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Power transformers #1 1.0 Introduction You have been introduced to transformers in EE 303. In these notes, I will review some of the main ideas related to transformers, but I do expect that you have come across these ideas previously. 2.0 Ideal Transformer and referring quantities (Section 5.1) The circuit model for the ideal transformer is given Fig. 1 below, with turns ratio n=N2/N1. I1 V1 I2 E1 N1 E2 Z2 V2 N2 Fig. 1 The dots mean that the currents flowing into the dotted terminal produces fluxes φm1 and φm2 in the core which add, based on the right-hand rule (see Fig. 5.1 of text). 1 A more “circuits” way of understanding the dots is [1, pg. 215] If the two currents I1 and I2 are either both entering or both leaving the dotted terminals of the coils, then the signs of the mutual terms are the same as the signs of the self-terms. If one current is entering and the other is leaving the dotted terminals, then the signs of the mutual terms are opposite to the signs of the self-terms. One other essential idea is to remember that a self-induced voltage across an inductor is positive at the terminal into which the current enters. The voltages E1 and E2 result from the composition of self-induced voltage and mutual induced voltage in each coil. That is, (a) E1 L1 I1 MI 2 (b) E2 L2 I 2 MI1 2 Let’s eliminate I2 from the above equations. Solving the (b) for I2 results in: M 1 L2 I 2 MI1 E2 I 2 I1 E2 (c) L2 L2 Substituting this expression for I2 back into (a) results in: M2 M E1 L1 I1 MI 2 L1 I1 I 1 E2 L2 L2 (d) 2 M M L1 I1 E2 L2 L2 Most circuits textbooks (in the chapter on mutual inductance) will use Faraday’s law and magnetic circuit theory to derive the relationship between mutual inductance and self inductance for two coupled coils, resulting in the definition of the coefficient of coupling k: M k L1 L2 (e) Two coils that are not coupled at all have k=0. If two coils are perfectly coupled 3 (which means no leakage flux, i.e., the flux linking one coil completely links with the other coil), then k=1. This of course is impossible, but the practice reality is that if you wind the coils on some ferromagnetic material such as iron, k will be almost 1. So let’s consider the “ideal” case, when k=1, which results in what we have called the ideal transformer. M k 1 M 2 L1 L2 L1 L2 (f) Substitute (f) into (d) to get: LL LL LL E1 L1 1 2 I 1 1 2 E2 1 2 E2 L2 L2 L2 (g) Multiply top and bottom of (g) by L2 : E1 L2 L2 L1 L2 L L L1 E2 2 1 E21 E2 L2 L2 L2 L2 Dividing both sides by E2, we obtain: E1 E2 L1 L2 (h) 4 Using L1 N12 A , l N 22 A for the two coils L2 l of equal cross section and equal length, we see that eq. (h) becomes: E1 N1 (i) E2 N2 The text derives this relation and also one for current, resulting in: E2 N 2 1 n E1 N 1 a I2 N1 1 a I1 N 2 n (1) (2) (Same as eq. 5.2 and 5.6 in text) Example 5.1 uses ABCD parameters to determine the relation for Z1 and Z2 that make the two circuits of Fig. 2 have identical input/output characteristics. I1 Ideal network I2 Z1 V1 E1 Network Network 2 1 E2 N1 N2 Fig. 2a 5 V2 I1 V1 Ideal network E1 N1 I2 E2 N2 Z2 V2 Network 2 Fig. 2b I think it is not so clear. Let’s go through it. Recall that the ABCD parameters are coefficients of the terms in the equations that express input voltage and current as a function of output voltage and current as follows: V1 A B V2 I C D I (3) 2 1 For each circuit, we will obtain two sets of ABCD parameters: The set for the ideal transformer The set for the impedance (network 1, 2). Then we will combine them together by recalling the great thing about ABCD parameters for cascaded circuits – they can just be multiplied together! 6 Ideal Network: The ABCD parameters for the ideal transformer are obtained by assuming E1 and E2 are the input and output voltages, respectively. We simply manipulate eqs. (1) and (2) to get: E1 aE2 (4) I1 1 I2 a (5) Therefore: E1 a 0 E 2 I 0 1 I (6) 1 a 2 So we see that A=a, B=0, C=0, D=1/a, and a 0 1 Tideal 0 (7) a 7 Network 1 (Fig 2a): We get the ABCD matrix for the network of the left-hand dotted box in Fig. 2a. Here, the input quantities are V1 and I1, and the output quantities are E1 and I1. In this case: V1 E1 Z1 I1 (8) I1 I1 (9) Therefore: V1 1 Z 1 E1 I 0 1 I (10) 1 1 and Tnetwork1 1 Z 1 0 1 8 (11) Network 2 (Fig. 2b): Now let’s get the ABCD matrix for network of the right-hand dotted box in Fig. 2b. Here, the input quantities are E2 and I2, and the output quantities are V2 and I2. In this case: E2 V2 Z2 I 2 (12) I2 I2 (13) Therefore: E 2 1 Z 2 V2 I 0 1 I (14) 2 2 and Tnetwork 2 1 Z 2 0 1 (15) Composite ABCD Matrices: The ABCD Matrix for Fig. 2a is: TFig 2 a Tnetwork1Tideal 1 Z 1 a 0 1 0 9 0 a 1 a 0 Z1 a 1 (16) a The ABCD matrix for Fig. 2b is: TFig 2b TidealTnetwork 2 a 0 1 Z a aZ 2 2 1 1 0 a 0 1 0 a (17) If these two-port networks are to be equivalent, then it must be the case that: TFig 2a TFig 2b (18) Or a 0 Z1 a aZ 2 a 1 1 0 a a (19) Eq. (19) says that equivalence requires Z1 aZ 2 a (20) which means that: 1 Z 1 a Z 2 and Z 2 2 Z1 a 2 (21) Equation (21) is important for transformers. It says that you can obtain equivalent networks by shifting impedances from one side to another according to these relations. 10 The process of moving impedances from one side of the transformer to another, while retaining an equivalent network, is called “referring” impedances. So we can “refer” impedances from primary to secondary. But how do you remember whether to multiply by a or 1/a? Remember that: a N1 N2 (22) Rewriting eqs. (21) we get: 2 2 N1 N2 Z 1 Z1 Z2 Z 2 and N2 N1 (23) we see that the side to which we refer corresponds to the multiplying turns (with the dividing turns being the side from which we refer). Looked at another way, we see: Z1 N 1 Z2 N 2 2 (24) 11 which indicates that an impedance (in ohms) has a larger numerical value on the side with the larger turns. So if you are looking to refer an impedance from the 13.8 kV side to a 161 kV side, that impedance will get larger, so you multiply by (161/13.8)2. 3.0 “Exact” model The circuits we have been addressing are actually approximate models of single phase (or per-phase) transformers, where we use an ideal transformer together with a series impedance Z1 (or Z2). The series impedance represents two things: Winding resistance Leakage flux, i.e., flux from the coil that “leaks” out of the transformer core. To better understand the leakage flux, write the total flux linkages with coil 1 as: 12 1 l 1 N 1 m (25) where λl1 is the leakage flux linkage from coil 1 with coil 1, and φm is the mutual flux that links all turns of both coils. The voltage induced in coil 1 will be, by Faraday’s law, d1 dl 1 dm N1 dt dt dt (26) To account for the voltage drop due to winding resistance, we insert an “r1”. Then the total voltage across the winding terminals is given by dl 1 dm v1 r1i1 N1 dt dt (27) Note that only the third term affects the voltage seen at the other (secondary) coil. 13 The second term has most of its flux in air, a linear medium, and therefore we have: l 1 Ll 1i1 (28) So that: di1 dm v1 r1i1 Ll 1 N1 dt dt (29) A similar situation is observed on the secondary side with respect to winding resistance and leakage reactance. The third term also is caused by a current, but not i1. This is evident by the fact that if we impress a voltage across the primary winding but leave the secondary open circuited (so that i1=(N2/N1)i2=0 because i2=0), we will still be able to measure a voltage across the secondary winding. 14 In reality, we can also measure a current into the primary winding, but a small current. We call this current the magnetization current, denoted im, the current necessary to set up the magnetic field. We model the impedance associated with this open-circuit condition with a large shunt inductance denoted as Lm in parallel with the (ideal) winding, so that the corresponding voltage in coil 1 is given by dm dim N1 Lm dt dt (30) Substituting eq. (30) into eq. (29), we obtain: di1 dim v1 r1i1 Ll 1 Lm dt dt The circuit conforming to eq. (31) is: 15 (31) I1 V1 Ideal network Z1 Lm E1 N1 I2 Z2 E2 V2 N2 Fig. 3 Note that we have also modeled the impedance Z2 on the secondary side. The two impedances Z1 and Z2 are given by: Z1 r1 jLl 1 Z 2 r2 jLl 2 (32) We will see how to REFER the secondary impedance to the primary side in the next set of notes. [1] K. Su, “Fundamentals of Circuits, Electronics, and Signal Analysis,” Houghton Mifflin Company, 1978. 16