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Electromagnetics ENGR 367

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Presentation on theme: "Electromagnetics ENGR 367"— Presentation transcript:

1 Electromagnetics ENGR 367

2 Introduction Question: 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 circuit Mathematical Expression:

4 Types of Inductance Self-Inductance (L):
whenever the flux linkage of a conductor or circuit couples with itself Mutual 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 only Units:

6 Inductance of Coaxial Cable
Magnetic Flux Inductance (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 of Energy Vector magnetic potential (A) Estimate by Curvilinear Square Field Map method

10 Inductance of a Long Straight Solenoid
Energy Approach Inductance

11 Internal Inductance of a Long Straight Wire
Significance: an especially important issue for HF circuits since Energy approach (for wire of radius a)

12 Internal Inductance of a Long Straight Wire
Expressing Inductance in terms of energy Note: 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 of a) a 3.5 m length of coax cable with a = 0.8 mm and b = 4 mm, filled with a material for which r = 50.

14 Example of Calculating Self-Inductance
Exercise 1 (continued) Find: the self-inductance of b) 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 of c) 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 Inductance Significant when current in one conductor produces a flux that links through the path of a 2nd separate one and vice versa Defined 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 long 1st: dia. D1= 2 cm, N1=1500 turns, core r=75 2nd: dia. D2=3 cm, N2=1200 turns, outside 1st Find: a) L1=? for the inner solenoid

20 Example of Calculating Mutual Inductance
Exercise 3 (continued) Find: b) L2 = ? for the outer solenoid Note: 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 Summary Inductance results from magnetic flux (m) generated by electric current in a conductor Self-inductance (L) occurs if it links with itself Mutual inductance (M) occurs if it links with another separate conductor The amount of inductance depends on How much magnetic flux links How many loops the flux passes through The amount of current that generated the flux

23 Summary Inductance formulas may be derived from
Direct application of the definition Energy approach Vector Potential Method The self-inductance of some common structures with sufficient symmetry have an analytical result Coaxial cable Long straight solenoid Toroid Internal 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 material Approximation based on a curvilinear square method if axial symmetry exists (uniform cross section) and a magnetic field map is drawn

25 References Hayt & 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.

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