ECE430 Power Circuits and Electromechanics Dr Nam Nguyen-Quang Fall 2009 http://www4.hcmut.edu.vn/~nqnam/lecture.php Lecture 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  C  C H  dl  J S E  dl     J  n da   B  n da  f S S S  n da Ampere’s law B  n da t Faraday’s law f Conservation of charge Gauss’s law Lecture Static magnetic circuits  There are no moving components in static magnetic circuits  Toroid: N uniformly wound turns r0 and r1 are inner and outer radii Consider the contour corresponding to the mean radius r = (r0 + r1) / 2, assuming magnetic field intensity Hc is uniform inside the core Using Ampere’s circuital law, it can be determined that Hc(2r) = Ni Or, H c l c  Ni where lc = 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 Bc  H c   Ni Wb /m lc Lecture 3 Static magnetic circuits (cont.) Flux is given by  c  Bc Ac  Ni Ni Ac  Wb lc l c Ac where  is the magnetic permeability of the core material, Ac is the core cross-sectional area Define Ni as magneto motive force (mmf), reluctance can be defined as l Ni mmf   c  R (At/Wb) c flux Ac P = 1/R is called permeance Flux linkage is now defined as  = Nc = PN2i By definition, self inductance L of a coil is given by L  Lecture  N2  PN  i R Static magnetic circuits (cont.)  There are similarities between electrical and magnetic circuits mmf flux reluctance permeance     voltage current resistance conductance  Toroid with air gap (no fringing): There is magnetic field intensity H in both the air gap and the iron portion lg – length of the air gap, lc – mean length of the iron portion Applying ACL around the contour c Ni  H g l g  H clc  Bg 0 lg  Bc lc r 0 where 0 = 4 x 107 H/m is the air’s permeability, and r is the relative permeability of the core material Lecture Static magnetic circuits (cont.) Applying Gauss’s law on the closed surface s covering one magnetic pole, BgAg = BcAc For the case of no fringing, Ag = Ac Hence, Bg = Bc Divide the mmf by the flux to calculate equivalent reluctance lg l Ni   c  R g  Rc   Ag Ac Where Rg and Rc are 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, Ag > Ac, i.e., effective air gap area increases This can be accounted for empirically, Ac  ab, Ag  a  l g b  l g  Lecture 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  47.7  103 At/Wb 7 4 10 4  10 10 0.001 Rg   7.23  106 At/Wb 7 4 4  10 1.1 10 Rc           Bg Ag  0.51.1 10 4   5.5 10 4 Wb Hence, Ni  R c  R g   47.7  7230  10  5.5  10 5  400 At Lecture 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  4  10 4  10   1.989  10 2 R1  R  R  R  7 4 At/Wb 2500 In the equivalent circuit, positive directions for 1, 2, and 3 are shown The algebraic sum of the fluxes at node a must be zero Let mmf of node a wrt b be F, then 500 b 1 R a 2 1500 2500  F 500  F F  1500   0 R R R Hence, R R 3 F  500, 1  10 3 Wb,   0,   10 3 Wb Lecture 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 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 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 is excited whilst coil is open The total flux linking coil is 11  l1   21 where l1 (called leakage flux) links to coil only; whereas, 21 is the mutual flux linking to both the coils, also the flux linking coil due to current in coil The order of subscripts is important  Since coil is open circuited, the flux linkage of this coil is 2  N 221 Lecture 10 Mutual inductance (cont.)  21 is linearly proportional to the current i1, hence 2  N 221  M 21i1  The induced voltage v2 (due to the change of flux linkage) is given by v2  d di  M 21 dt dt M21 is called the mutual inductance between the coils Similarly, induced voltage v1 in coil can also be determined as follows 11 is proportional to i1, hence 1  N111  L1i1 , then v1  d1 di  L1 dt dt with L1 is the self inductance of coil 1, as you may know Lecture 11 Mutual inductance (cont.)  Consider now the case where coil is open and coil is excited The same procedure can be used to calculate induced voltages 22  l  12 1  N112  M 12 i2 v2  2  N 222  L2i2 v1  d1 di  M 12 dt dt d2 di  L2 dt dt where L2 is 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    21   l  12   21   22 Lecture 12 Mutual inductance (cont.)  Noting that M21 = M12 = M 1  N111  N112  L1i1  Mi2 2  N 2 21  N 2 22  Mi1  L2 i2  By differentiating those, induced voltages can be calculated v1  L1 di1 di M dt dt v2  M di1 di  L2 dt dt  Coefficient of coupling between the two coils is defined by k   It can be shown that  k  1, or equivalently,  M  M L1 L2 L1 L2  Most air core transformers are loosely coupled (k < 0.5), whilst iron core transformers are tightly coupled (k > 0.5, can approach 1) Lecture 13 Example  Ex 3.4: Given reluctances of three air gaps in the magnetic circuit Draw equivalent circuit and compute flux linkages and inductances N 1i1  R3 1     R11 N i  R 2  R 1    100i1  51  2   10 100i   21  4   10 1 Solving these equations for 1 and 2 N1i1 R1 R3 6 1  25i1  12.5i   10 6   12.5i1  31.25i2   10 From R2 1  N11  25i1  12.5i2  10 4 2  N 22  12.5i1  31.25i2  10 4 It can be seen that L1  25  10 4 L2  31.25  10 4 H  3.125 mH N2i2 2 H  2.5 mH M  12.5  10 4 H  1.25 mH Lecture 14 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 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 16 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 17 Writing equations with mutually coupled coils  Given 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 di M dt dt di di v  i R  L2  M dt dt v1  i1 R1  L1 Lecture R1 i1 v1 R2 M i2 v2 18 Example  Ex 3.6: Write loop equations for a circuit with mutually coupled coils Assuming zero initial voltage on capacitor  L1 0 R1 i1 v1  i1 R1  i1  i2 R2 L2 R2 v1 d i1  i2   M di2 dt dt (i1 – i2) C M i2 L1 di t d d i2 dt  L2  M i1  i   L1 i2  i1   C dt dt dt di  M  i2  i1 R2 dt Lecture 19 Lecture 20 In-class quiz  Problem 3.15 10