LECTURE 31: Basics of Magnetically Coupled Circuits Part 1

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1 LECTURE 31: Basics of Magnetically Coupled Circuits Part 1 1 Question 1: What is a magnetically coupled circuit? Answer: voltages and currents of two nearby inductors affect each other when excited by a sinusoidal current. Question 2: Right! Can you be more rayspecific Professor. Answer: OK then. A changing current (non-constant derivative) entering inductor L 1 induces a voltage across its neighboring inductor L 2 according to a formula v 2 = M di 1 where M is a measure of their coupling a measure of dt how neighborly they really are. Question 3: Am taking marriage and the family 101 so would these two inductors become co-dependent? Answer: exactly, but they still maintain their own independent identities unlike the co-dependency illness you learn about in M&F 101. Question 4: Sew what? Answer: changing currents in neighboring and neighborly inductors induce voltages in each other in proportion to their degree of coupling, denoted by

2 Lecture 31 Sp 15 2 R. A. DeCarlo M, with polarities determined by how the coils of wire are wound relative to each other. Question 5: And how are we supposed to know how the coils are wound relative to each other? Answer: Using the dot convention which unfortunately does not take place in LasVegas, but on your HW and test problems?? Question 6: What are some applications of magnetically coupled circuits? (a) Transformers hanging on poles with high voltage lines attached; they step voltages down for household use. (b) Isolation between circuits; they were once used in audio to isolate speakers from tube amplifiers, back in the time of Edison, maybe Columbus. (c) Electric drives or motors that haul coal across the rockies. (d) Tuning in AM radios broad and narrow band bandpass filter designs. Question 7: Some former 202 students told me I better learn the dot convention. Can you tell me how to connect the dots? Ha ha. Well, what is the dot convention? Answer 1: A CURRENT entering the dotted terminal of one coil (say P) induces an open circuit VOLTAGE at the terminals of the neighboring coil (say S) whose positive voltage reference direction is with respect to the dotted terminal (on S).

3 Lecture 31 Sp 15 3 R. A. DeCarlo Now let s flip the polarity on the secondary voltage measurement which we take independent of Mr. Coupled Inductors. Ahhhh, the value of the measurement becomes negative. Answer 2 (Property Definition): A CURRENT entering the UNdotted terminal of one coil (say P) induces an open circuit VOLTAGE at the terminals of the neighboring coil (say S) whose positive voltage reference direction is with respect to the UNdotted terminal (on S). Exhibit 1 for the Defense:

4 Lecture 31 Sp 15 4 R. A. DeCarlo Exhibit 2 for the Offense: Question 8: OK, enough of the dot your eyes stuff what are the equations you are always telling us we have to know the equations? Answer: Consider the two neighborly inductors, L 1 and L 2, in the figure having a coupling inductance M which is hiding behind a white appearance, but it is there. One salient characteristic of the circuit is that the dots on the

5 Lecture 31 Sp 15 5 R. A. DeCarlo secondary (i.e., L 2 ) are labeled as A or B meaning that the equations are going to be of the what if form: what if dot is at A and what if dot is at B. SEW: (i) The unmarried relationships: v k = L k di k dt (assuming passive sign convention) hold (each inductor maintains its own identity). (ii) The induced voltage by the neighborly neighbor inductor is: v 1 = ± M di 2 dt and v 2 = M di 1 dt. (III) CONCLUSION: By the Principle of Superposition the two effects combine (Oh Deere) L 1 and L 2 in which case: di v 1 = L 1 1 dt ± M di 2 dt v 2 = ± M di 1 dt + L 2 di 2 dt

6 Lecture 31 Sp 15 6 R. A. DeCarlo Question 9: And in the s-domain Professor Ray????? ASSUMING ZERO IC s and the passive sign convention on each inductor, Laplacetransform the time domain equations to obtain: V 1 = L 1 si 1 ± MsI 2 = ± MsI 1 + L 2 si 2 having the matrix form V 1 = L 1 s ± Ms ± Ms L 2 s I 1 I 2 5. Question 10: Well professor Ray, maybe it s time for an example??? It s not that we don t like theory, but examples are helpful. You do understand that don t you Professor Ray? Raysponse: Hmmmm. Smart donkeys you are. Example 1: Consider the circuit below in which there is no internal stored energy. Let, L 1 = 1 H, L 2 = 2 H, M = 1 H, and i 1 (t) = cos(2t)u(t) A in which case I 1 = s s Find v 2 (t).

7 Lecture 31 Sp 15 7 R. A. DeCarlo Step 1. Coupled inductor equations. Clearly, V 1 = s s s 2s I 1 I 2 Step 2. Terminal constraint. The 4 Ω resistor constrains the voltage and current of the secondary. Specifically, = 4I 2 or equivalently I 2 = Step 3. Hence, After some algebra, in which case = si 1 + 2sI 2 = = 2s 2 (s + 2)(s 2 + 4) = 1 s s2 s s s s s 2 + 4

8 Lecture 31 Sp 15 8 R. A. DeCarlo v 2 (t) = e 2t u(t) + cos(2t)u(t) sin(2t)u(t) V Exercise. Identify the steady state and transient parts of this zero-state response. Example 2: Consider the circuit of example 1 above in which there is no internal stored energy. Again let L 1 = 1 H, L 2 = 2 H, and M = 1 H. Find Z in = V 1 I 1. Step 1. Equation for V 1 is: V 1 = si 1 + si 2. Step 2. Find I 2 in terms of I 1. This time we use the terminal condition differently. This time: implies = 4I 2 = si 1 + 2sI 2 I 2 = 0.5s s + 2 I 1 Step 3. Combine ingredients and blend. imples V 1 = si 1 + si 2 = si 1 0.5s2 s + 2 I 1

9 Lecture 31 Sp 15 9 R. A. DeCarlo Z in = V 1 = 0.5s2 + 2s I 1 s + 2 = 0.5 s(s + 4) s + 2

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12 Lecture 31 Sp R. A. DeCarlo Worksheet 2. Find Z in = for the configuration below. Step 1. Write the coupled inductor equations in the s-world. Insert appropriate signs in the equations below: V 1 = L 1 si 1 MsI 2 = MsI 1 L 2 si 2 Step 2. (i) = (ii) Determine I 2 in terms of I 1 using second equation of step 1. = = MsI 1 L 2 si 2 implies I 2 = I 1 Step 3. Find V 1 in terms of I 1 by substituting. Then determine Z in.

13 Lecture 31 Sp R. A. DeCarlo Worksheet 3. Find Z in = V in I in. 1. V in = V 1 ±?? 2. I 1 = ±?? I in 3. I 2 = ±?? I in 4. V 1 = ±?? L 1 si 1 ±?? MsI 2 = 5. = ±?? MsI 1 ±?? L 2 si 2 = si in si in 6. V in = si in 7. Z in =

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