the current filament. The magnetic field is perpendicular
to the electric field at every point in space and has both a
component and an component. If we are close to the source
relative to a wavelength ( When we are far from the source ( (10) and . (11) Again we note that, in the far field,
are perpendicular to each other and to the direction of propagation, Hr.
The fields are in phase with each
other and the ratio of their amplitudes is η_{0}.## Fields Produced by Electrically Small CircuitsNow let's apply what we know about the radiation from current
filaments and current loops to estimate the radiation from an electrically
small circuit. We'll start by considering the simple circuit illustrated in
Figure 3. This circuit has an ideal voltage source and resistor connected by
wire forming a loop with dimensions
If the resistor has a very small value, we might expect this circuit to radiate much like a current loop. The current in the loop would be, (12) where . (13) Since we are normally interested in the maximum
radiated field, regardless of orientation, we can replace the . (14)
If . (15) Therefore, the current flowing in the vertical section on the left side of the circuit in Figure 4 is approximately, (16) If we substitute the above equation for . (17) We'll simplify this expression by noting that η.
We'll also take the maximum value of the magnitude of this expression as we did
for the low-impedance circuit, resulting in the following simple estimate of
the maximum radiated field from a high impedance circuit,_{0}(18) Note the similarity between the expression for high-impedance
circuits (18) and the expression for low-impedance circuits (14). Both are
proportional to the source voltage and the loop area. Both are proportional to
the square of the frequency and inversely proportional to the distance from the
source. The only difference between the two expressions is that the
low-impedance circuit expression has an additional . (19) ## Example 1: Estimating the Radiated Field from an Electrically Small Circuit
First we need to determine whether or not the circuit is electrically small at the frequency of interest. At 80 MHz, the wavelength in free space is 3.75 meters. Since the maximum dimension of the circuit is much less than a wavelength, we can use Equation (19) to estimate the maximum radiated field from this circuit. The impedance is 500 ohms, which is greater than the 377-ohm intrinsic impedance of free space, so we use the upper equation in (19), . (20) The FCC Class B limit at 80 MHz is 100 μV/m or 40 dB(μV/m), suggesting that this circuit would be 2.5 dB over the limit. However, the field calculated above is in free space and the FCC EMI testing is done in a semi-anechoic environment (above a ground plane). Reflections off the ground plane may add in-phase or out-of-phase with the radiation directly from the circuit. Since we are calculating the maximum radiation (and since the FCC scans the antenna height looking for the maximum), we should double our calculated field strength (i.e. add 6 dB) to account for the presence of the ground plane. In this case, our estimate of the maximum emissions from the circuit above a ground plane becomes 48.5 dB(μV/m) or 8.5 dB above the FCC Class B limit. As the example above indicates, the presence of a ground plane complicates a radiation calculation. If the ground plane is infinite (or at least very large relative to a wavelength), the amplitude of the radiated field can be as much as twice its value without the ground plane. How about the planes on a printed circuit board or the walls
of a metal enclosure? Do they have the same effect? Generally speaking, if the
planes are much larger than a wavelength and much larger than the dimensions of
the source, we can model the plane by placing an Figure 6 illustrates some simple current configurations and their images in a perfectly conducting plane. The image currents flowing perpendicular to the plane will be in the same direction as the source currents. The image currents flowing parallel to the plane are in the opposite direction of the source currents. This suggests that the fields from current sources parallel to and near the plane are decreased by the plane, while fields from current sources perpendicular to the plane are enhanced by the plane.
## Example 2: Estimating the Radiated Field from a Small, Low-Impedance Circuit
The 50-ohm load resistance is less than the 377-ohm intrinsic impedance of free space, so we use the lower equation in (19). We also need to determine whether or not the inductance of the circuit is limiting the current. To calculate the inductance of a half-loop above a ground plane, we can replace the ground plane with the image of the half-loop. In this case, we have a virtual loop that is 5 cm x 3 cm as illustrated in Figure 8. Figure 8. The wire half-loop and its image. Applying our equation for the inductance of a rectangular wire loop, we note that the inductance of the 5 cm x 3 cm loop is loop is 114 nH. The inductance of the half-loop above the plane is then half this value or 57 nH. This corresponds to a reactive loop impedance of 29 ohms at 80 MHz. Although we are able to apply image theory to calculate the inductance of the loop, we cannot use image theory to calculate the radiation from the circuit. The ground plane's dimensions are small relative to a wavelength and therefore, the plane looks more like a wide conductor than an infinite plane. We don't have an expression for the radiation from circuits with wide conductors, but if we just want to get a rough estimate, we can apply Equation 19, . (21) Note that we use the actual loop area of the circuit and do not adjust our calculation to account for the circuit's plane. If we add 6 dB to this estimate to account for making the measurement in a semi-anechoic environment, we estimate the radiation to be about 62 dB(μV/m) or about 22 dB above the FCC Class B limit. ## Input Impedance and Radiation ResistanceIn general, if a time-varying voltage appears between any two conducting objects in an open environment, time-varying currents will flow on these conductors and radiation will result. The electrically small circuits described in the previous section are relatively inefficient sources of electromagnetic radiation. Larger resonant structures can produce radiated fields that are many orders of magnitude stronger when driven with the same source voltages.
Consider the basic antenna structure illustrated in Figure 9. A sinusoidal voltage source
connected between two metal bars draws charge off one bar and pushes it onto
the other bar when the voltage is positive. One half cycle later, the polarity
is reversed and the charge distribution is inverted. The moving charge results
in a current. The ratio of the voltage to the current through the source is the
. (22) At low frequencies, the amount of charge the bars can hold for a given voltage is determined by the mutual capacitance between the bars. In this case, the imaginary part of the input impedance is, (23) where f is the source frequency and C is the mutual capacitance. If the bars are good conductors, at low frequencies and very little real power is delivered by the source. However, as the frequency increases (and the bar looks longer relative to a wavelength), several factors combine to change the antenna input impedance: - The inductance associated with the currents flowing in the bars (and the associated magnetic fields) begin to affect the reactive part of the input impedance;
- The resistive losses increase due to the skin effect;
- Power is lost to radiation which contributes to the real part of the input impedance.
It is convenient to express the real part of the input impedance as the sum of two terms, (24) where radiation resistance of the
antenna and R is the
loss resistance. The power radiated can then be calculated as,_{diss}(21) and the power dissipated as heat can be calculated as, . (26) The ratio of the power radiated to the total power delivered to the structure is called the radiation efficiency and can be calculated using the following equation, . (27) ## Example 3: Radiation Efficiency of an Electrically Small Circuit
We'll start by calculating the power dissipated. If we assume that the power is primarily dissipated in the load resistor (as opposed to the wires), the dissipated power is simply, . (28) To estimate the power radiated, we note that the maximum
electric field strength at 3 meters is 134 μV/m (as calculated in Example 1). The
maximum . (29) This is the maximum power density radiated in any direction, so we can calculate an upper bound on the radiated power by assuming that this power density is radiated in all directions and integrating over a sphere with a 3-meter radius, (30) Therefore, the radiation efficiency of the circuit is, . (31) Note that the input impedance of an antenna structure may depend on the antenna environment as well as the size and shape of the antenna. For example, the radiation resistance and the radiated power of any antenna will drop to zero if the antenna is operated in a fully-shielded resonant enclosure. ## The Resonant Half-Wave DipoleAn antenna consisting of two simple conductors driven
relative to each other by a single source is called a
At very low frequencies (where The reactance is zero when the total length
of the wire is slightly less than one-half wavelength. A dipole antenna with
this length has a real input impedance of approximately 70 ohms and is called a
This is a very simple calculation, since both the input resistance and the radiation resistance are about 72 ohms. The correct solution is, . (32) To find the maximum radiated field strength at 3 meters from this antenna, we first determine the maximum radiated power density, (33) where is the average power density and The maximum radiated electric field can then be calculated using Equation (29) in reverse, . (34) Comparing this to the field strength radiated by the electrically small circuit in Figure 5, we can gain an appreciation for just how important the size and shape of the antenna can be. In this case, if we assume both structures were driven at 80 MHz, the maximum dimension of the circuit was 5 cm while the maximum dimension of the dipole is 37.5 cm (half a wavelength at 80 MHz). This is a factor of 7.5. However, the radiated emissions increased by a factor of or 66 dB. ## Example 5: Radiation Efficiency of a Half-Wave Dipole
The power radiated by resonant half-wave dipole is simply, (35) where I is the current at the source. To calculate the power dissipated, we start by determining the resistance per unit length of the copper wire at 100 MHz. . (36) The total power dissipated in the half-wave dipole is then, . (37) Therefore the efficiency of this resonant half-wave dipole is, . (38) Compare this to the efficiency of the small circuit in Example 3. Resonant length wire antennas tend to be very efficient compared to electrically small antennas. They can easily be 4 to 6 orders of magnitude more efficient. ## Quarter-Wave MonopolesHalf-wave dipoles make good antennas for many applications,
but they are large at low frequencies and may not operate as intended close to
a large metal surface. A
Cables driven relative to large metallic enclosures can often be modeled as monopole antennas. Since resonant monopole antennas are very efficient radiation sources, it is important to ensure that voltages between cables and enclosures be held to very small values at frequencies that might be near cable resonances.
The exact answer depends on the orientation of the wire, the cross section of the wire, the structure's size and shape, and other factors. However, 25 cm is a quarter wavelength at 300 MHz. The cable is likely to resonate and become an efficient antenna near this frequency. ## Dipoles Driven Off-CenterWhen a wire antenna is driven off-center, it will still exhibit a resonance near the frequency where it is a half-wavelength long. However, the radiation resistance at resonance will be a function of the source location. Figure 12 plots the radiation resistance of a resonant half-wave dipole as a function of source location. Note that the resistance quickly increases as the source is moved away from the center. A voltage source located near the end of the wire cannot deliver power effectively to the antenna even at resonance.
## Characteristics of Efficient and Inefficient AntennasMost of the unintentional radiation sources that an EMC engineer encounters can be modeled as simple dipole antennas. There are basically 3 conditions that have to be met in order for these antennas to radiate effectively: - The antenna must have two parts;
- both parts must not be electrically small;
- something must induce a voltage between the 2 parts.
The first condition is important to remember when trying to track down the source of a radiation problem. It is not correct to say that a particular wire or a particular piece of metal is "the antenna". A single conductor will not be an effective antenna unless it is driven relative to something else. The "something else" is an equally important part of the antenna. Once identified, options for reducing radiated emissions normally become clearer. Locating the two parts of an effective antenna becomes much easier when we recognize the second condition. For example, if we are looking for the "antenna" responsible for radiated emissions at 50 MHz (λ = 6 meters), then we are looking for 2 conducting objects on the order of a meter long. It is unlikely that these antenna parts are located on the printed circuit board. Most table-top size products can only radiate effectively at low frequencies by driving one cable relative to another. At frequencies below a few hundred MHz, the number of possible antenna parts is limited and often readily apparent without a detailed examination of the entire design. The third condition above suggests a method for controlling radiated emissions. Once the possible antenna parts have been identified, a device will not generate significant radiated emissions if the voltage between these parts is kept low. This is best accomplished by locating these parts near each other and ensuring that no high-frequency circuits come between them. Tying them together electrically with a good high-frequency connection will further ensure that they are held to the same potential.
The key to answering this question is to examine each antenna and identify the two antenna parts that are driven by the indicated voltage source. Since at least one part of every antenna is electrically small, it is the smallest part that will limit the overall effectiveness of the antenna. In Figure 13(a), the short upper wire on the antenna on the right is most limiting; therefore the antenna on the left is more effective. In Figure 13(b), an extra piece of wire has been added to each of the antennas in Figure 13(a). The short upper wire on the antenna on the right is again the most limiting, so the antenna on the left is still more effective. Note that adding to the shorter half of an electrically small antenna has a much greater impact on the antenna efficiency than adding to the larger half. ## Slot AntennasSlot antennas are another potentially efficient type of
antenna that EMC engineers should be familiar with. As illustrated in Figure 14, a slot antenna is formed by a long
thin aperture in a conducting surface. An electric field distribution appearing
in the slot (e.g. due to a surface current that is disrupted by the slot)
produces a radiated field in the same way that a current distribution on a wire
does. In fact, slot antennas are generally analyzed by replacing the electric
field distribution with an equivalent (but fictitious) magnetic current and
solving for the fields radiated by these magnetic currents. The fields radiated
by a resonant half-wave slot have the same form (with the role of
## Receiving AntennasGenerally speaking, the same structures that make good radiating antennas also make good receiving antennas. For this reason, many of the same techniques used to identify or prevent radiated emission problems can be applied to radiated susceptibility problems. However unlike radiation, where the source impedance is almost always low relative to the antenna input impedance, the devices that exhibit susceptibility problems often have high impedance inputs. Because of this, it is not necessarily true that higher antenna input impedances correspond to poorer antenna performance. The power received by a device connected to a dipole antenna can be calculated using the following formula, (39) where is the power density of the incident wave, is the effective aperture of the antenna, is the factor accounting for the impedance mismatch between the antenna and the receiver, and is the voltage reflection coefficient at the receiver. This formula may be difficult to apply in many situations however, because it requires significant information about both the antenna and the receiver. If an order-of-magnitude approximate solution is good enough, it is convenient to estimate the maximum voltage dropped across a higher impedance input as, (40) where ## Example 6: Estimating the Maximum Voltage Coupled to a Half-Wave Dipole Antenna
If we assume that the receiver is positioned at the point on the dipole where its impedance is matched to the radiation resistance, the expression for the received power becomes, . (41) The received voltage is, . (42) We can compare this to the estimated value, (43) and note that the estimate was accurate to within 2 dB in this case. ## Example 7: Estimating the Maximum Voltage Coupled to an Electrically Short Dipole Antenna
The radiation resistance of an electrically short dipole antenna is approximately, . (44) The directivity, . (45) The received voltage is therefore, . (46) We can compare this to the estimated value, . (47) In this case, the voltage is over-estimated by a factor of 4 or 12 dB. |