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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 L1 induces a voltage across its neighboring inductor L2 according to
a formula v2 = M
di1dt
where M is a measure of their coupling—a measure of
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
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).
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:
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, L1 and L2 , 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
Lecture 31 Sp 15 5 © R. A. DeCarlo
secondary (i.e., L2 ) 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: vk = Lk
dikdt
(assuming passive sign
convention) hold (each inductor maintains its own identity).
(ii) The induced voltage by the neighborly neighbor inductor is:
v1 = ±M
di2dt
and v2 = M
di1dt
.
(III) CONCLUSION: By the Principle of Superposition the two effects
combine (Oh Deere) L1 and L2 in which case:
v1 = L1
di1dt
± Mdi2dt
v2 = ±M
di1dt
+ L2di2dt
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, Laplace-
transform the time domain equations to obtain:
V1(s) = L1sI1(s) ± MsI2(s)V2(s) = ±MsI1(s)+ L2sI2(s)
having the matrix form
V1(s)
V2(s)
⎡
⎣⎢⎢
⎤
⎦⎥⎥=
L1s ±Ms
±Ms L2s
⎡
⎣⎢⎢
⎤
⎦⎥⎥
I1(s)
I2(s)
⎡
⎣⎢⎢
⎤
⎦⎥⎥
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, L1 = 1 H, L2 = 2 H, M = 1 H, and i1(t) = cos(2t)u(t) A in which
case I1(s) = s
s2 + 4. Find v2(t) .
Lecture 31 Sp 15 7 © R. A. DeCarlo
Step 1. Coupled inductor equations. Clearly, ☺
V1(s)
V2(s)
⎡
⎣⎢⎢
⎤
⎦⎥⎥= s s
s 2s⎡
⎣⎢
⎤
⎦⎥
I1(s)
I2(s)
⎡
⎣⎢⎢
⎤
⎦⎥⎥
Step 2. Terminal constraint. The 4 Ω resistor constrains the voltage and
current of the secondary. Specifically, V2(s) = −4I2(s) or equivalently
I2(s) = −0.25V2(s) .
Step 3. Hence,
V2(s) = sI1(s)+ 2sI2(s) = s2
s2 + 4− 0.5sV2(s)
After some algebra,
V2(s) = 2s2
(s+ 2)(s2 + 4)= 1
s+ 2+ s
s2 + 4− 2
s2 + 4
in which case
Lecture 31 Sp 15 8 © R. A. DeCarlo
v2(t) = e−2tu(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 L1 = 1 H, L2 = 2 H, and M = 1 H. Find
Zin(s) =
V1(s)I1(s)
.
Step 1. Equation for V1(s) is: V1(s) = sI1(s)+ sI2(s) .
Step 2. Find I2(s) in terms of I1(s) . This time we use the terminal condition
differently. This time:
V2(s) = −4I2(s) = sI1(s)+ 2sI2(s)
implies
I2(s) = − 0.5s
s+ 2I1(s)
Step 3. Combine ingredients and blend.
V1(s) = sI1(s)+ sI2(s) = sI1(s)− 0.5s2
s+ 2I1(s)
imples
Lecture 31 Sp 15 9 © R. A. DeCarlo
Zin(s) =
V1I1
= 0.5s2 + 2ss+ 2
= 0.5 s(s+ 4)s+ 2
Lecture 31 Sp 15 10 © R. A. DeCarlo
Lecture 31 Sp 15 11 © R. A. DeCarlo
Lecture 31 Sp 15 12 © R. A. DeCarlo
Worksheet 2. Find
Zin(s) = for the configuration below.
Step 1. Write the coupled inductor equations in the s-world. Insert
appropriate signs in the equations below:
V1(s) = L1sI1 MsI2
V2(s) = MsI1 L2sI2
Step 2. (i) V2(s) =
(ii) Determine I2(s) in terms of I1(s) using second equation of step 1.
V2(s) = = MsI1 L2sI2 implies I2(s) = I1(s)
Step 3. Find V1(s) in terms of I1(s) by substituting. Then determine Zin(s) .
Lecture 31 Sp 15 13 © R. A. DeCarlo
Worksheet 3. Find Zin(s) =
VinIin
.
1. Vin =V1 ±?? V2
2. I1 = ±?? Iin
3. I2 = ±?? Iin
4. V1 = ±?? L1sI1 ±?? MsI2 = ⎡
⎣⎤⎦sIin
5. V2 = ±?? MsI1 ±?? L2sI2 = ⎡
⎣⎤⎦sIin
6. Vin = ⎡
⎣⎤⎦sIin
7. Zin(s) = ______________
Lecture 31 Sp 15 14 © R. A. DeCarlo