Magnetically coupled circuits Examples + The rest of the circuit i(t) L R v(t) – Electrically ect ca y coupled coup ed Magnetically coupled Inductance Inductance occurs when current flows through a (real) conductor. conductor The current flowing through the conductor sets up a magnetic field that is proportional to the current. The voltage difference across the conductor is proportional p opo t o a to the t e rate ate of o change c a ge of o the t e magnetic ag et c field. ed The proportionality constant is called the inductance, denoted L. Units of Henrys (H) - V V·s/A. s/A. 5 Self-inductance Self inductance in AC An inductor is a two-terminal device that consists of a coiled conducting wire around a core. A current flowing through the device produces a magnetic flux Φ forms closed loops threading its coils. Total flux linked by N turns of coils, coils flux linkage λ=NΦ For a linear inductor, λ=Li (L is the inductance). Self-inductance Self inductance and induced voltage Faraday’s law of electromagnetic induction: Th EMF (Electromotive The (El t ti Force) F ) induced i d d in i a magnetic ti circuit i it is i equal to the rate of change of flux linked with the circuit. d d ( N ) d e N dt dt dt Li N e dLi di L dt dt λ is total flux linkage and L the self inductance. Mutual inductance in AC Two coil arrangement: When the flux produced by coil 1 links with coil 2, a voltage is generated across the terminals of that coil. The arrangement shown below an AC voltage applied to the terminals of coil 1 that produced an AC current which produces an AC flux. flux This flux links the second coil through the centre from bottom to top. This gives rise to V2(t). 1 i2 i1 N1 turns N2 turns v2(t) v1(t) Coil 1 Coil 1 Coil 2 Mutual inductance in AC Consider the coupled coils: coil1 and coil 2 The total flux Φ1 threading coil 1 is the sum of components due to i1 and i2: Φ1 = Φ11 + Φ12 The net flux linkage for coil 1 is λ1= N1Φ1 = N1Φ11 + N1Φ12 The first term is due to coil’s own current L1 is i called ll d the th self lf inductance i d t The second term is due to current in the other coil N1Φ12 =M12i2 M12 is called the mutual inductance The sign is determined by the direction of the flux By Faraday’s law: d 1 di1 di2 v1 L1 M 12 dt dt dt Energy analysis Similar analysis, we have d 1 di di d 2 di2 di1 L1 1 M 12 2 v L M 2 2 21 dt dt dt dt dt dt The instantaneous power delivered to the coils: di di di di p1 v1i1 L1i1 1 M 12i1 2 p2 v2i2 L2i2 2 M 21i2 1 dt dt dt dt v1 The energy stored in the coils: 1 2 1 2 L i t L2i2 (t ) M 12i1 (t )i2 (t ) ( ) 2 11 2 w(t ) 1 1 L2i22 (t ) L1i12 (t ) M 21i1 (t )i2 (t ) 2 2 apply i1 first, then i2 apply i2 first, then i1 Initial and final conditions are the same. The mutual inductances are identical: M 12 M 21 M Formulas for coupled coils i-v laws for coupled coils: di1 di M 2 dt dt di2 di1 v2 L2 M dt dt v1 L1 The energy stored w in a pair of coupled coils: 1 2 1 2 w L1i1 L2i2 Mi1i2 2 2 Various flux definitions 1 2 i1 12 N1 turns ell1 v1(t) 21 Coil 1 i2 N2 turns v2(t) ell2 C il 2 Coil 2 Note that ell1 and ell2 "ell" means l as little but will only confuse if it is in the diagram as l. These flux terms represent flux that is produced in coils 1 (ell1) and 2 (ell2) which do not link coils 2 and 1 respectively. They can be seen as leakage terms. But they are assumed negligible for the purposes of the analysis. analysis Various flux definitions Φ11 is the flux produced in coil 1 by the current flowing in coil 1 Φ11 = Φell1 + Φ21 Φ22 is the flux produced in coil 2 by the current flowing in coil 2 Φ22 = Φell2 + Φ12 Φ21 is the flux linking coil 2 produced by coil 1 Φ12 is the flux linking coil 1 produced by coil 2 Φ1 - The total flux threading coil 1 is the sum of components due to i1 and i2: Φ1 = Φ11 + Φ12 d 1 d1 d11 d12 di1 di2 v1 N1 N1 ( ) L1 M 12 dt dt dt dt dt dt Phasor format From the time-domain equations: di1 di2 M v1 L1 dt dt di2 di1 v2 L2 M dt dt In phasor analysis (used for AC steady-state steady state response to sinusoidal excitations) V1 j L1 I1 j MI 2 V2 j L2 I 2 j MI1 Coefficient of coupling The total flux Φ11 resulting from i1 through N1 turns consists of leakage flux Φell1 and coupling flux Φ21. The coupling coefficient, k, is defined as the ratio of l k linking fl flux to totall fl flux. Also, l the h coupling l is b bilateral: l l 21 12 k 11 22 Coefficient of coupling Coefficient of coupling k A measure off the th degree d off coupling li Defined by k M L1 L 2 If there is no coupling between the coils, M=k=0 There is equivalent to two simple, uncoupled coils Since M21<L1 and M12<L2, we have M2 L1 L2 0≤ K ≤ 1 If k is close to 1, the coils are said to be tightly coupled Self evaluation When one coil of a magnetically coupled pair has a current of 5.0A, the resulting fluxes ell1 and 21 are 0.2mWb and 0.4mWb, respectively. If the turns are N1 = 500 and N2 = 1500,, find L1, L2, M and the coefficient of coupling k. [60mH, 540mH, 120mH, 0.667] Dot convention The dot convention is used to determine whether the fluxes add and therefore the mutual inductance adds. adds The rules are: Assume positive M Place a dot where i1 enters coil 1 Determine the direction of the flux produced in coil 1 due to i1 Consider coil 2 and the direction of the current i2. If i2 provides a flux in the same direction as the flux due to coil 1 and i1, a dot is placed on coil 2 where i2 enters. If the flux in coil 2 opposes the flux due to coil 1 and i1, a dot is placed on coil 2 where i2 leaves. Examples • a i1 – 1 • a b •b + v21 2 – Positive M • i1 1 + v21 – Negative M 2 Self evaluation The physical construction of two mutually coupled coils. From a consideration of the direction of magnetic flux produced by each coil, it is shown that dots may be placed either on the upper terminal of each coil or on the lower terminal of each coil. • • Example 1 Two circuits magnetically coupled with positive M i1 M + • v1 L1 _ R1 v1 I1R1 L1 i2 + • L2 v2 _ R2 di1 di M 2 dt dt di1 di2 v2 I 2 R2 M L2 dt dt + V1 I1 j M • j L1 I2 + • j L2 V 2 _ _ R1 R2 V1 I1R1 jLI 1 1 jMI2 V2 I2R2 jMI1 jL2I2 Example 2 Two circuits magnetically coupled with negative M i1 + • u1 L1 _ R1 i2 M L2 • + u2 _ R2 V1 I1R1 jLI 1 1 jMI2 V2 I2R2 jMI1 jL2I2 Example 3 i Series adding R + v – i vR v 1 + • v L1 – + • v2 L2 R M – L . V VR +V1 V2 (j L I j MI) ( j L I j MI) =IR 1 2 RR L L1 L2 2 M Example 4 Series opposing i + + R1 v1 v – – . v2 R2 i – + • L1 M L2 • + R v – L V 2 j MI j ( L1 L2 ) I I( R1 R2 ) R R1 R2 L L1 L2 2 M Mutual inductance measurement If you do practical tests on the coils for series adding and opposing you can calculate the mutual inductance if the self-inductances are known. LA L1 L2 2 M Series adding LB L1 L2 2 M S i opposing Series i LA LB M 4 Mutual inductance k LA LB M L1 L 2 4 * L1L 2 Coupling p g coefficient Example 5 Mutual inductance in parallel and dotted terminals at the same side M i + u – i1 * L1 * i2 L2 d i1 u L1 M dt di u L2 2 M dt d i2 dt d i1 dt i = i1 +i2 Solve the relation of u, i: ( L1L2 M 2 ) di u L1 L2 2 M dt ( L1L2 M 2 ) Leq L1 L2 2 M 0 Example 5 cont cont’d. d. M i º+ u _ º i1 L1 * * di1 di 2 u L1 M dt dt i2 di1 ( L1 M ) M di dt dt L2 Draw equivalent q circuit: i º + u _ º M i1 L1-M i2 = i - i1 i2 L2-M di 2 di1 u L2 M d dt dt d i1 = i - i2 di 2 ( L2 M ) M di dt dt T circuit i1 12 N1 turns ell1 v1(t) N2 turns v2(t) ell2 i2 21 Coil 2 Coil 1 di di V1 L1 1 M 2 dt dt i1(t) v1(t) V 2 L2 d i2 di M 1 dt dt L2 ‐ M L1 ‐ M M i1(t) + i2(t) i2(t) v2(t) T circuit V1 ( L1 M ) d i1 d( i1 i2 ) di di M L1 1 M 2 dt dt dt dt V 2 ( L2 M ) i1(t) v1(t) d i2 d( i1 i2 ) di di M L2 2 M 1 dt dt dt dt L2 ‐ M L1 ‐ M M i1(t) + i2(t) i2(t) v2(t) Mutual inductance and transformers If we have two coils wound on the same magnetic (iron) core, a changing current in one coil would cause a voltage in both coils. The effect of changing current in one coil inducing a voltage in another is mutual inductance and is the basis of transformer. Transformers are one of the most common devices to convert from one voltage to another with high (>99%) efficiency. Since the current must continuously change for these devices to operate, they are generally only used with AC supplies. This ease of voltage conversion is one of the main reasons why the majority of power is distributed in AC rather than DC. Review of example 2: Two circuits magnetically coupled with negative M, ignore leakage fluxes Two coils with opposing fluxes (Φ21- Φ12) and N1≠N2. i1 12 N1 turns N2 turns ell1 v1(t) () v2(t) ell2 i2 21 Coil 2 Coil 1 v1 (t ) N1 d [21 12 ] dt Dividing the two equations: v1 (t ) N1 v2 ( t ) N 2 d v2 (t ) N 2 [21 12 ] d dt q defines the ideal transformer. This equation