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Chapter 13Magnetically Coupled Circuits

13.1 What is a transformer?13.2 Mutual Inductance

13.3 Energy in a Coupled Circuit13.4 Linear Transformers13.5 Ideal Transformers

13.6 Ideal Autotransformers13.8 Multisim Analysis13.9 Applications

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13.1 What is a transformer? (1) It is an electrical device designed on the basis of the

concept of magnetic coupling It uses magnetically coupled coils to transfer energy from

one circuit to another It is the key circuit elements for stepping up or stepping

down ac voltages or currents, impedance matching ,isolation , etc.

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Mutual Inductance When two inductors (or coils) are in close proximity

of each other, the magnetic flux caused by currentin one coil links with the other coil, thereby inducing

voltage in the latter.

3

13.2 Mutual Inductance (1)

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First consider a single inductor, a coil with N turns:

According to Faradays law, the voltage v inducedin the coil is proportional to the number of turns Nand the time rate of change of magnetic flux :

4


When current i flows through thecoil, a magnetic flux isproduced around it.

t d

d N v

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Voltage induced in the coil: But the flux is produced by current i

so that any change in is caused by achange in the current:

Recall the voltage-current relationshipfor an inductor.

The inductance L of the inductor (Self inductance) is given by:

Self-inductance L relates the voltageinduced in a coil by a time-varying

current in the same coil. 5

t d

id Lv

t d

id

id

d N v


t d

d

N v

id

d N L

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Now consider two coils with self-inductances L 1 and L 2 that are in close proximity of each other:

The total magnetic flux 1

emanating from coil 1has two components:

11 links only coil 1 12 links both coils

6


Coil 1 has N 1 turns

Coil 2 has N 2 turnsAssume coil 2 carries no current.

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Although the two coils are physically separated,they are magnetically coupled.

The voltage induced in each coil is proportional tothe flux in each coil.

7


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t d

d N v

111

t d

d N v

1222

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The voltage equations can be rewritten as follows:

8


ECE 202 Ch 13

t d

d N v

1

11

t d

d N v

12

22

t d

id

id

d N v

1

1

111

t d

id Lv

111

t d

id

id

d N v

1

1

1222

t d

id M v

1212

L1 is the self-inductance of coil 1 M21 is the mutual inductance ofcoil 2 with respect to coil 1

v 2 is the open-circuit mutualvoltage (or induced voltage)across coil 2

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Suppose now we let current i 2 flow in coil 2 while

coil 1 carries no current:

The magnetic flux 2 emanating from coil 2comprises flux 22 that links only coil 2 and flux 21 that links both coils:

The resulting symmetry is true:

9

222


ECE 202 Ch 13

t d

id Lv

222

t d

id M v

2121

2

21112

id

d N M

L2 is the self inductance of coil 2M12 is the mutual inductance ofcoil 1 with respect to coil 2

v 1 is the open-circuit mutualvoltage (or induced voltage)across coil 1

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10

M12 = M21 = M - Mutual inductance between thecoils

As with self-inductance L, mutual inductance M ismeasured in Henrys

Mutual inductance only exists when inductors orcoils are in close proximity and the circuits aredriven by time-varying sources

Although mutual inductance M is always apositive quantity, the mutual voltage M di/dt maybe negative or positive, just like the self-inducedvoltage L di/dt

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11

Self-induced voltage polarity is determined by the

reference direction of the current and the referencepolarity of the voltage

The polarity of the mutual voltage is not as easy todetermine (depends on the winding direction of the coils)

We use the dot convention to determine

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13.2 Mutual Inductance (10)Dot Convention

12

If a current enters the dottedterminal of one coil, the referencepolarity of the mutual voltage inthe second coil is positive at thedotted terminal of the second coil.

Alternatively, if a current leaves thedotted terminal of one coil, thereference polarity of the mutualvoltage in the second coil isnegative at the dotted terminal ofthe second coil.

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13

Replace the dot with a dependent voltage source. I1 induces a voltage in the second coil (j MI1 term) while I 2 induces a

voltage in the first coil (j MI2 term). I1 enters L1 at the dotted end, then the polarity of the mutual voltage

(j MI1) in L2 is positive at the dotted terminal on top and negative at thebottom. Therefore, the controlled voltage source has a plus sign on top andminus sign at the bottom.

I2 leaves L2 at the dotted end, then the polarity of the mutual voltage in L 1 isnegative at the dotted terminal on top and positive at the bottom.Therefore, the controlled voltage source (j MI2) has a minus sign on topand plus sign at the bottom.

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13.2 Mutual Inductance (12)Recommended

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Watch these videos illustrating solving mutualinductance problems:http://www.youtube.com/watch?v=tD35a-uzd34 http://www.youtube.com/watch?v=hzU4XKQYTWw

http://www.youtube.com/watch?v=OqSvesTtnUo
http://www.youtube.com/watch?v=tD35a-uzd34http://www.youtube.com/watch?v=hzU4XKQYTWwhttp://www.youtube.com/watch?v=OqSvesTtnUohttp://www.youtube.com/watch?v=OqSvesTtnUohttp://www.youtube.com/watch?v=hzU4XKQYTWwhttp://www.youtube.com/watch?v=tD35a-uzd34http://www.youtube.com/watch?v=tD35a-uzd34http://www.youtube.com/watch?v=tD35a-uzd34

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15

)connectionaiding-(series

221 M L L L

)connectionopposing-(series

221 M L L L

Dot convention for coils in series; the sign indicates thepolarity of the mutual voltage; (a) series-aiding connection,(b) series-opposing connection.

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16

Example 13.1 (Textbook)

Calculate the phasor currents I 1 and I 2 in the circuit shownbelow.

A04.1491.2I A;39.4901.13I 21 Ans:ECE 202 Ch 13

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13.3 Energy in a Coupled Circuit (1)

17

The instantaneous energy w stored in the circuit is:

21

2

22

2

11

2

1

2

1i Mii Li Lw

The positive sign is selected for the mutual term if bothcurrents enter or leave the dotted terminals; the negativesign is selected otherwise.

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18

The coupling coefficient, k, is a measure of the magneticcoupling between two coils; 0 k 1.

21 L Lk M K = 1 coils perfectly coupledK < 0.5 coils loosely coupledK > 0.5 coils tightly coupled

1211

12

K

2221

21

K

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19

Example 13.3 (Textbook)Consider the circuit below. Determine the couplingcoefficient. Calculate the energy stored in the coupledinductors at time t = 1s if v=60cos(4t +30) V.

Ans: k =0.56; w(1) =20.73J

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13.4 Linear Transformers (1)

20

A transformer is generally a four-terminal device comprising two (ormore) magnetically coupled coils

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The coil directly connected to the voltage source is called the primarywinding.

The coil directly connected to the load is called the secondary winding. Resistances R1 and R2 are included to account for the losses in the coils. A transformer is said to be linear if the coils are wound on a magnetically

linear material for which the permeability is constant. Air, plastic, Bakelite, wood, etc. Most materials are magnetically linear.

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21

Obtain input impedance Z in as seen from the source:

ECE 202 Ch 13

Mesh 1:

Mesh 2:

2111 )( MI j I L j RV

2221 )(0 I Z L j R MI j L

1

22

2 IZ

I L L j R

M j

1

22

22

111 Z

)( I L j R

M I L j RV

L

From Mesh 2: Substituting into Mesh 1 gives:

11 I Z I Z V reflected primary

ZZ

22

22

L

R

L j R

M

Reflected Impedance

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22

Obtain input impedance Z in as seen from the source:

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Mesh 1:

Mesh 2:

21 MI j I Z V primary

21 )(0 I Z MI j Secondary

12II

Secondary Z

M j 1

22

1 I

Z

M I Z V

Secondary

primary

From Mesh 2: Substituting into Mesh 1 gives:

11 I Z I Z V reflected primary Z

22

Secondary

R

Z

M

Reflected Impedance

ZSecondary = the total seriesimpedance in the secondary loop

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23

The input impedance Z in can be broken into two parts as

follows:

ECE 202 Ch 13

reflected primaryin Z Z Z

L

in L j Z

M L j Z Z Z

22

22

11

Note: Z in will be the same if the dot on L2 is switched

Secondary

primaryin Z

M Z Z

22

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24

The input impedance Z in can be broken into two parts as

follows:

ECE 202 Ch 13

reflected primaryin Z Z Z

Note: Z in

will be the same if the dot on L2 is switched

Secondary

primaryin Z

M Z Z

22

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25

Example 13.4 (textbook)In the circuit below, calculate the input impedance and current I 1.

Take Z 1=60- j 100, Z 2=30+ j 40, and Z L=80+j60.

A1.1135.0I ;1.5314.100Z 1in Ans:

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26

Analyzing magnetically coupled circuits issomewhat challenging, so it is sometimesconvenient to replace a magneticallycouple circuit with an equivalent circuitwith no magnetic coupling.

ECE 202 Ch 13

We want to replace the linear transformer with anequivalent T or circuit that has no mutual inductance.

Circuit: T Circuit:

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13.4 Linear Transformers (8)Equivalent T Circuit:

27ECE 202 Ch 13

2

1

2

1

2

1

I

I

jwL M j

M j L j

V

V

Linear Transformer Circuit

Equivalent T Network

2

1

2

1

)(

)(

I

I

L L jw jwL

jwL L L j

V

V

cbc

cca

Equating the terms gives the following relationships:

La = L1 - M; Lb = L2 - M; Lc = M

Mesh Analysis of this transformer circuitresults in the following:

Mesh Analysis of the equivalent T networkresults in the following:

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13.4 Linear Transformers (9)Equivalent T Circuit (Swapped Dots):

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2

1

2

1

2

1

I

I

jwL M j

M j L j

V

V

Linear Transformer Circuit

Equivalent T Network

2

1

2

1

)(

)(

I

I

L L jw jwL

jwL L L j

V

V

cbc

cca

Equating the terms gives the following relationships:

La = L1 + M; Lb = L2 + M; Lc = -M

Mesh Analysis of this transformer circuitresults in the following:

Mesh Analysis of the equivalent T networkresults in the following:

13 4 Li T f (10)

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13.4 Linear Transformers (10)Equivalent Circuit:

29ECE 202 Ch 13

2

1

2

1

111

111

V

V

L j L j L j

L j L j L j

I

I

C BC

C C A

Similarly, for the network nodal analysis provides:

Equivalent Network

By equating terms in admittance matrices, for the equivalentnetwork we obtain:

M M L L; L

M L M L L ; L

M L M L L L C B A

2

21

1

2

21

2

2

21

M

M L L; L

M L

M L L ; L

M L

M L L L C B A

2

21

1

2

21

2

2

21

13 4 Li T f (11)

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13.4 Linear Transformers (11)Equivalent T or Circuits Summary:

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T Circuit

Circuit

T Circuit

Circuit

13 4 Li T f (12)

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13.4 Linear Transformers (12)Equivalent Circuit:

31ECE 202 Ch 13

Find I1 and I 2 using the T equivalent circuit:

0324012 2111 I I j I j I j

01233 2212 I I j I I j

Mesh I 1:

Mesh I 2:

0123 21 I j jI

06123 21 I j I j

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13.5 Ideal Transformers (1)

32

An ideal transformer has perfect coupling (k=1). It consists of two or more coils with a large number of

turns wound on a common core of high permeability.

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Because of the high permeability of the core, the fluxlinks all the turns of both coils, thereby resulting in aperfect coupling.

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13.5 Ideal Transformers (2)(Dots the same polarity)

33

Recall the coupled circuit:

ECE 202 Ch 13

2111 MI j I L jV

2212 I L j MI jV

(1)

(2)

Solving for I 1 in (1):

11

1

2

1

221

1

121

222 nV V L

L

L

I L L j

L

V L L I L jV

(3)

Substituting (3) into (2):

1

211I

L j

MI jV

21 L L M For perfect coupling (k=1):

1

2

2

1

1222

L

I M j

L

MV I L jV

12 nV V Therefore: where 12 L Ln = turns ratio

13 5 Id l T f (3)

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13.5 Ideal Transformers (3)(Dots opposite each other)

34

Mesh equations give the following:

ECE 202 Ch 13

2111 MI j I L jV

2212 I L j MI jV

(1)

(2)

Solving for I 1 in (1):

11

1

2

1

221

1

121

222 nV V L

L

L

I L L j

L

V L L I L jV

(3)

Substituting (3) into (2):

1

21

1I L j

MI jV

21 L L M For perfect coupling (k=1):

1

2

2

1

1

222 L

I M j

L

MV I L jV

12 nV V Therefore: If dot is swapped at output

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A transformer is said to be ideal if it has thefollowing properties:1. Coils have very large reactances (L 1, L2, M )2. Coupling coefficient is equal to unity (K=1)

3. Primary and secondary coils are lossless (R 1 = R2 = 0) An ideal transformer is a unity-coupled lossless

transformer in which the primary and secondarycoils have infinite self-inductances.

Iron core transformers are close approximations toideal transformers and are used in power systemsand electronics.

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36

When a sinusoidal voltage is applied to the primarywinding, the same magnetic flux goes through bothwindings.

ECE 202 Ch 13

Using the phasor voltages rather than theinstantaneous voltages:

t d

d N v

11

n N

N

v

v

1

2

1

2

t d

d

N v

22

;

Turns ratio ortransformation ratio

n N

N

V

V

1

2

1

2

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37

Power conservation:

The energy supplied to the primary must equal theenergy absorbed by the secondary, since there are nolosses in an ideal transformer.

In phasor form:

n=1 isolation transformer (V 2 = V1) n>1 step-up transformer (V 2 > V1) n

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38

Transformer ratings are usually specified as V 1 / V2

Power companies often generate at some convenientvoltage and use the step-up transformer to increasethe voltage so that the power can be transmitted atvery high voltage and low current over transmissionlines, resulting in significant cost savings. Nearresidential consumers, step-down transformers areused to bring the voltage down to 120 V.

It is important to get the proper polarity of thevoltages and the direction of the currents for thetransformer.

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39ECE 202 Ch 13

13.5 Ideal Transformer (8) Two rules to be followed for voltage polarity and current

direction:1. If V1 and V 2 are both positive or both negative at the

dotted terminals, use the +n. Otherwise, use the n.2. If I1 and I 2 both enter into or both leave the dotted

terminals, use n. Otherwise, use +n.

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40ECE 202 Ch 13

Expressing V 1 in terms of V 2 and I 1 in terms of I 2 or viceversa:

Complex Power is:

Complex power supplied to the primary is delivered to thesecondary without loss.

The ideal transformer is lossless and absorbs no power .

n

V V 21 12 nV V

n

I I 12

21 nI I

2*

22

*

22*

111 S I V nI n

V I V S

Complex Conjugate of I 2

13 5 Id l T f (10)

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13.5 Ideal Transformers (10)(Reflected Impedance)

41

The input impedance as seen by the source is:

The input impedance is also called the reflected

impedance since it appears as if the load impedanceis reflected to the primary side. The ability of the transformer to transform a given

impedance to another allows impedance matching toensure maximum power transfer.

ECE 202 Ch 13

2

2

2

2

1

1

n

Z

I n

V

I

V Z Lin

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42

In analyzing a circuit containing an ideal transformer,it is common practice to eliminate the transformer byreflecting impedances and sources from one side ofthe transformer to the other.

Suppose we want to reflect the secondary side of the

circuit to the primary side:

ECE 202 Ch 13

2

2

n

Z Z R

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43

We find the Thevenin equivalent of the circuit to the right of a-b: Obtaining V th from open circuit voltage:

Obtaining Z Th (remove the voltage source in the secondary andinsert a unit source at a-b terminals.)

ECE 202 Ch 13

n

V

n

V V V

V V

I I

s

Th

s

22

1

22

21 0 Since a-b is open

Equivalent CircuitReflecting the secondary to the primary

2

2

2

2

2

1

1

n

Z

I n

V

I

V Z Th

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44

The general rule for eliminating thetransformer and reflecting the secondary

circuit to the primary side: Divide the secondary impedance by n 2 Divide the secondary voltage by n Multiply the secondary current by n

The general rule for eliminating thetransformer and reflecting the primarycircuit to the secondary side:

Multiply the primary impedance by n 2 Multiply the primary voltage by n Divide the primary current by n

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13.5 Ideal Transformer (14)


An ideal transformer is rated at 2400/120V, 9.6 kVA, and has 50turns on the secondary side.

Calculate:(a) the turns ratio,(b) the number of turns on the primary side, and(c) the current ratings for the primary and secondary windings.

45

Ans:(a) This is a step-down transformer, n=0.05(b) N 1 = 1000 turns(c) I1 = 4A and I 2 = 80A

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13.5 Ideal Transformer (15)


For the ideal transformer, find: (a) the source current I 1, (b) theoutput voltage V 0, and (c ) the complex power supplied by thesource.

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13.6 Ideal Auto-Transformers (1)

47ECE 202 Ch 13

An autotransformer is atransformer in which both theprimary and the secondaryare in a single winding

A connection point called atap is separates the primaryand secondary.

The tap is often adjustable toprovide a desired turns ratio.

An adjustable tap provides a

variable voltage to the load A disadvantage of the

autotransformer is it providesno electrical isolation

Step Down Auto-Transformer

Step Up Auto-Transformer

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13.6 Ideal Auto-Transformers (2)

48ECE 202 Ch 13

The voltage / current relationships for the lossless ideal

autotransformer are as follows:

Step Down Auto-Transformer Step Up Auto-Transformer

1

21

22 V

N N

N V 1

1

212 V

N

N N V

1

21

12 I

N N

N I 1

2

212 I

N

N N I

Lin Z N

N N Z

2

2

21

Lin Z

N N

N Z

2

21

1

Similar toVoltage Divider

equation

Inverse Relation

Derive from V/I

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13.8 Multisim

ECE 202 Ch 13 49

REMEMBER:k=M/sqrt(L 1L2)-add ground!-freq=omega/2pi

The dot is always on the left-hand

terminal of the inductor when theinductor component L is placed(horizontally) without rotation onthe schematic. Thus the dot willbe at the top after 90 clockwiserotation and at the bottom after90 counterclockwise rotation.

For Magnetically Coupled Coils or Linear Transformer use:Multisim TRANSFORMER / INDUCTOR_COUPLING

For Ideal Transformer use:Multisim TRANSFORMER / 1P1S

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13.9 Applications (1)

50

Transformer as an Isolation Device to isolate ac supply

from a rectifier

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51

Transformer as an Isolation Device to isolate dc betweentwo amplifier stages.

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52

Transformer as a Matching Device

Using an ideal transformer to mat chthe speaker to the amplifier

Equivalent circuit

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Practice Problem 13.16 (Textbook)

Calculate the turns ratio of an ideal transformer

required to match a 400 load to a source withinternal impedance of 2.5k. Find the loadvoltage when the source voltage is 30V.

53

Ans: n = 0.4; V L = 6V

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54

A typical power distribution system

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Chapter 13 - Equation / Analysis Summary Series Aiding / Opposing Dot Convention Model Coupling coefficient k Linear Transformer

Input Impedance:

Reflected Impedance:

Equivalent T or Circuits:

Ideal Transformer K = 1, L1, L2, Lossless (S 1 = S2) Voltage / Current Relationship:

Dots same = +n, Dots diff = -n Complex Power:

Autotransformer Adjustable tap No electrical Isolation Voltage Divider like relationship

221 M L L L 221 M L L L

21 L Lk M

Secondary

primaryin Z

M Z Z

22

Secondary

reflected Z

M Z

22

12 nV V n

I I 12 2

n

Z Z Lin

1

21

22 V

N N

N V

1

2

212 I

N

N N I

2

*

22

*

111 S I V I V S Lrmsave

R I P 2

2

Documents

ECE202 Ch13-Fall2014