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Signal Integrity & EMC

Signal Integrity & EMC

Unit 3: Electrostatics, EMC & SI

Electrostatic fields can cause electrostatic discharges which are covered by the Electrostatic Discharge (ESD) standard in the EMC directive. Electrostatic discharges can cause product malfunction and even be damaged.

Electric fields stress insulation material and can cause electrical breakdown, which is covered by the Low Voltage Directive (LVD).

An understanding of the concepts of electrostatics helps in designing protective shielding for electronic based products.

The concept of capacitance is explained by electrostatics, which in turn is useful in explaining cross-talk and degradation of digital signal rise and fall times.

Capacitors are also useful electronic components and are used extensively in virtually all electronic based products.

Computer simulations of field problems rely heavily on the equations we shall develop in this this unit. In order to setup a field simulation and interpret the results of the simulation an understanding of electrostatics is essential

Unit Contents

3.1 The Inverse Square Law

Consider a positive electrical charge q, separated from a small positive test charge qt by distance r in free space, as shown in Fig. 3.1

Figure 3.1 The Inverse Square Law



The charges, which are both positive, repel each other and if we imagine the charge q is fixed in space, the small test charge experiences a real "mechanical" force that tends to push it away from q along the radius vector r. Note that the test charge is assumed to be small enough not to disturb the field produced by q.

The magnitude of the force is given by:-

$F = {\mathbf{r}}\frac{{qq_t }} {{4\pi \varepsilon _0 r^2 }}$ .......................3.1

The variables in equation 3.1 have the following units:-

Variable

Units

Comments
F
 newtons Force between the two charges
q
 coulombs Electrical charge
r
 metres Distance between the two charges
$\varepsilon _0 = 8.85 \times 10^{ - 12} $
Farads/metre Permittivity of free space and air
r
 metres Unit vector along the radius vector r

If the test charge was free to move it would be accelerated along the radius vector r. We shall assume that both charges are fixed in space.

We account for the forces between the two charges by assuming they are surrounded by an electric field. The strength of the field produced by the charge q, at distance r, is defined as:-

$E = \frac{F} {{q_t }} = \mathop r\limits^ \wedge \frac{q} {{4\pi \varepsilon _0 r^2 }}$ ............................3.2

The field strength has the units of force/unit charge, that is newtons/coulomb. The force is a vector quantity having both magnitude and direction.

It is important to realise that the electric field extends around the charge q in all directions and its strength drops off as the distance squared. The circle S in Fig. 3.1 represent a circle of constant force, no matter where we position the test charge at distance r, it will experience the same force. In reality, the force on the test charge will be the same on any point on a sphere, of radius r, centered on the charge q.

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3.2 Electrical Potential- Voltage

Now imagine the test charge moves along the radius vector to a new position, as shown in Fig. 3.2.

Figure 3.2 The Definition of Voltage



We wish to calculate the work done on the charge- note that the dimensions of work are energy.

\[ \begin{gathered} W = \int\limits_{r_1 }^{r_2 } {F.dr} \hfill \\ W = \int\limits_{r_1 }^{r_2 } {FCos\theta dr} \hfill \\ W = \int\limits_{r_1 }^{r_2 } {Fdr} {\text{ because }}\theta {\text{ = 0 degrees, that is the direction of the force is along r}} \hfill \\ W = \int\limits_{r_1 }^{r_2 } {\frac{{qq_t }} {{4\pi r^2 }}dr} \hfill \\ W = \frac{{qq_t }} {{4\pi r^2 }}\left[ { - \frac{1} {r}} \right]_{r_1 }^{r_2 } \hfill \\ W = \frac{{qq_t }} {{4\pi r^2 }}\left[ { - \frac{1} {{r_1 }} - \left[ { - \frac{1} {{r_2 }}} \right]} \right] \hfill \\ W = \frac{{qq_t }} {{4\pi r^2 }}\left[ {\frac{1} {{r_2 }} - \frac{1} {{r_1 }}} \right] \hfill \\ \end{gathered} \]

Because r2 > r1, the work done is negative, that is work done by the field is negative, work done against the field is positive.

The electrical potential difference between the surfaces S2 and S1 is defined as:-

$V = \frac{W} {{q_t }} = \frac{q} {{4\pi r^2 }}\left[ {\frac{1} {{r_2 }} - \frac{1} {{r_1 }}} \right]$ ..............................3.3

The electrical potential has the dimensions of voltage, so voltage has the dimensions of energy/per unit charge, that is joules/coulomb. Note that the energy is potential energy and energy is a scalar quantity.

The physical interpretation of the above derivations is important. There are two related fields associated with the charge q, a force field, with field strength E, directed along any radius vector r and an electrical potential field, V. The electric field and the potential field are orthogonal, that is they are always a right angles to each other. The surface S1 is an equipotential surface, no matter where we are on the surface the potential is the same. If we move to another surface, S2 say, the potential changes but the potential on S2 is everywhere the same. Equation 3.3 defines the potential difference between the surface S1 and S2.

We can move a test charge over any equipotential surface without expending energy- no force is required to move the charge over the surface because it is always traveling at right angles to the direction of the force along the radius vector. The gravitational field surrounding the earth also obeys an inverse square law and it is possible to develop analogous equations to 3.2 and 3.3, for the gravitational field. We are all familiar with gravitational equipotential surfaces when we walk along a level surface- because the surface is level we are doing no work against the gravitational field and walking is easy. As soon as we walk uphill, that is up a gradient, we are doing work against the field and we expend energy, walking is difficult. Walking downhill , that is down a gradient, produces the opposite affect because we are gaining energy from the field.

Referring again to the fields produced by the charge q, we should be clear that the fields extend throughout space. There really is a potential difference, measured in volts, between any two equipotential surfaces. This is important when we consider EMC: fields from external charged surfaces can penetrate a product with possible detrimental affects.

We shall now consider some other important aspects of the electrostatic field. We can re-write equation 3.3 as:-

 

$V = \frac{W} {{q_t }} = \frac{q} {{4\pi r^2 }}\left[ {\frac{1} {{r_2 }} - \frac{1} {{r_1 }}} \right]$ .................................3.4

V21 is the potential difference between surface 2 and surface 1. V2 and V1 are the absolute potentials of surface 2 and 1, that is the potentials of the surfaces with respect to a surface at infinity. In engineering we are always concerned with potential differences, that is the voltage at one point relative to another.

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3.3 Electrical Field- Voltage Gradient

Now assume that the difference between the two surfaces S2 and S1 is small and equal to r. We can calculate the gradient of the electrical potential field as follows:-

$\begin{gathered} \Delta V = E\Delta r \hfill \\ \frac{{\Delta V}} {{\Delta r}} = E \hfill \\ \frac{{\Delta V}} {{\Delta r}}{\text{ is called the gradient of the potential field}} \hfill \\ \end{gathered} $

It is important to note that the electrical field E has the dimensions of volts/metre as well as newtons/coulomb. We shall frequently use volts/metre as the units of the electric field in this module. The insulation strength of material is specified in volts/metre.

The gradient we have developed above is along the radius vector and this is the maximum value it can have. The gradient of the electrical potential field is a vector quantity.

In vector notation we may write the gradient as:-

$E = - \mathop r\limits^ \wedge \frac{{\partial V}} {{\partial r}}$ ...................................3.5

The negative sign indicates that in moving our test charge against the field results in positive work.

We can express 3.5 in rectangular, or Cartesian co-ordinates, as:-

$\begin{gathered} {\mathbf{E}} = - \left( {{\mathbf{x}}\frac{{\partial V}} {{\partial x}} + {\mathbf{y}}\frac{{\partial V}} {{\partial y}} + {\mathbf{z}}\frac{{\partial V}} {{\partial z}}} \right) \hfill \\ {\mathbf{E}} = - grad(V) \hfill \\ {\mathbf{E}} = - \nabla V \hfill \\ \hfill \\ \end{gathered} $

For a more detailed development of the equations for grad go to topic 0001.

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3.4 Electric Flux

We shall now introduce the concept of electrical flux and use it to develop an equation for the capacitance of a simple electrode structure.

Figure 3.3 Electrical Flux Crossing an Elemental Area dS

Fig 3_3 Electrical Flux Crossing an Elemental Area dS

We imagine the charge q to be surrounded by an electrical flux and the electrical flux density D is related to the field strength E:-

$D = \varepsilon _0 E$ .............................................3.6

The amount of flux crossing the small area dS in Fig.3.3, is:-

$d\psi = {\mathbf{D}}.d{\mathbf{S}}$ ...........................................3.7

The total flux crossing the surface of a sphere, at radius r centered on q, is:-

\[ \begin{gathered} \psi = \oint\limits_S {{\mathbf{D}}.d{\mathbf{S}}} \hfill \\ {\text{Performing this integral over a spherical surface gives}} \hfill \\ \hfill \\ \psi = q \hfill \\ \end{gathered} \]

So far the concepts we have introduced are familiar, once we have accepted that the fundamental electric quantity is charge. Charge is surrounded by a force field and an energy field and we are familiar with both force and energy. So what is electric flux? The flux of a vector was introduced in fluid mechanics where D would represent the surface density of fluid crossing the surface dS. The the integral then yields the total mass of fluid crossing the area of a sphere of radius r. In the case of the electrostatic field, the field is static, but we can still use the concept of flux and flux density- it enables us, for example, to relate aspects of field theory to aspects of circuit theory.

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3.5 The Parallel Plate Capacitor

We shall now use the concept of electrical flux to derive the equation for a parallel plate capacitor. We shall assume the plates are separated by a vacuum, as shown in Fig 3.4

Figure 3.4 A Parallel Plate Capacitor

Fig 3_4 A Parallel Plate Capacitor

In Fig 3.4, we consider 2 plates of conducting material each of surface area A, separated by a distance d. The medium between the plates is "free space". A voltage V is applied across the plates as shown.

When the voltage is applied to the plates, the top plate becomes positively charged and the negative plate becomes negatively charged. What happens is that the free electrons in the top plate are attracted to the positive terminal of the battery and driven, by chemical action, to the bottom plate. The top plate becomes positively charged and the bottom plate becomes negatively charged.

The voltage across the plates is V and the electric field between the plates is E = V/d. The electric flux density in the region between the plates is D.

Note we have assumed the plates are relatively close together and the field E is uniform.

We can now calculate the capacitance of the plates as follows:-

$\begin{gathered} D = \varepsilon _0 E \hfill \\ {\text{D = }}\varepsilon _0 \frac{V} {d} \hfill \\ {\text{D = }}\frac{{{\text{d}}\psi }} {{{\text{dA}}}} = \frac{\psi } {A}{\text{, assuming the flux density is constant between the plates}} \hfill \\ \frac{\psi } {{\text{A}}} = \varepsilon _0 \frac{V} {d} \hfill \\ \psi = \varepsilon _0 \frac{{VA}} {d} \hfill \\ {\text{But }}\psi {\text{ = the total flux between the plates = Q}} \hfill \\ {\text{Q = }}\varepsilon _0 \frac{{VA}} {d} \hfill \\ Q = CV{\text{, where C = }}\varepsilon _0 \frac{A} {d},{\text{ }} \hfill \\ {\text{C is called the capacitance of the plates and has the dimensions of farads}} \hfill \\ \end{gathered} $

In the development of the equation for the parallel plate capacitor we have assumed that the plates are close together and that the electrical field between the plates is constant. We have also ignored the field associated with the rest of the circuit- that is the battery and the wires connecting the battery to the plates. In reality the field at the edge of the plates spreads out- a phenomena called fringing and a field is associated with the wires and the battery.

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3.6 Fields Associated with a Parallel Plate Capacitor

The field plot of the of the parallel plate capacitor is shown in Fig 3.5.

Figure 3.5 The Field Plot of an Idealised Parallel Plate Capacitor



The idealised and realistic field plots of a parallel plate capacitor are shown in Fig 3.5. Actually the field plots are for a section through the plates of the capacitor, in the x-y plane and it has been assumed that the field patterns are the same in the z direction, that is into the page.

For the idealised case, lines of electric intensity are vertical and equipotential lines are horizontal. The horizontal equipotential lines are equally spaced, indicating that the voltage gradient is constant. The vertical field lines are also equally spaced indicating that the electric flux density is constant. The cells marked 1,2,3 and 4 are the edge view of a rectangular tube and the flux density through the whole tube is constant. If we wanted a more detailed plot of the fields we would draw more horizontal and vertical lines- that is the cells would become smaller.

For the more realistic field plot that shows fringing, the tubes become distorted at the edges of the capacitor plates. Where the cells become larger, field strengths and flux densities become weaker. We can think of cells as curvilinear squares that posses curved sides but all intersections are at right angles.

Note that for the parallel plate capacitor lines of force originate on positive charges on the top plate and terminate on negative charges on the bottom plate. Lines of force always originate and terminate on charges. When we consider a charge in isolation we are assuming that the lines of force associated with it terminate on charges at an infinite distance from the charge. Note also that charge density has the dimensions of coulombs/square metre. The parallel plate is an example of a surface charge and the surface charge density is expressed as coulombs/square metre. We may consider a charged wire as a line charge.

The charge on the the plates of the capacitor really does reside on their surfaces- it is not possible for static charge to exists for any length of time inside a conductor. We should remember that like charges repel and any static free charge inside a conductor is forced outwards onto the surfaces of the plates. The thickness of the charge layer approaches the diameter of a single atom, so we can safely assume it is a surface charge.

We shall consider field plotting in more detail in the exercises at the end of this unit, and throughout the module.

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3.7 Relative Permittivity

So far we have considered charges and conductors in free space. If we insert an insulating material between the plates of the capacitor the electric field acts on the electrons and protons in the atoms (or molecules) that make up the material and cause them to be slightly displaced from their normal positions. This produces millions of atomic dipoles in the material all aligned with the electric field. The effect is to increase the electric flux in the material.

The effect is illustrated in Fig 3.6:-

Figure 3.6 Illustrating Electric Flux Enhancement in an Insulator



The increase in electric flux density is measured relative to the flux density produced in free space by the same field. The relative permittivity is used to define the increase:-

$\begin{gathered} D = \varepsilon E \hfill \\ \varepsilon = \varepsilon _0 \varepsilon _r \hfill \\ \varepsilon _r = {\text{ the relative permittivity of the material}} \hfill \\ \end{gathered} $

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3.8 Divergence

So to convert all the equations we have developed so far to include an insulating material, we multiply the permittivity of free space by the relative permittivity of the material.

Finally, in our development of electrostatics we shall introduce the concept of divergence. We have already indicated that the total flux crossing the spherical surface surrounding a charge is equal to the charge. This may be generalised as follows:-

$\begin{gathered} \mathop{{\int\!\!\!\!\!\int}\mkern-21mu \bigcirc} {{\mathbf{D}}.d{\mathbf{S}} = q} \hfill \\ {\text{Now assume that the charge q is distributed in}} \hfill \\ {\text{a volume dV with a volume distribution of }}\rho \hfill \\ {\text{Then }}q = \mathop{{\int\!\!\!\!\!\int\!\!\!\!\!\int}\mkern-31.2mu \bigodot} {\rho dV{\text{ = }}} \mathop{{\int\!\!\!\!\!\int}\mkern-21mu \bigcirc} {{\mathbf{D}}.d{\mathbf{S}} = q} \hfill \\ {\text{In general terms, the total electric flux crossing a closed surface surrounding a charged region}} \hfill \\ {\text{ = the charge enclosed}} \hfill \\ \end{gathered} $

Now consider a region in an electrostatic field as shown in Fig 3.7. We are going to calculate the net electric flux leaving a small cube in the field.

 

Figure 3.7 Illustrating Divergence of Electrical Flux

Fig 3_7 Illustrating Divergence of Electrical Flux

\[ \begin{gathered} {\text{The flux entering the left hand face of the cube is: - }} \hfill \\ D_x dydz \hfill \\ {\text{The flux leaving the right hand face of the cube is: - }} \hfill \\ \left( {D_x + \frac{{\partial D_x }} {{\partial x}}dx} \right)dydz \hfill \\ {\text{The net flux leaving the cube in the }}x{\text{ diection is: - }} \hfill \\ {\text{d}}\psi _x {\text{ = }}\left( {D_x + \frac{{\partial D_x }} {{\partial x}}dx} \right)dydz - {\text{D}}_{\text{x}} dydz \hfill \\ {\text{The total flux leaving the small volume is: - }} \hfill \\ d\psi = \left( {\frac{{\partial D_x }} {{\partial x}} + \frac{{\partial D_y }} {{\partial x}} + \frac{{\partial D_z }} {{\partial x}}} \right)dv \hfill \\ \frac{{d\psi }} {{dv}} = \left( {\frac{{\partial D_x }} {{\partial x}} + \frac{{\partial D_y }} {{\partial x}} + \frac{{\partial D_z }} {{\partial x}}} \right) \hfill \\ {\text{The term }}\frac{{d\psi }} {{dv}} = {\text{ the charge density }}\rho {\text{ inside the cube}} \hfill \\ \left( {\frac{{\partial D_x }} {{\partial x}} + \frac{{\partial D_y }} {{\partial x}} + \frac{{\partial D_z }} {{\partial x}}} \right) = \rho \hfill \\ {\text{The term }}\left( {\frac{{\partial D_x }} {{\partial x}} + \frac{{\partial D_y }} {{\partial x}} + \frac{{\partial D_z }} {{\partial x}}} \right){\text{ is called the divergence of D or just }}div{\mathbf{D}} \hfill \\ \hfill \\ \end{gathered} \]

For a more detailed development of the equations for div go to topic 0002

Note that divD is a scalar quantity. We can express the divergence in a form that is particularly suited for computer simulations, as:-

\[ \begin{gathered} {\text{Note that }}D_x = \varepsilon E = \varepsilon \frac{{\partial V_x }} {{\partial x}} \hfill \\ {\text{So }}\frac{{\partial D_x }} {{\partial x}} = \frac{\partial } {{\partial x}}\left( {\varepsilon \frac{{\partial V_x }} {{\partial x}}} \right) = \varepsilon \frac{{\partial ^2 V_x }} {{\partial x^2 }} \hfill \\ {\text{The same process can be repeated for the other terms in the div equation: - }} \hfill \\ div{\mathbf{D}} = \varepsilon \frac{{\partial ^2 V_x }} {{\partial x^2 }} + \varepsilon \frac{{\partial ^2 V_y }} {{\partial y^2 }} + \varepsilon \frac{{\partial ^2 V_z }} {{\partial z^2 }} \hfill \\ {\text{Or}} \hfill \\ \rho {\text{ = }}\varepsilon \left( {\frac{{\partial ^2 V_x }} {{\partial x^2 }} + \frac{{\partial ^2 V_y }} {{\partial y^2 }} + \frac{{\partial ^2 V_z }} {{\partial z^2 }}} \right) \hfill \\ \hfill \\ {\text{If }}\rho \ne {\text{0, the equation is called Poissions equation}} \hfill \\ \hfill \\ {\text{If }}\rho = {\text{0, the equation is called Laplaces equation}} \hfill \\ \end{gathered} \]

Both La place's equation and Poisson's equation are used extensively by computer based simulation software packages.

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3.9 Unit Summary

In this unit we have covered the main concepts of electrostatics and developed some important equations using vector calculus.

The main concepts we have introduced in this unit are:-

We have stressed that the electrostatic field extends throughout space and that lines of electric force originate on positive charge and terminate on negative charge.

The electric field and the potential field are related to each other: equipotential surfaces are orthogonal to lines of force.

Throughout this module we shall be using field plotting software to illustrate concepts and solve problems relevant to both EMC and signal integrity. This is a practical approach because the field equations can only be solved in limited number of simple cases.

In order to use field plotting software effectively and efficiently an understanding of the concepts of electrostatics is essential.

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Problems

 

Problem 3.1

This problem requires you to obtain the capacitance/unit length of a PCB microstrip transmission line and is used as the first walk-through for the finite difference field plotting software Comsol. Click here to go to the walk-throughs.

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Updated RJH 18.07.08

 

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