Bài giảng Power Circuits and Electromechanics - Lecture 3

ECE430

Power Circuits and Electromechanics

Dr. Nam Nguyen-Quang

Fall 2009

http://www4.hcmut.edu.vn/~nqnam/lecture.php

Lecture 3

1

Introduction

 Electromagnetic theory: basis for explaining the operation of all

electrical and electromechanical systems.

 There are magnetic field and electric field systems, the discussion is

restricted to magnetic field systems.

 Integral form of Maxwell’s equations

 n da

Ampere’s law

f



^{H  dl } ^J

C

S

B

t

Faraday’s law

 n da



^{E  dl  }

C

S

 n da  0

 ^J

Conservation of charge

Gauss’s law

f

S

 ^{B  n da  0}

S

Lecture 3

2

1

Static magnetic circuits

 There are no moving components in static magnetic circuits.

 Toroid: N uniformly wound turns. r₀and r₁are inner and outer radii.

Consider the contour corresponding to the mean radius r = (r₀+ r₁) / 2,

assuming magnetic field intensity H_cis uniform inside the core. Using

Ampere’s circuital law, it can be determined that H_c(2r) = Ni. Or,

H_cl_c Ni

where l_c= 2r is the mean length in core. Assuming

B is a linear function of H in the core, the flux

density in the core is

Ni

B_c H_c 



Wb

/m²



l_c

Lecture 3

3

Static magnetic circuits (cont.)

Flux is given by

Ni

l_c

Ni

_c B_cA_c

A_c

Wb

l_cA_c

where  is the magnetic permeability of the core material, A_cis the core

cross-sectional area.

Define Ni as magneto motive force (mmf), reluctance can be defined as

l_c

Ni mmf



 R (At/Wb)

_c

flux A_c

P = 1/R is called permeance. Flux linkage is now defined as  = N_c=



N ²

L   PN ²



PN²i. By definition, self inductance L of a coil is given by

i

R

Lecture 3

4

2

Static magnetic circuits (cont.)

 There are similarities between electrical and magnetic circuits

mmf  voltage

flux  current

reluctance  resistance

permeance  conductance

 Toroid with air gap (no fringing): There is magnetic field intensity H in

both the air gap and the iron portion. l_g– length of the air gap, l_c– mean

length of the iron portion. Applying ACL around the contour c

B_g

B_c

Ni  H_gl_g H_cl_c

l_g

l_c

₀

_r₀

where ₀= 4 x 10^7H/m is the air’s permeability, and _ris the relative

permeability of the core material.

Lecture 3

5

Static magnetic circuits (cont.)

Applying Gauss’s law on the closed surface s covering one magnetic

pole, B_gA_g= B_cA_c.For the case of no fringing, A_g= A_c. Hence, B_g= B_c.

Divide the mmf by the flux to calculate equivalent reluctance

l_g

l_c

Ni





 R_g R_c



₀A_gA_c

Where R_gand R_care reluctances of the air gap and the core,

respectively. In the equivalent magnetic circuit, these are in series.

 Suppose there is “fringing”, i.e., not all the flux is confined to the area

between the two faces of iron portion. In this case, A_g> A_c, i.e., effective

air gap area increases. This can be accounted for empirically,







A_c ab, A_g a  l_gb  l_g

Lecture 3

6

3

Class examples

 Ex. 3.1: Find the required mmf to produce a given flux density. Air

gap and core length and area are known.

0.06

R_c

R_g

 47.710³At/Wb

 7.2310⁶At/Wb



10⁴4 10^710^4



0.001



4 10^71.110^4

  B_gA_g

0.5



1.110^4



 5.510^4Wb

Hence,

3

5









Ni  R_cR_g  47.7  7230 10 5.510  400 At

Lecture 3

7

Class examples (cont.)

 Ex. 3.2: Find the flux through the coils. All air gaps are the same in

length and area. Iron’s permeability is infinite and ignore fringing.



0.110^2



R₁ R₂ R₃ R 

1.98910⁶At/Wb



4 10^7410^4



2500

In the equivalent circuit, positive directions

for ₁, ₂, and ₃are shown. The algebraic

sum of the fluxes at node a must be zero.

R

₁

500

R

b

a

₂

Let mmf of node a wrt. b be F, then

1500

R

2500  F 500  F F 1500





 0

₃

R

Hence,

F  500,₁ 10^3Wb,₂ 0,₃ 10^3Wb

Lecture 3

8

4

In class quiz

 Problem 1: A toroid has a mean length with a radius of 500 mm, the

working flux density in the air gap is 0.6 Wb/m², creating by a coil of 100

turns. An air gap with the length of 2 mm is made. Given a = 20 mm.

Ignore the reluctance of the core.

a) Find the required excitation current

b) Determine the self inductance of the coil

 Special question: Suppose you were asked to build a linear variable

inductor. Describe your solution, considering fringing effect and

reluctance of the core (if exists)?

Lecture 3

9

Mutual inductance

 Mutual inductance: parameter related induced voltage in one coil with

time varying current in another coil.

 Consider two coils wound on the same magnetic core, coil 1 is

excited whilst coil 1 is open. The total flux linking coil 1 is

₁₁ _l1₂₁

where  (called leakage flux) links to coil 1 only; whereas, ₂₁is the

l1

mutual flux linking to both the coils, also the flux linking coil 2 due to

current in coil 1. The order of subscripts is important.

 Since coil 2 is open circuited, the flux linkage of this coil is

₂ N₂₂₁

Lecture 3

10

5

Mutual inductance (cont.)

 ₂₁is linearly proportional to the current i₁, hence

₂ N₂₂₁ M₂₁i₁

 The induced voltage v₂(due to the change of flux linkage) is given by

d₂

dt

di₁

dt

v₂

 M₂₁

M₂₁is called the mutual inductance between the coils. Similarly,

induced voltage v₁in coil 1 can also be determined as follows.

  N   L i

₁₁is proportional to i₁, hence

_{1 1}, then

1

1 11

d

dt

di₁

1

v₁

 L

1

dt

with L₁is the self inductance of coil 1, as you may know.

Lecture 3

11

Mutual inductance (cont.)

 Consider now the case where coil 1 is open and coil 2 is excited. The

same procedure can be used to calculate induced voltages.

d

dt

di₂

dt

1

v₁

 M₁₂

₂₂ _l2₁₂

₁ N₁₁₂ M₁₂i₂

v₂

d₂

dt

di₂

dt

 L₂

₂ N₂₂₂ L₂i₂

where L₂is the self inductance of coil 2, as you may know.

 From energy considerations, it can be shown that M₂₁= M₁₂= M.

 Finally, consider now the case where both the coils are excited.

₁ _l1 ₂₁₁₂ ₁₁₁₂

₂ ₂₁_l2 ₁₂ ₂₁ ₂₂

Lecture 3

12

6

Mutual inductance (cont.)

 Noting that M₂₁= M₁₂= M

₁ N₁₁₁ N₁₁₂ L₁i₁ Mi₂

₂ N₂₂₁ N₂₂₂ Mi₁ L₂i₂

 By differentiating those, induced voltages can be calculated

di₁

dt

di₂

dt

di₁

dt

di₂

dt

v₁ L₁

 M

v₂ M

 L₂

M

k 

 Coefficient of coupling between the two coils is defined by

L₁L₂

0  M  L L

 It can be shown that 0  k  1, or equivalently,

1

2

 Most air core transformers are loosely coupled (k < 0.5), whilst iron

core transformers are tightly coupled (k > 0.5, can approach 1).

Lecture 3

13

Example

 Ex. 3.4: Given reluctances of three air gaps in the magnetic circuit.

Draw equivalent circuit and compute flux linkages and inductances.

N₁i₁ R₃



₁₂



 R₁₁

10⁶

N₂i₂ R₂₂ R₃



₁₂



₁

100i₁ 5₁ 2₂





100i₂  2₁ 4₂





10⁶

N₁i₁

R₁

Solving these equations for ₁and ₂

R₃

25i₁12.5i₂

10^6



₂ 12.5i₁ 31.25i₂





10^6

₁



R₂

N₂i₂

₁ N₁₁



25i₁12.5i₂

12.5i₁ 31.25i₂

L₁ 2510^4H  2.5 mH

L₂ 31.2510^4H  3.125mH



10^4

From

₂ N₂₂





₂

It can be seen that

M  12.510^4H  1.25 mH

Lecture 3

14

7

Polarity markings (dot convention)

 Lenz’s law: the voltage induced is in such a direction that the current

due to it opposes the flux causing the voltage.

 Signs of mutually induced voltages are monitored by a dot marking

convention. A current i entering a dotted (undotted) terminal in one

winding induces a voltage Mdi/dt with positive polarity at the dotted

(undotted) terminal of the other winding.

 Two problems: (1) given the coil configuration, determine the dot

markings. (2) given the dot markings, how they are used in writing

equations.

Lecture 3

15

Determining polarity markings

 Steps:

. Arbitrarily select one terminal of a coil and assign a dot in one coil.

. Assume a current flowing into the selected dotted terminal and

determine the flux flowing in the core.

. Select an arbitrary terminal of the second coil and assign a positive

test current to it.

. Determine flux direction due to this current.

. Compare directions of two fluxes. If both is additive, then a dot is

placed in the second coil where the test current enters.

. If the fluxes are in opposite directions, then a dot is placed in the

second coil at the terminal where the current leaves.

Lecture 3

16

8

Practical ways of determining dot markings

 For a device such as a transformer, there is no way of knowing how

the coils are wound, therefore a practical way is adopted:

A DC source is used to excite one

coil of the transformer.

+

_

Place the dot on the terminal to

which the + side of DC source is

connected.

Close the switch: up-scale kick in voltmeter => the dot on the other coil is

on the + side of the voltmeter. Down-scale momentary deflection in

voltmeter => the dot is placed the – side of the voltmeter.

Lecture 3

17

Writing equations with mutually coupled coils

 Given 2 mutually coupled coils and dot markings, write loop equations.

Choose arbitrary direction for currents.

Rule: Reference current enters a dotted (undotted) terminal, induced

voltage in the other coil is positive at the dotted (undotted) terminal.

Reference current leaves a dotted (undotted) terminal, induced voltage at

the dotted (undotted) terminal of the other coil is negative.

di₁

dt

di₂

dt

R₁

R₂

M

v₁ i₁R₁ L₁

 M

i₁

i₂

v₂

v₁

di₂

di₁

v₂ i₂R₂ L₂

 M

dt

Lecture 3

18

9

Example

 Ex 3.6: Write loop equations for a circuit with mutually coupled coils.

Assuming zero initial voltage on capacitor

L₂

R₁

i₁

C

R₂

v₁ i₁R₁



i₁ i₂R₂



v₁

M

i₂

di₂

dt

d

 L₁

dt





i₁ i₂ M

L₁

(i₁– i₂)

di₂

1

d

0 

^ti₂dt  L₂

 M

i₁ i₂



 L₁

i₂ i₁





0

C

dt

di₂

dt

 M





i₂ i₁



R₂

Lecture 3

19

In-class quiz

 Problem 3.15.

Lecture 3

20

10

Bài giảng Power Circuits and Electromechanics - Lecture 3 - Nguyễn Quang Nam