MME 467: Ceramics for Advanced Applications

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1 MME 467: Ceramics for Advanced Applications Lecture 26 Dielectric Properties of Ceramic Materials 2 1. Barsoum, Fundamental Ceramics, McGraw-Hill, 2000, pp Richerson, Modern Ceramic Engineering, Dekker, 1992, pp Prof. A. K. M. B. Rashid Department of MME, BUET, Dhaka Topics to discuss...! q Dielectric strength q Dielectric loss and dielectric breakdown q Capacitance q Dielectric properties of BaTiO 3 q Problems 1

2 Dielectric Strength q Besides dielectric constant k, dielectric strength is the second most important properties of dielectric materials. q It is the capacity of materials to withstand an electric field without breaking down and allowing electric current to pass. It has units of volts per unit of thickness of dielectric material. Example: V/cm, V/mil (1 mil is thousandth of an inch) q A phenolic resin has dielectric strength of about 2000 V/mil. This means that a thin film of inch thick can block the passage of an electric current under an applied voltage gradient of 2000 V. q Materials of high dielectric strength are used as electric insulation. 2

3 Dielectric Loss q An ideal dielectric would allow no flow of electric charge, only a displacement of charge via polarization. q When an alternating electric field is applied to an ideal capacitor, the current leads the voltage by a phase angle of 90º. Under this circumstance, no power is absorbed by the dielectric and the capacitor has zero loss. q Real materials always have some loss. The phase angle between current and voltage is not exactly 90º; the current lags slightly. q The angle of lag is defined as δ and the amount of lag or the dissipation factor (a.k.a. loss tangent) becomes tan δ. q The dielectric loss is a measure of energy dissipated in the dielectric in unit time when an electric field acts on it. Loss Tangent = tan δ = k!! / k! k!! = k! tan δ k = rela,ve dielectric constant k = rela,ve loss factor q The total power loss can be expressed as ( ) E 2 f V P L = 5.56x10 11 k! tanδ E = applied electric field (V/m) f = frequency of applied field (Hz) V = volume of dielectric (m 3 ) q To reduce power loss, it is imperative that [1] Use solids that are highly insulating (σ dc è 0), i.e., use pure material with large band gap such that number of carrier charge is as low as possible. [2] Reduce k (or, δ) to reduce loss tangent. 3

4 q A dielectric loss represents [1] a wastage of energy, as well as [2] attendant heating of the dielectric. q If the rate of heat generation is faster than it can be dissipated, the dielectric will heat up, which could lead [1] dielectric breakdown [2] change in dielectric constant k (as T is increased), which can cause severe problem to finely tuned electric circuits [3] reduction in the sharpness of the resonance frequencies of tuning circuit. q Dielectric loss results from several mechanisms [1] ion migration (the most important mechanisms) [2] ion vibration and deformation [3] electronic polarization Factors Affecting Dielectric Loss [1] Temperature and Impurity Content They tend to increase conductivity of ceramics. So their effect on dielectric loss can be substantial. Varia,on of loss factor with temperature Effect of frequency and impurity content on loss factor 4

5 [2] Frequency of Applied Field Ion migration is strongly affected by temperature and frequency. The loss due to ion migration increases at low frequencies. Frequency at which a dielectric is to be used must be as far removed from a resonance frequency as possible. Because at near resonance, k can increase substantially. [3] Crystal Structure and Crystallinity Crystal structure has an effect on k and loss tangent. For close-packed ionic solids, dielectric loss is quite small, while for loosely packed structures, dielectric loss is higher. For example, for α-alumina (densely packed), loss tangent, tan δ < For γ-alumina (loosely packed), loss tangent, tan δ >

6 Dielectric Breakdown q When a dielectric is subjected to an ever-increasing electric field, at some point a short circuit develops across it. q Dielectric breakdown is defined as the voltage gradient or electric field sufficient to cause the short circuit, or electricity to pass. q Depends on many factors: [1] sample thickness [2] temperature [3] electrode composition and shape [4] porosity in structure q Two basic types of dielectric breakdown: [1] Intrinsic v Electrons in the conduction band are accelerated to such points that they start to ionize lattice ions v As more ions are ionized, number of free electron increases, and an avalanche effect is created v Higher the electric field, faster the electrons will be accelerated, and the more likely the breakdown will occur. [2] Thermal breakdown v Rate of heat generation is higher than the rate of heat dissipation v Dielectric will be heated up, conductivity increases, causing further heating, and eventually thermal breakdown 6

Capacitance q A capacitor is an electrical device used to store charge which is received from a circuit in a dielectric material between two conductor plates received from a circuit.

7 Capacitance q A capacitor is an electrical device used to store charge which is received from a circuit in a dielectric material between two conductor plates received from a circuit. q The dielectric material [1] must easily be polarizable (should contain high dielectric constant k ) [2] should have high electrical resistivity (so that charge cannot pass through from one conductor to the other) [3] must have high dielectric strength (so that it can operate at high voltage and yet made as small as possible) Single layered capacitor Mul2- layered capacitor q To improve capacitor performance, the layer of dielectric or the number of plates is increased. q For capacitor containing n parallel plates, the capacitance C = k! ε 0 ( n 1) A d 7

Dielectric Properties of BaTiO 3 q Results from its crystal structure (Perovskite) q The unstable Ti atom in the octagonal position has 6 minimum-energy positions off-centre along six oxygen atoms q

8 Dielectric Properties of BaTiO 3 q Results from its crystal structure (Perovskite) q The unstable Ti atom in the octagonal position has 6 minimum-energy positions off-centre along six oxygen atoms q Ti atom positions randomly in one of these 6 positions, resulting spontaneous polarization q Since Ti has 4+ charge, degree of polarization is very high. q When an electric field is applied, Ti ions shift from random to aligned positions, resulting high bulk polarization and high k q Temperature has strong effect on polarization as it changes the crystal structure of BaTiO 3. Curie Temperature Different polymorphs of BaTiO 3 and accompanying changes in la>ce constants and dielectric constants. 8

9 q Grain size of BaTiO 3 has strong influence on k.! q Finer grains produce higher dielectric constant [1] due to the presence of internal stress [2] due to the increase in the number of domain walls q Larger grains show extremely high k at Curie point [1] due to the formation of multiple domain in a single grain q Effect of doping v Depends upon size of A and B substitution (ABO 3 ) v Substitution of Ba with smaller ion (Ca/Mg) reduces the sensitivity to temperature by broadening k T curve v Substitution of Ba by Sr and Zr reduces Curie point, while substitution of Ba by Pb increases it. 9

10 Problems 1. An average displacement of electrons rela,ve to nucleus in a copper atom is 1x10 8 Å when an electric field is imposed on a copper plate. Calculate the polariza,on. Polarization, P = Nqδ M Cu = 29 (atomic number) Number of electrons in each copper atom = 29 Lattice parameter of copper = Å No. of dipole formed ( 4 atoms / cell) x 29 electrons / atom N = x10 10 ( ) ( ) 3 m 3 / cell = 2.46x10 30 electrons / m 3! P = 2.46x10 30 electrons $! # & 1.16x10 19 C $ # & 1x10 8 x10 10 m " m 3 %" electron % P = 3.94x10 7 C / m 2 ( ) 2. The ionic polariza,on observed in NaCl crystal is 4.3x10 8 C/m 2. Calculate the displacement between Na + and Cl ions. Here, there is one electric charge on each Na+ ion. In NaCl unit cell, a = 5.5 Å, and there are 4 Na+ ions. ( ) x ( 1 charge/na + ion) ( 5.5x10 10 ) 3 m 3 / cell N = 4 Na+ ions / cell δ = P Nq = 4.3x10 8 C/m 3 2.4x10 28 charge/m 3 ( )( 1.6x10 19 C/charge) δ =11.2x10 18 m =11.2x10 8 Å = 2.4x10 28 charge/m 3 10

11 3. Suppose NaCl has a polariza,on of 4.3x10 8 C/m 2 in an electric field of 1000 V/m. Calculate the dielectric constant for NaCl. k! 1= P ε 0 E k! 1= k! = x10 8 C/m 2 ( 8.85x10 12 F/m) 1000 V/m ( ) = Calculate the maximum polariza,on per cubic cen,meter and the maximum charge that can be stored per square cen,meter for barium,tanate. (a) The oxygen ions are at face centers, Ba +2 ions are at cube corners and Ti +4 is at cube center in cubic BaTi0 3. (b) In tetragonal BaTi0 3,the Ti +4 is off- center and the unit cell has a net polariza,on. 11

12 In BaTiO 3, the separations are the distances that the Ti 4+ and O 2- ions are displaced from the normal lattice points. The charge on each ion is the product of q and the number of excess or missing electrons. Thus, the dipole moments are: Each oxygen ion is shared with another unit cell, so the total dipole moment in the unit cell is: The polarization per cubic centimeter is: The total charge on a BaTiO 3 crystal 1 cm 1 cm is: 12

13 5. We want to make a simple parallel plate capacitor that can store 4x10 5 C at a poten,al of V. The separa,on between the plates is to be 0.2 mm. Calculate the area of the plates required if the dielectric is (a) a vacuum, (b) polyethylene, (c) water, and (d) BaTiO 3. C = Q /V = 4x10 5 C V = 4x10 9 F C = εa d = k! ε 0A d ( )( 0.2x10 3 m) ( ) A = Cd = 4x10 9 F k! ε 0 k! 8.85x10 12 F/m = 0.09 k! (a) For vacuum, k = 1: A = 0.09 m 2 (b) For polyethylene, k = 2.26: A = 0.04 m 2 (c) For water, k = 78.3: A = 1.15x10 3 m 2 (d) For BaTiO 3, k = 3000: A = 3x10 5 m 2 m 2 6. A mica capacitor 250 mm 2 area and 2.5 mm thick is to have a capacitance of mf. (a) How many plates are needed? (b) What is the maximum allowable voltage? For mica, k = 7, and dielectric strength = 40x10 6 V/m. (a) For multilayer capacitor ( ) A d C = k! ε 0 n 1 ( n 1) = Cd k! ε 0 A = x10 6 F x10 12 F/m ( )( 2.5x10 6 m) ( )( 250x10 6 m 2 ) = 4 n = 5 (Five plates with four dielectric layers) (b) Since E = V/d V = (40x10 6 V/m) (2.5x10 6 m) = 100 V 13

14 Next Class Lecture 27 Superconductivity 14

Chap. 7. Dielectric Materials and Insulation

Chap. 7. Dielectric Materials and Insulation - The parallel plate capacitor with free space as an insulator: - The electric dipole moment for a pair of opposite changes +Q and -Q separated by a finite