Chapter 39: Dielectrics and Dielectric Loss [15898]
39.1 Electric Field, Capacitance and Permittivity
In AC circuits with capacitors, loss occurs when the direction of charging and discharging changes, resulting in heat [15899][15901][15902]. The study of dielectrics is concerned with the storage of electrical and magnetic energy [15914][15915]. This chapter covers loss, loss angle, and dissipation factors [15916][15917].
The fundamental relationships in an electric field for a capacitor are given by:
Where D is the electric flux density. The relationship between flux density and electric field strength E is [15923][15926]:
The insulating material separating capacitor plates is the dielectric. Compared with conductors, dielectric materials have very high resistance [15934][15935][15936]. Because of this, they are used to separate conductors which are at different potentials, such as in overhead lines or parallel plate capacitors [15939][15940][15942].
39.2 Polarization
In an insulator, the electrons cannot freely move [15953][15957]. When an insulator is put between two plates, a slight separation of the oppositely charged bodies in the nucleus and the electrons results in a dipole [15958][15960]. This insulator dipole acts in a direction against the electric field that causes it, resulting in a reduction in voltage (V) and an increase in capacitance [15961][15962].
39.3 Dielectric Strength
Dielectric strength is the maximum applied electric field strength a dielectric can withstand without melting or breaking down [16011][16021]. Leakage current occurs through the dielectric separating the two plates [16022][16024]. The resistance R relates to the thickness and area of the capacitor plates [16026][16034].
A measure of quality for the capacitor is related to the product of C and R [16041][16043]. Reducing the thickness of a dielectric film increases the capacitance, but it also reduces the voltage the capacitor can withstand before breaking down, since E = V/d [16054][16056][16059].
39.4 Thermal Effects and Mechanical Properties
The insulation resistance of most dielectrics falls with an increase in heat [16098]. This causes the leakage current to increase, generating further heat. If the heat is not dissipated faster than generated, this will cause deterioration of essential properties [16099][16104][16108].
Mechanical properties include tensile strength, transverse strength, shearing, and compressive strength [16121][16122]. It is often necessary to consider factors such as compressibility, deformation under bending stresses, impact strength, and extensibility [16124][16125].
39.6 Types of Practical Capacitors
Capacitors are designated by the material used for the dielectric [16128][16129]. Main types include paper, plastic, ceramic, mica, and electrolytic, as well as super capacitors [16132][16133].
39.8 Dielectric Loss and Loss Angle
In capacitors with solid dielectrics, losses are caused by two main things [16163][16168]:
- Dielectric Hysteresis: A loss analogous to magnetic hysteresis loss. Electrical energy is converted into heat as a result of the reversal of electrostatic stress in a dielectric subjected to an alternating field [16169][16172].
- Leakage: Currents may flow through the dielectric or along paths between the terminals [16175][16176].
Series Representation
The total dielectric loss may be represented by an additional resistance connected in series with an ideal capacitor [16179][16181]. From the phasor diagram, the loss angle δ is derived [16192][16193]:
Dissipation Factor (D)
This is defined as an indication of the quality of the dielectric [16215][16222]:
Parallel Representation
For there to be equivalence between series and parallel representations [16233][16234]:
Dielectric power loss is given by [16244]:
Problems
Problem 1: A 1.5 Ω resistance is in series with a 400 pF capacitor. f = 8 kHz [16248][16251].
δ = tan-1(Rs × ω × Cs) = tan-1(1.5 × 16π × 103 × 400 × 10-12) [16256]
δ = 1.727° [16257]
Problem 2: Given δ = 1.433°, f = 50 Hz, V = 5 kV, Power = 20 W [16271][16272].
20 W = (5 × 103)2 × 100π × C × tan(1.433°) [16279]
C = 1.0179 × 10-7 F [16281]
Problem 3: C = 2000 pF, V = 20 V, Power loss = 500 μW, f = 10 kHz [16296][16298][16309].
500 × 10-6 = 202 × 2π × 104 × 2000 × 10-12 × tan δ [16310]
tan δ = 0.57 [16315]
Rs = tan δ / (ω Cs) [16305]
Rp = 1 / (tan δ × ω Cp) [16318]