Understanding Capacitors in Electrical Engineering

1. Lecture 7 Capacitors and dielectric permittivity Capacitance of parallel-plate capacitor Capacitance per meter of coaxial cable Magnetic permeability Inductance Inductance of Solenoid Inductance of Toroid Mutual Inductance Coefficient of Coupling ECE342 Lecture 7

2. Capacitor • Formed whenever two conductors are separated by an insulator (dielectric) material, such as plastic, mica, or ceramic. • Capacitor stores electric charge. • +Q charge is stored on one conductor. • -Q charge is stored on the other conductor. • Charge stored “Q” is proportional to the voltage impressed across the capacitor’s conductors. ECE342 Lecture 7

3. Capacitance is a function of the geometries of the conductors and the permittivity “ε = ε0εR” of the dielectric. ε0 = permittivity of free space (or air) = 8.854 X 10-12units of C2/N*m2 = Farads/meterεR = relative permittivity of the medium (> 1.0)It indicates the ability of the material to polarize, where the positive and negative charge centers within its atoms separate slightly, under the influence of an external E field. This sets up an opposing internal field that acts to make the overall E field in the material less. EEXT Polarized Atom of dielectric - + EINT ECE342 Lecture 7

4. Definition of Capacitance of a Capacitor The capacitance of a capacitor indicates its ability to store charge. Definition of capacitance: If a voltage source of strength “V” is placed across a capacitor, a positive charge “Q” will be forced onto one conductor, and at the same time, an equal amount of negative charge “-Q” will necessarily be forced onto the other conductor). Then C = Q / Vwith units of Coulombs / Volt = Farads (F) (Note: capacitance is “Charge per unit volt”) ECE342 Lecture 7

5. Parallel-Plate Capacitor Each conducting plate is of cross sectional area “A” (Transient) Charging current i(t) i(t) +Q + + + + + + + + V - d E field - - - - - - - - -Q z y i(t) (Transient) Charging current i(t) x C = Q / V ECE342 Lecture 7

6. (Since E field remains vertical and approximately constant as one moves away from a horizontal sheet of uniform charge density) Thus E = V / d, and the electric field between the plates (E) will be proportional to the voltage across the plates. Since Q = CV, and E is also proportional to the charge “Q” stored on the plates. ECE342 Lecture 7

7. Intuitive observations • For a constant voltage “V” across the capacitor plates, the magnitude of the charge on either plate “Q” doubles as the separation distance “d” is halved, since the electric field between the plates E = V / d becomes twice as strong, attracting twice as many charges onto the plates. • For a constant voltage across the capacitor plates, Q increases as the plate area “A” increases, since there is physically more room for the charges to spread out on the plate. ECE342 Lecture 7

8. Capacitor current-voltage relationship • C = Q / V (by definition) • Q = C V (Solve for Q) • The current “i(t)” flowing into the top conductor of the capacitor or away from the bottom conductor of the capacitor is the rate at which charge is transported onto the conductor and hence is the rate at which the stored charge “Q” changes: • i(t) = dQ/dt = d(CV)/dt Assuming C is a constant, i(t) = C dV/dt i(t) C + v(t) - ECE342 Lecture 7

9. Relative Permittivity of Dielectric • Recall what happens when some material other than a vacuum with εR > 1.0 is placed between the capacitor plates. • The net E field between the plates is diminished, since the material polarizes (atomic charge centers separate) to some extent. • Imagine a parallel-plate capacitor with an air dielectric is charged to a specific voltage. Then with the voltage source disconnected, a polystyrene dielectric (with εR = 2.5) is slid between the plates, with d held constant. This makes the voltage across the capacitor, V = E / d, decrease, since E becomes 2.5 times smaller. Because C = Q/V, the capacitance of the parallel-plate capacitor, must increase by a factor of 2.5! ECE342 Lecture 7

10. Capacitance of a parallel-plate capacitor • We have shown, from Gauss Law, as well as from elementary application of Coulomb’s Law, that the E field due to an infinite sheet of uniform surface charge density ρso is given by E = ρso / (2ε)inWhere inis the unit vector normal to the surface of the sheet of charge. • Thus if the parallel plates of the capacitor carry equal and opposite charges Q and –Q, then the E field set up by the two plates reinforces in the region between the plates, and the E field largely cancels elsewhere if the separation distance d is considerably less than the length or width of either plate. d ρso = -Q/A Ex ρso = +Q/A Where A is the plate area In between the plates, E = 2(ρso / (2ε))ix = (ρso/ε)ix = Q/(Aε)ix x ECE342 Lecture 7

11. Minor E field fringing, d << W,L. => E ≈ Ex ix d Severe E field fringing, since d is not << W,L E≠ Ex ix d ECE342 Lecture 7

12. Capacitance of Coaxial Cable • A coaxial cable consists of an inner conductor and an outer concentric cylindrical conductor (the shield), with an insulating dielectric (air or polystyrene) in between. • Let us calculate the capacitance of an L-meter length of coaxial cable. • A voltage V is placed across the inner and outer conductors, depositing charge “Q” on the inner conductor and “-Q” on the inner conductor. • A closed concentric Gaussian cylindrical surface (pill bottle indicated by the dotted line) is positioned as shown (r = constant), and Gauss’ law is applied around its surface. Because Er is a constant along the Gaussian cylinder wall, and is tangential to the front and rear “end caps” of the cylinder, • Gauss’ Law => The electric displacement flux emanating from the pillbox = εEr*(2πrL) = Q => Er = Q / (2πεrL) for a < r < b Let the z axis be directed into the paper and located at the center of the circles. Then by symmetrythe electric field has only a radial outward component: E = Er(r)ir ix b r iy a iz Q -Q ECE342 Lecture 7

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14. Magnetic Permeability • A material’s magnetic permeabilityμ = μRμ0is a measure of • The existence of magnetic dipoles (loops of spinning and orbiting charges) inside the atoms of the material. • How easy it is for these magnetic dipoles to re-orient themselves along the direction of an externally applied magnetic field. • A nonmagnetic material (such as air, polystyrene, glass) has no available, moveable magnetic dipoles. For these materials μR = 1.0Thus in non-magnetic materials, such as air B = μ0H, where μ0 = 4π*10-7 with units of Wb/(A*m) = H/m • Due to random thermal vibration, a magnetic material (a material with μR > 1.0) will normally (with no magnetic field present) have its microscopic magnetic dipoles randomly oriented in all directions, so there is no net macroscopic magnetic field set up in the material. • But if an external magnetic field intensity “H” is applied across such a magnetic material, it will cause the randomly oriented dipoles to “line up” to some extent, producing an enhanced magnetic flux field B in that material. (Think of “H” as the stimulus and “B” as the response, or resulting flow field.) B = μR μ0H where μR > 1.0 ECE342 Lecture 7

15. Current Carrying Loops Are Magnetic Dipoles: B B i(t) The magnetic flux flows upward through the center of the loop, then it curls back downward outside of the loop ECE342 Lecture 7

16. Ferromagnetic Materials In ferromagnetic materials (Fe, Co, Ni, Gd (Gadolinium), and Dy (Dysprosium) and their alloys) a special effect called exchange coupling occurs between groups of neighboring atoms (domains), that causes all of their magnetic dipole moments to line up in parallel and point in a single direction. The domains in a ferromagnetic material can be seen under a microscope using a colloidal suspension of finely powered iron oxide. With no magnetic field present, each domain points in a random direction due to thermal agitation of the molecules, so there is no net magnetic field in the material. When an H field is applied, the domain boundaries (walls) shift, allowing the domains pointing along the direction of the applied H field to get bigger, while the domains pointing in other directions shift. This results in a VERY BIG enhancement in the resulting magnetic flux (B), or flow of the magnetic field through a ferromagnetic material, thus μR can be quite large (on the order 100 – 300) for ferromagnetic materials. ECE342 Lecture 7

17. Ferromagnetic domains in an unmagnetized sample of iron with no applied H field With no applied magnetic field, the net macroscopic magnetic field = 0, since the microscopic domain magnetic dipole moments cancel out. With applied magnetic field, the domain walls shift such that the domains pointing along the magnetic field get bigger, resulting in an enhancement of the magnetic flux in the region. If the applied magnetic field is strong enough, the domains can be “rotated into alignment” ECE342 Lecture 7

18. Steps in Magnetization of Ferromagnetic Material • No applied H field (unmagnetized state) • Weak applied H field (domain wall motion) • Strong applied H field (domain rotation) ECE342 Lecture 7

19. Nonlinear B vs H curve (Hysteresis Curve) of a ferromagnetic material If the applied H field is varied over a large range, we can no longer say that B = μH except over a small range of H variation, since the relationship between B and H is not linear. This is because eventually all of the domains get rotated into alignment with the applied H field, and “that’s all there is, and their aint no more!” The material is said to be “magnetically saturated.” Thus the slope of the B vs. H curve levels off at higher H values, meaning that there is less B field enhancement for higher H values. ECE342 Lecture 7

20. Magnetized Sample (Permanent Magnet) Note as H is increased from 0 in the initially unmagnetized sample (Point a) up to a maximum value at Point b, the material approaches magnetic saturation. Then, as the H field is decreased back to zero, B remains at a nonzero positive B field value (Point c). This is because the domain walls move with friction. The walls do NOT move back all the way in order to return to their original unmagnetized value. Instead, the sample retains a residual B field. The sample has been “magnetized”. The same principle allows the sample to be magnetized in reverse. (See Point f). This is the principle used in magnetic recording media, such as a PC’s hard drive read/write head! ECE342 Lecture 7

21. Principle used in digital magnetic recording media (tape / magnetic strip on credit cards / hard drive / floppy drive, etc.) READ: a left-to-right or right-to-left pointing magnetic field is picked up at air gap and transmitted through the ferrite core and pickup coil. By Faraday’s law of induction, v(t) = -d/dt(Flux cutting coil), the domain transitions pick up a positive or negative-going voltage pulse WRITE: Magnetic domains in magnetic recording medium (iron-oxide coated plastic surface) are left aligned either left-to-right (Logic 1), or right-to-left (Logic 0), depending on direction of the current pulse through read/write head coil, which results in a magnetic field across the narrow air gap ECE342 Lecture 7

22. Use of ferromagnetic materials to conduct and guide magnetic flux • In much the same way that copper wire, which has a high conductivity compared to air, is used to conduct electrical current in an electrical circuit, ferromagnetic material, which has a relatively high permeability compared to air, is used to conduct and guide magnetic field flux lines in a magnetic device such as a hard drive read/write head, toroidal inductor core, electric motor, loudspeaker, D’Arsonval meter movement, etc. ECE342 Lecture 7

23. Inductance • If two coils are near each other, a time-varying current in one coil will set up a time-varying magnetic flux that cuts the surface enclosed by the second coil, and hence by Faraday’s Law of Induction, a time- varying voltage will be induced in the second coil. This is called “transformer action”. • However, two coils are not needed to demonstrate Faraday’s Law of Induction. If a single coil carries a time-varying current, a self-induced magnetic flux will be set up that cuts the surface enclosed by this coil, which by Faraday’s Law of Induction, will set up a time-varying voltage across that coil that is given by ECE342 Lecture 7

24. We have found that the magnetic field intensity due to current flowing in a wire Hø = i(t)/2πris always proportional to the current that creates it, i(t). • Thus the resulting magnetic flux Φ must be proportional to i(t), since B = µH = µ0µRH (for a linear magnetic medium) and Φ is the integral of B·ds • Exception: ferromagnetic (nonlinear) medium, where the the B vs. H curve is nonlinear, unless i(t) varies over a suitably small range, in which case an approximate µR can still be assigned. • Let this proportionality constant be defined as the “self-inductance” of the coil, “L” where We will take this as the definition of the inductance (L) of a coil, where the units are in Weber-turns / Ampere Volt*second / Ampere = Henrys ECE342 Lecture 7

25. Inductor voltage – current relationship • Applying Faraday’s law to our defining equation for inductance, we may determine the voltage across the inductor v(t) in terms of the current through the inductor i(t). v(t) - i(t) + X B v(t) now referenced positive at the terminal where the current enters the inductor. ECE342 Lecture 7

26. Inductance of a long, thin solenoidal coil • Consider a long, thin solenoidal coil with length “l”, N turns, cross sectional area A, carrying a current i. • By definition, L = N Ф / i • From symmetry considerations, the B field inside the thin solenoid is directed along the axis of the solenoid, and is approximately uniform, as can be seen from Ampere’s circuital law, picking the Amperian contour shown below b Amperian contour normal to surface of coil, with top and bottom sides parallel to axis, and half inside and half outside the solenoidal coil. a Js=Ni/l H = Hziz z axis Length l ECE342 Lecture 7

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28. Inductance of a toroidal inductor Powdered Iron or ferrite core (μR>>1) Amperian contour of radius r b a r B Thickness “h” z iФ i(t) N Turns ECE342 Lecture 7

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30. Numerical example ECE342 Lecture 7

31. Mutual Inductance • When two coils are near each other, some of the magnetic flux created by one coil links (passes through the surface enclosed by the other coil). • The two coils are said to be “magnetically coupled” • A time-varying current in one coil will induced a time-varying voltage in the other coil by Faraday’s Law of Induction. • When the two coils are tightly coupled (with nearly every flux line produced by one coil linking the other coil as well, we have a transformer. • Mutual inductance is a measure of the degree of magnetic coupling between two coils. ECE342 Lecture 7

32. Mutual Inductance B flows out of paper inside 2nd coil i1(t) B -v2(t)+ Coil 1N1 turns Coil 2N2 turns The mutual inductance between Loop 1 and Loop 2 “M12” is defined as the constant of proportionality between the magnetic flux linking Coil 2 due to the current flowing in neighboring Coil 1. Note that Coil 2 is open-circuited. N2Ф2= M12*(i1) => M12 = N2Ф2 / i1 ECE342 Lecture 7

33. Experimental measurement of Mutual Inductance between two coils • Measure the self-inductance of each of the two inductors, L1 and L2 • Connect a sinusoidal ac voltage source across L1 • Measure the amplitude of v1(t), Vm1, and the open-circuit value of v2(t), Vm2. • From the expressions for self-inductance and mutual inductance we find that • Thus we may calculate M12 = (Vm2/Vm1)L1 • The roles of coils L1 and L2 may be reversed and the steps repeated to find M21. With the sinusoidal voltage source placed across L2, • M21 = (Vm1/Vm2)L2 • However, due to the symmetry of the situation, you should find that M12 = M21 = “M”. (There is no need to retain subscripts) ECE342 Lecture 7

34. Coefficient of Coupling Consider two magnetically coupled coils where currents are flowing in both coils (the second one is no longer open-circuited). Now the voltage across each coil has a self-induced and a mutually induced component: ECE342 Lecture 7

35. If the second coil is short-circuited (RL = 0), then v2(t) = 0, allowing us to solve for di2/dt in terms of di1/dt => Now the voltage across Coil 1 can be written purely in terms of di1/dt, And the equivalent inductance “LEQ” across Coil 1 can be found: ECE342 Lecture 7

36. But the stored energy in an inductor: can never be allowed to go negative, since that would violate the passive sign convention and allow the inductor to act as a source of electrical energy. Therefore LEQ must remain positive all times. ECE342 Lecture 7

37. Thus the maximum possible value of M is found by setting LEQ = 0 This maximum value of mutual inductance occurs when the two coils are placed in such close proximity, or linked together by a magnetic material of very high µR, such that all of the flux produced by one coil completely links the other coil as well. The coefficient of coupling “k” is defined to indicate the degree of mutual coupling: 100% magnetic coupling between two coils (M = MMAX, or k = 1) is not possible in practice, however values of k as high as 0.98 are attainable. ECE342 Lecture 7