2 IntroductionQuestion: What physical parameters determine how much inductance a conductor or component will have in a circuit?Answer: It all depends on current and flux linkages!
3 Flux Linkage Definition: Mathematical Expression: the magnetic flux generated by a current that passes through one or more conducting loops of its own or another separate circuitMathematical Expression:
4 Types of Inductance Self-Inductance (L): whenever the flux linkage of a conductor or circuit couples with itselfMutual Inductance (M):if the flux linkage of a conductor or circuit couples with another separate one
5 Self-Inductance Formula by Definition Applies to linear magnetic materials onlyUnits:
6 Inductance of Coaxial Cable Magnetic FluxInductance(as commonly used in transmission line theory)
7 Inductance of Toroid Magnetic Flux Density Magnetic Flux If core small vs. toroid
8 Inductance of Toroid Inductance Result assumes that no flux escapes through gaps in the windings (actual L may be less)In practice, empirical formulas are often used to adjust the basic formula for factors such as winding (density) and pitch (angle) of the wiring around the core
9 Alternative Approaches Self-inductance in terms ofEnergyVector magnetic potential (A)Estimate by Curvilinear Square Field Map method
10 Inductance of a Long Straight Solenoid Energy ApproachInductance
11 Internal Inductance of a Long Straight Wire Significance: an especially important issue for HF circuits sinceEnergy approach (for wire of radius a)
12 Internal Inductance of a Long Straight Wire Expressing Inductance in terms of energyNote: this result for a straight piece of wire implies an important rule of thumb for HF discrete component circuit design:“keep all lead lengths as short as possible”
13 Example of Calculating Self-Inductance Exercise 1 (D9.12, Hayt & Buck, 7th edition, p. 298)Find: the self-inductance ofa) a 3.5 m length of coax cable with a = 0.8 mm and b = 4 mm, filled with a material for whichr = 50.
14 Example of Calculating Self-Inductance Exercise 1 (continued)Find: the self-inductance ofb) a toroidal coil of 500 turns, wound on a fiberglass form having a 2.5 x 2.5 cm square cross section and an inner radius of 2.0 cm
15 Example of Calculating Self-Inductance Exercise 1 (continued)Find: the self-inductance ofc) a solenoid having a length of 50 cm and 500 turns about a cylindrical core of 2.0 cm radius in which r = 50 for 0 < < 0.5 cm and r = 1 for 0.5 < < 2.0 cm
16 Example of Estimating Inductance: Structure with Irregular Geometry Exercise 2: Approximate the inductance per unit length of the irregular coax by the curvilinear square method
17 Mutual InductanceSignificant when current in one conductor produces a flux that links through the path of a 2nd separate one and vice versaDefined in terms of magnetic flux (m)
18 Mutual Inductance Expressed in terms of energy Thus, mutual inductances between conductors are reciprocal
19 Example of Calculating Mutual Inductance Exercise 3 (D9.12, Hayt & Buck, 7/e, p. 298)Given: 2 coaxial solenoids, each l = 50 cm long1st: dia. D1= 2 cm, N1=1500 turns, core r=752nd: dia. D2=3 cm, N2=1200 turns, outside 1stFind: a) L1=? for the inner solenoid
20 Example of Calculating Mutual Inductance Exercise 3 (continued)Find: b) L2 = ? for the outer solenoidNote: this solenoid has inner core and outer air filled regions as in Exercise 1 part c), so it may be treated the same way!
21 Example of Calculating Mutual Inductance Exercise 3 (continued)Find: M = ? between the two solenoids
22 SummaryInductance results from magnetic flux (m) generated by electric current in a conductorSelf-inductance (L) occurs if it links with itselfMutual inductance (M) occurs if it links with another separate conductorThe amount of inductance depends onHow much magnetic flux linksHow many loops the flux passes throughThe amount of current that generated the flux
23 Summary Inductance formulas may be derived from Direct application of the definitionEnergy approachVector Potential MethodThe self-inductance of some common structures with sufficient symmetry have an analytical resultCoaxial cableLong straight solenoidToroidInternal Inductance of a long straight wire
24 Summary Numerical inductance may be evaluated by Calculation by an analytical formula if sufficient information is known about electric current, dimensions and permeability of materialApproximation based on a curvilinear square method if axial symmetry exists (uniform cross section) and a magnetic field map is drawn
25 ReferencesHayt & Buck, Engineering Electromagnetics, 7/e, McGraw Hill: Bangkok, 2006.Kraus & Fleisch, Electromagnetics with Applications, 5/e, McGraw Hill: Bangkok, 1999.Wentworth, Fundamentals of Electromagnetics with Engineering Applications, John Wiley & Sons, 2005.