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# Magnetic circuits

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#MAGNETIC-CKTS

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### Magnetic circuits

1. 1. BITS Pilani Pilani Campus EEE G529T MAGNETIC CIRCUITS 13.1-13.5 EEE Department BITS Pilani, Pilani
2. 2. BITS Pilani, Pilani Campus MAGNETIC FIELDS  Ferromagnetic Material – that has the property of attracting other pieces of such materials  A permanent magnet is a ferromagnetic material  Just as we have the Electric Field associated with a stationary charge, we have magnetic field associated with a moving charge (current)  Magnetic Lines of Force – Imaginary closed loops that origin from the north pole and end at the south pole of the magnet  The tangent at a point on these lines gives the direction of the magnetic field. Bar Magnet Magnetic Lines of Force
3. 3. BITS Pilani, Pilani Campus MAGNETIC FLUX AND FLUX DENSITY  Magnetic flux (denoted as Φ), is the amount of magnetic field passing through a surface (such as a conducting coil).  The SI unit of magnetic flux is the weber (Wb) More magnetic lines pass through the circular ring
4. 4. BITS Pilani, Pilani Campus LORENTZ FORCE When a charge q moves with a velocity v in the presence of a Magnetic Field B, a force is exerted on the charge known as the Lorentz Force. The direction of this force is given by Right Hand Rule
5. 5. BITS Pilani, Pilani Campus MAGNETIC FIELD DUE TO A CONDUCTOR CARRYING CURRENT A conductor carrying a current I produces a magnetic field B around it in a cylindrically radial pattern as shown The direction of the magnetic field is given by the right hand rule which states that if you hold the conductor in your right hand with thumb pointing in the direction of the current, then the direction in which the fingers curl gives you the direction of the magnetic field. 𝑩 = 𝝁𝑰 𝟐𝝅𝒅 Magnetic Field B at a radial distance d from the conductor carrying current I is given by μ is permeability of the medium
6. 6. BITS Pilani, Pilani Campus MAGNETIC FIELD DUE TO A SOLENOID Solenoid : Cylindrical coil of wire wound on an air-core, an iron core or any other core 𝐵 = 𝜇𝑁 𝑙 I Consider a cylindrical coil wound on a soft iron core fed by a DC current of magnitude I. A magnetic field will be setup in the core that will be almost constant provided that the length of the cylinder (l) is much larger than its radius. The magnetic field inside the Iron Core is given by The magnetic flux is given by B.A = B.(∏r2) Permeability defines the easy with which magnetic field can be set up in the material
7. 7. BITS Pilani, Pilani Campus MAGNETIC FIELD DUE TO A TOROID  If we bend the iron core in the form of a ring, we get a toroid  The magnetic field at the center of the toroid is given by RELATIVE PERMEABILITY  It is the ratio of permeability of the medium to the permeability of free space  Ferromagnetic materials have high values of relative permeability (>1000) 𝐵 = 𝜇𝑁𝐼 2∏𝑟
8. 8. BITS Pilani, Pilani Campus MAGNETIC CIRCUITS  Magnetic circuits are analogue of electrical circuits.  The magneto motive force of N-turn current carrying coil is  The reluctance R of a magnetic path depends on the mean length l, the area A, and the permeability μ of the material.  Magnetic flux is analogous to current in electrical circuit and is related to F and R in a similar way as Ohm’s law A l  R NiF  RF
9. 9. BITS Pilani, Pilani Campus ANALOGY BETWEEN ELECTRIC AND MAGNETIC CKTS
10. 10. BITS Pilani, Pilani Campus IRON CORES WITH AN AIR GAP Core with an air gap Equivalent magnetic circuit The air gap will have some reluctance that will be in series with the reluctance of the iron core.  Fringing of the flux lines occur when the air gap length is somewhat large  More flux is concentrated in the inner portion of the core than in the outer portion
11. 11. BITS Pilani, Pilani Campus MAGNETIZATION CURVES H B B H Linear knee saturation  A plot of B v/s H is a magnetization curve  As 𝑩 = ∅/𝑨 and ∅ = 𝑴𝑴𝑭 𝑹 we get 𝑩 = 𝝁𝑯 where H is the Magnetic Field Intensity given by  𝑯 = 𝑴𝑴𝑭 𝒍  H is independent of core material Practical B-H Curve Ideal Curve
12. 12. BITS Pilani, Pilani Campus HYSTERESIS Demagnetized Material : The Magnetic Flux Density (B) is 0 when no external Magnetizing Field (H) is applied Consider a demagnetized material and let us apply a magnetizing field to it to magnetize it  For smaller values of H, B-H curve is almost linear  For larger values of H, B-H curve saturates and B is almost constant  If we increase the magnetizing field even more, no increase in B is observed (point b)  The material is said to be saturated now  If we reduce H now, B does not reduce the same way as it increased  There is some amount of magnetism left over when H reduces to 0  This is called Residual Magnetism (point c)  Permanent Magnets are made of materials having high values of residual magnetism
13. 13. BITS Pilani, Pilani Campus  If H is increased again but in the –ve direction, then at a particular value of H, the Residual Magnetism goes away and B becomes 0  This value of H is called the coercive force (point d)  Permanent magnets have high coercivity  If H is made –ve enough, material saturates but in the opposite direction (point e)  On increasing H from its max –ve value, we reach point f that indicates negative residual magnetism  The resulting loop is called a hysteresis loop and the phenomenon Hysteresis HYSTERESIS
14. 14. BITS Pilani, Pilani Campus If the magnetizing force applied to the demagnetized material is less than the required to produce saturation, a hysteresis loop as shown is produced. HYSTERESIS SQUARE LOOP MATERIALS  Materials having hysteresis loop approximately rectangular are called square loop materials  Slopes of the sides of the hysteresis loop are quite large. A small change in H can lead to a large change in B  Small cores made from such materials are used as binary memory devices in switching circuits and digital computers
15. 15. BITS Pilani, Pilani Campus HYSTERESIS LOSS  When a magnetizing material is periodically magnetized and demagnetized using an alternate current, Energy is absorbed by the material that gets converted to heat  Energy lost per cycle – (Area under the B-H curve)  Power loss due to Hysteresis –  Kh and n depends on the core material  loss is directly proportional to frequency HdB Kh(Bm)nf Use cores made of materials that have thin hysteresis loop
16. 16. BITS Pilani, Pilani Campus ELECTROMAGNETIC INDUCTION  EMF (Electro-Motive-Force) is induced in a multi-turn coil when the magnetic flux passing through the coil varies with time  First discovered by Michael Faraday in 1831  The induced EMF depends on the rate of change of total flux linkages with the coil 𝑒 = −𝑁 𝑑Φ 𝑑𝑡 Induced emf Lenz law The polarity of the voltage induced by a changing flux tends to oppose the change in flux that produced the induced voltage Circuit is closed and thus current flows
17. 17. BITS Pilani, Pilani Campus ELECTROMAGNETIC INDUCTION The principle of electromagnetic induction is also involved in power generation when the armature of a DC Machine is rotated quite fast in the presence of a radial magnetic field POWER GENERATION The current reverses its direction whenever there is a change in the direction of motion of the magnet
18. 18. BITS Pilani, Pilani Campus v= 𝐿 𝑑𝑖 𝑑𝑡 = 𝑁 𝑑∅ 𝑑𝑡 Consider an inductor fed by a time varying current. An EMF is induced across the inductor governed by the equation On solving the above equation, we get a relation 𝐿 = 𝑁 ∅ 𝑖 As Φ=Ni/R we get 𝐿 = 𝑁2 𝑅 Reluctance Number of flux linkages per ampere 1 Henry = 1 Weber per ampere INDUCTANCE OF AN INDUCTOR
19. 19. BITS Pilani, Pilani Campus  Consider an iron core that has a primary coil and a secondary coil  AC Sine wave is fed through the primary coil  The current in the primary coil produces a magnetic field and hence flux lines  The magnetic flux has a sinusoidal nature and is this variable flux travels through the soft iron core  This variable flux cuts the secondary coil and induces and EMF in it that follows the Lenz rule  If a load is connected across the secondary, time varying current flows in the secondary coil MAGNETICALLY COUPLED COILS N1 N2
20. 20. BITS Pilani, Pilani Campus  Suppose the primary winding having Inductance L1 has N1 turns and secondary winding having Inductance L2 has N2 turns  Since a time varying current in the primary induces a voltage across the secondary, we say that the 2 coils are magnetically coupled  The flux that is setup in the core on account of the current in the primary is given by  Neglecting flux leakages, the same flux links the secondary coil inducing an EMF across it Φ=N1i/R Reluctance 𝑒 = 𝑁2 𝑑Φ 𝑑𝑡 𝑒 = 𝑁1 𝑁2 𝑅 𝑑𝑖 𝑑𝑡 On solving M=N1N2/R Mutual Inductance between the coils MUTUAL INDUCTANCE
21. 21. BITS Pilani, Pilani Campus DOT CONVENTIONS
22. 22. BITS Pilani, Pilani Campus TRANSFORMERS  A transformer is a magnetic circuit consisting of 2 coils wound on a common iron core  More than 2 windings can also be used  Used in efficient transfer of Electric Power from the Generating station to our homes  2 types – Step Up and Step down  Step Up : Steps up the voltage at lower currents ( v x I = constant ) ( Neglecting leakage flux)  Step Down : Steps down the voltage but at a higher current  Voltages are stepped up prior to transmission so that the Copper losses are minimal  Used in Electronic, Control and Communication systems  Used for isolating 2 circuits as there is a magnetic coupling between the two and no physical contact  Used for impedance matching to have maximum power transfer from source to load
23. 23. BITS Pilani, Pilani Campus TRANSFORMERS Circuit symbol  There may be connections to both windings so i1 and i2 both can be non zero  As a result, i1 that passes through L1 produces a voltage L1 𝑑𝑖1 𝑑𝑡 and i2 that passes through L2 induces a voltage M12 𝑑𝑖2 𝑑𝑡 across the primary  Total voltage across the primary : v1 = 𝐋 𝟏 𝒅𝒊 𝟏 𝒅𝒕 + 𝐌 𝟏𝟐 𝒅𝒊 𝟐 𝒅𝒕 M12 i1 i2 L1 L2 M21
24. 24. BITS Pilani, Pilani Campus v2 = 𝐋 𝟐 𝒅𝒊 𝟐 𝒅𝒕 + 𝐌 𝟐𝟏 𝒅𝒊 𝟏 𝒅𝒕 Similarly the voltage across the secondary winding is  𝐌 𝟐𝟏 = 𝐌 𝟏𝟐 = M = N1N2 / R  The Energy stored in the form of magnetic field in the transformer is given by 𝒘 𝒕 = 𝟏 𝟐 𝑳 𝟏 𝒊 𝟏 𝟐 𝒕 + 𝟏 𝟐 𝑳 𝟐 𝒊 𝟐 𝟐 𝒕 + 𝑴𝒊 𝟏 𝒕 𝒊 𝟐(𝒕)  As a transformer works on AC, the currents and voltages are all phasors, we can represent the transformer equations as follows 𝑽 𝟏 = 𝒋𝝎𝑳 𝟏 𝑰 𝟏 + 𝒋𝝎𝑴𝑰 𝟐 𝑽 𝟐 = 𝒋𝝎𝑳 𝟐 𝑰 𝟐 + 𝒋𝝎𝑴𝑰 𝟏 𝑣 = 𝐿 𝑑𝑖 𝑑𝑡 corresponds to 𝑽 = 𝑗𝜔𝐿1 in the frequency domain TRANSFORMERS
25. 25. BITS Pilani, Pilani Campus TRANSFORMER CIRCUIT REPRESENTATION  The transformer can be represented by 3 uncoupled inductors as shown here TRANSFORMER LOSSES HYSTERESIS LOSS EDDY CURRENT LOSSES Energy dissipation in the form of heat in the core of the t/f on account of rapid magnetization and demagnetization As a core is a conductor and a time varying magnetic flux will pass through it, an EMF hence circulating currents are generated in the core that lead to I2R losses (core heating)
26. 26. BITS Pilani, Pilani Campus COUPLING COEFFICIENT  It is the measure of the magnetic coupling between the 2 coils  Denoted by k  0<k<1  𝑘 = 𝑀 √𝐿1 𝐿2  Coupling coefficient depends upon  Permeability of the core material  Number of turns in each coil  Relative position and the dimensions of the 2 coils  Loosely Coupled T/F -> k=0 (almost) (Air Core T/F)  Tightly Coupled T/F -> k=1 (almost) (Iron Core T/F)
27. 27. BITS Pilani, Pilani Campus IDEAL TRANSFORMER k=1 perfect coupling L1 , L2 = ∞ no losses 𝑽 𝟏 = 𝑗𝜔𝐿1 𝑰 𝟏 + 𝑗𝜔𝑀𝑰 𝟐 𝑽 𝟐 = 𝑗𝜔𝐿2 𝑰 𝟐 + 𝑗𝜔𝑀𝑰 𝟏 i1 i2 L1 L2 k=1 V1 V2 I1 I2  Figure shows the circuit symbol for an ideal t/f (k=1)  The phasor relationship is as follows  As 𝑀 = √𝐿1 𝐿2 , we can write V2 in terms of V1 as  Turns Ratio (N) Ratio of secondary to primary turns 𝑁 = 𝐿2 𝐿1 = 𝑁2 2 /𝑅 𝑁1 2 /𝑅 = 𝑁2 𝑁1 𝑽2 = 𝐿2 𝐿1 𝑽1 N  N>1 : Step Up T/F  N<1 : Step Down T/F  N=1 : Isolation T/F
28. 28. BITS Pilani, Pilani Campus IDEAL TRANSFORMER MODEL A transformer with perfect coupling is said to be ideal if L1 and L2 approach ∞ and the turns ratio remains constant For an ideal t/f 𝑽2 = 𝑁𝑽1 𝑰2 = −𝑰1/𝑁 + +- -- + 𝑰1 𝑰2 𝑽1 𝑽2 𝑽2/𝑁 𝑰1/𝑁+ - -- ++ 𝑰1 𝑰2 𝑽1 𝑽2𝑁𝑰2 𝑁𝑽1 IDEAL TRANSFORMER MODEL ALTERNATE IDEAL TRANSFORMER MODEL
29. 29. BITS Pilani, Pilani Campus IDEAL TRANSFORMER AS A LOSSLESS DEVICE Instantaneous power absorbed by the primary winding : 𝒑 𝟏 = 𝒗 𝟏 𝒊 𝟏 Instantaneous power absorbed by the secondary winding : 𝒑 𝟐 = 𝒗 𝟐 𝒊 𝟐 Total Instantaneous power absorbed by the T/F: p = p1 + 𝑝2 𝑝 = 𝑣1 𝑖1 + 𝑣2 𝑖2 𝑝 = 𝑣1 𝑖1 + 𝑁𝑣1 −𝑖1 𝑁 𝐩 = 𝟎 Since the instantaneous power is 0, the average power and the energy stored = 0 IDEAL TRANSFORMER IS A LOSSLESS DEVICE