Losses in TransformersIdeal / Real Transformer

Some common applications

Some common issues with transformers

ELEC 3105 BASIC EM AND POWER

ENGINEERING

TRANSFORMER LOSSES

There are two dominant loss mechanisms in transformers

Eddy currents Hysteresis

Lenz’s Law

ma g n etB

lo o pB

Move loop

dBvemf motional emf

The induced emfalways opposes the change in

flux

v

•Move loop towards magnet•B increases in loop•Flux increases in loop•Current induced•Current produces magnetic field in loop•This magnetic field in opposite direction to magnetic field of magnet

I

The changing flux induces an emf about the loop and as a result a current flow in the loop.Provided the loop resistance is not infinite “or zero” power loss will be observed through Joule’s heating. RI 2

The changing magnetic field on the conducting sheet will induce current in the sheet. These currents are known as Eddy currents. Energy loss occurs through Joule’s heating.

EDDY CURRENTS

B increasing on sheetI direction such as to create B in opposite direction to magnets B as expected from

Lenz’s law.

Magnet slides along sheet. In the direction of motion the electrons of the sheet are

first introduced to an increasing magnetic field. I induced opposes this increase. Once the magnet has passed over the

electrons, they will then be subjected to a decreasing magnetic field and as such will

induce a current which attempts to reinforce the dropping external magnet

field.

EDDY CURRENTS

The Eddy Current Story • Suppose a current i1 in the primary is increased• Magnetic field increases• Flux in the core increases• This induces an emf in the core• This sets up a counter current which tries to keep B constant• These currents are Eddy Currents.

The Eddy Currents heat the core. This heat energy must be extracted from the electric energy supplied by the primary.

EDDY CURRENTS

Eddy currents are reduced by lamination of the core.

Lamination breaks the Eddy Current paths

EDDY CURRENTS

Eddy current loss

Eddy current losses occur whenever the core material is electrically conductive. Most ferromagnetic materials contain iron: a metal that has fairly low resistivity (roughly 10-7 Ω m). The problem is intuitively obvious if you consider that the magnetic field is contained within a 'circuit' or loop formed by the periphery of the core in the same way as it is contained within a turn on the windings. Around that periphery a current will be induced in the same way as it is in an ordinary turn which is shorted at its ends. What is needed, then, is some method of increasing the resistance of the core to current flow without inhibiting the flow of magnetic flux. In mains transformers this is achieved by alloying the iron with about 3% of silicon. This lifts the resistivity to 4.5×10-7 Ω m. Dependingupon the amount of silicon this material is called'transformer iron', 'electrical iron' or 'armature iron'.The alternative name 'silicon steel' is a misnomerbecause steel is iron alloyed with carbon; and carbondoes no good in a transformer core. The silicon does,though, increase the mechanical hardness of iron in the same way as carbon - try sawing up a transformer core and you'll discover this quickly.

EDDY CURRENTS

Eddy current loss

In any resistive circuit the power is proportional to the square of the applied voltage. The induced voltage is itself proportional to f × B and so the Eddy losses are proportional to f2B2. The flux is also related to the size of the loop. Figure PLM shows how the idea of lamination is used to reduce the power losses caused by eddy currents in mains transformers. The same principle applies to motors and generators too. Using a solid iron core (as in cross-section B) results in a large circulating current. So, instead, the core is made up of a stack of thin (~0.5 millimeter) sheets (cross section C). Here I have shown only four laminations but there will normally be many more. The lines of magnetic flux can still run around the core within the plane of the laminations. The situation for the Eddy currents is different. The surface of each sheet carries an insulating oxide layer formed during heat treatment.This prevents current from circulating from one lamination across to its neighbors.

EDDY CURRENTS

Eddy current loss

Clearly, the current in each lamination will be less than the very large current we had with the solid core; but there are more of these small currents. So have we really won? The answer is yes, for two reasons.

Power loss (the reduction of which is our aim) is proportional to the square of induced voltage. Induced voltage is proportional to the rate of change of flux, and each of our laminations carries one quarter of the flux. So, if the voltage in each of our four laminations is one quarter of what it was in the solid core then the power dissipated in each lamination is one sixteenth the previous value. Hurrah!

But wait; it gets better. Look at the long thin path that the Eddy current takes to travel round the lamination. Suppose we made the laminations twice as thin (we halved d1). The path length of the current

isn't much changed; it's still about 2×d2. However, the width of the path has halved and therefore its

resistance will double and so the current will be halved. The bottom line is that Eddy current loss is inversely proportional to the square of the number of laminations.

EDDY CURRENTS

This idea of dividing up the iron into thin sections is carried a stage further in the iron dust cores. Here the iron is ground into a powder, mixed with some insulating binder or matrix material and then fired to produce whatever shape of core is required. These cores can function at several megahertz but their permeability is lower than solid iron.

Note on expression for loss

RI 2

JE2E

22Bf

When alternating field is present at frequency f

fBdt

dBE

Then222 BfE

As the frequency increases so do the power losses. Can fix this using ferrites 22BfkP

e

EDDY CURRENTS

FerritesEDDY CURRENTS

Ferrites

EDDY CURRENTS

FerritesEDDY CURRENTS

FerritesEDDY CURRENTS

The following are examples of induced currents which are normally not identified as Eddy Currents

• Opposing magnetic force due to induced currents in linear motors.

• Braking action due to Eddy Currents in a pendulum.

• Opposing force due to induced currents in MHD generator.

• Induced currents in plasma engine.

• Currents due to back emf in a motor.

• Currents produced by generator.

EDDY CURRENTS

• Opposing magnetic force due to induced currents in linear motors.

EDDY CURRENTS

• Braking action due to Eddy currents in a pendulum.

EDDY CURRENTS

• Opposing force due to induced currents in MHD generator.

MHD = magnetohydrodynamic

EDDY CURRENTS

• Induced currents in plasma engine.EDDY CURRENTS

• Currents due to back emf in a motor

EDDY CURRENTS

EDDY CURRENTS

• Currents produced by generator.

EDDY CURRENTS

When a magnet is dropped down a metallic tube, the changing magnetic field created by the falling magnet pushes electrons in the metal tube around in circular, eddy-like currents. These eddy currents have their own magnetic field that opposes the fall of the magnet. The magnet falls dramatically slower than it does in ordinary free fall in a nonmetallic tube.

Material•   A cow magnet or neodymium magnet.

•   A nonmagnetic object, such as a pen or a pencil.

•   One 3 foot (90 cm) length of aluminum, copper, or brass tubing (do not use iron!) with an inner diameter larger than the cow magnet and with walls as thick as possible.

•   One 3 foot (90 cm) PVC or other nonmetallic tubing.

•   Optional: 2 thick, flat pieces of aluminum (available at hardware and home-repair stores); cardboard; masking tape; rubber bands or cord.

Assembly • None required.

Eddy CurrentsA magnet falls more slowly through a metallic tube than it does

through a nonmetallic tube.

EDDY CURRENTS

To do and notice:Hold the metal tube vertically. Drop the cow magnet through the tube. Then drop a nonmagnetic object, such as a pen or pencil, through the tube. Notice that the magnet takes noticeably more time to fall. Now try dropping both magnetic and nonmagnetic objects through the PVC tube.

In addition to dropping these objects through the tubes, a very simple, visible, and dramatic demonstration can be done by merely dropping the magnet between two thick, flat pieces of aluminum. The aluminum pieces should be spaced just slightly farther apart than the thickness of the magnet. A permanent spacer can easily be made with cardboard and masking tape if you don't want to hold the pieces apart each time. Rubber bands or cord can hold the pieces all together. The flat surfaces need to be only slightly wider than the width of the magnet itself. Thickness, however, is important. The effect will be seen even with thin pieces of aluminum, but a thickness of about 1/4 inch (6 mm) will produce a remarkably slow rate of fall. Allow at least a 6 inch (15 cm) fall.

As the magnet falls, the magnetic field around it constantly changes position. As the magnet passes through a given portion of the metal tube, this portion of the tube experiences a changing magnetic field, which induces the flow of eddy currents in an electrical conductor, such as the copper or aluminum tubing. The eddy currents create a magnetic field that exerts a force on the falling magnet. The force opposes the magnet's fall. As a result of this magnetic repulsion, the magnet falls much more slowly.

What's going on

EDDY CURRENTS

Eddy currents are often generated in transformers and lead to power losses. To combat this, thin, laminated strips of metal are used in the construction of power transformers, rather than making the transformer out of one solid piece of metal. The thin strips are separated by insulating glue, which confines the eddy currents to the strips. This reduces the eddy currents, thus reducing the power loss.

With the new high-strength neodymium magnets, the effects of eddy currents become even more dramatic. These magnets are now available from many scientific supply companies, and the price has become relatively affordable. (An excellent source is Dowling Miner Magnetics Corp., P.O. Box 1829, Sonoma, CA 95476. )

EDDY CURRENTS

Hysteresis Losses

ELEC 3105 BASIC EM AND POWER ENGINEERING

H y s t e r e s i sIn a previous lecture it was stated that the area of the

hysteresis loop represents the energy dissipated as heat.

49

Ferromagnetic materialsSOFT and HARD Ferromagnetic materials

Soft ::: transformer cores, solenoids, ….Hard :: permanent magnets

Area of hysteresis loop is equivalent to energy lost in one cycle.

We want to find an expression for Wh.

The hysteresis power loss for one cycle can be expressed

as:

fWPhh

where f is the cycle frequency and Wh is the

energy dissipated in each cycle.

First consider the work which must be done by the power source supplying the primary to increase the magnetic field in the core by dB.

We shall use the model device a toroidal coil with an iron core.

dIII dHHH

dBBB

Increase the current in time interval dt

dttt

H y s t e r e s i s

Then by Lenz’s law an electromotive force will be

induced in the winding tending to oppose the increase in current.

To increase the current the generator must furnish energy in the amount of.

dt

dNV

NIdVIdtW

Taking B as constant over the cross-section of the toroid core, an expression for the flux can be written.

AdBd

BA

H y s t e r e s i s

dIII dHHH dBBB

dttt

From Ampere’s law we can obtain an expression for the magnetic

field inside the toroid core.

To increase the current the generator must furnish energy in the amount of.

dIII dHHH dBBB

dttt

r

NIH

2

rAHdBW 2

where the volume of the toroid is:

rAvol 2

H y s t e r e s i s

Now we can sum “integrate” over the initial magnetization part of the hysteresis curve in order to obtain the work done by the generator in establishing the maximum magnetic flux density in the core Bmax

max

0

B

HdBvolW

This integral is the shaded area of the above curve multiplied by the volume “vol” of the toroid’s core.

H y s t e r e s i s

The work done in one cycle, Wh, is the striped area of the hysteresis cycle. The cross hatched area is the work returned to the source.

Consider area 0abcIn order for the external source to increase the H

field from a to a value corresponding at b, it must do work on the ferromagnetic specimen equivalent to the area 0abc.

Consider area bcdWhen H is reduced back to zero, the ferromagnetic

material will do work on the source equivalent to the area bcd, because H remains positive while dB changes sign.

max

0

B

HdBwvol

W

H y s t e r e s i s

49

Ferromagnetic materialsSOFT and HARD Ferromagnetic materials

Soft ::: transformer cores, solenoids, ….Hard :: permanent magnets

Area of hysteresis loop is equivalent to energy lost in one cycle.

max

0

B

HdBwvol

W

h

W Area of Hysteresis loop

ffWPhh

Power loss

Area of Hysteresis loop

H y s t e r e s i s

FERROMAGNETIC MATERIALS

H y s t e r e s i s

START Losses in Transformers

again !!!!

ELEC 3105 BASIC EM AND POWER ENGINEERING

There are two dominant loss mechanisms in transformer

Eddy currents Hysteresis

Total core loss

22 fBkfWPheh

k constant that depends on geometry

22 fBkPe

fWPhh

TRANSFORMER LOSSES

Ideal / Real Transformer

ELEC 3105 BASIC EM AND POWER ENGINEERING

IDEAL TRANSFORMER

What would the ideal transformer look like?

• NO LOSS

• PERFECT FLUX COUPLING

• INFINITE CORE PERMEABILITY

MLL ,,21

• TRANSFORMER DRAWS NO CURRENT WHEN SECONDARY IS OPEN

• PERFECT IMPEDANCE TRANSFORMATION AT ALL FREQUENCIES

Sorry: We must deal with real transformers.

REAL TRANSFORMER

Consider:

11 Lk

1v

2v

Ideal transformer

2

1 Lk

1i 2

i

Leakage inductance

1kL

Magnetization inductance

ai

2

aii 2

1

2111

12

11111

MijiLjv

kLjaiiLkjiv

Equivalent circuit reproduces basic transformer equation

REAL TRANSFORMER

Add losses:

1

1 Lk

1v

2v

Ideal transformer

Resistance of primary

Eddy currents and Hysteresis losses in core

ai

2

Resistance of secondary

TRANSFORMER EFFICIENCY

Ideal transformer 𝑃𝑜𝑢𝑡=𝑃 𝑖𝑛 Efficiency = 100 % 𝜂=100 %

Real transformer 𝑃𝑜𝑢𝑡≠ 𝑃 𝑖𝑛 Efficiency < 100 % 𝜂<100 %

𝑃 𝑖𝑛=𝑃𝑜𝑢𝑡+𝑃𝑐+𝑃𝑒

100%Copper loss in winding resistance

Eddy current & core loss

Real transformer

VOLTAGE REGULATION

Observed voltage drop when we draw a current from the secondary

100%

Real transformer

42

ELEC 3105 BASIC EM AND POWER ENGINEERING

Transformer core saturation

43

CORE SATURATION

HysteresisIn Lecture 21 it was stated that the area of the hysteresis loop represents the energy dissipated as heat.

49

Ferromagnetic materialsSOFT and HARD Ferromagnetic materials

Soft ::: transformer cores, solenoids, ….Hard :: permanent magnets

Area of hysteresis loop is equivalent to energy lost in one cycle.

The hysteresis power loss for one cycle can be expressed as:

fWPhh

where f is the cycle frequency and Wh is the energy dissipated in each cycle.

We want to find an expression for Wh.

At a sufficiently high H the core saturates and B is essentially constant.

Recall the Hysteresis curve

B versus H

The usual design principle is to have B = Bsat at the voltage peaks in the primary

(minimizes the amount of iron needed in the core)

44

dt

dv 1

1

Recall for the primary that:

dt

dv max,1

max,1

Then at maximum

ABvsatpeak

,1

Frequency

Area of the core

CORE SATURATION

45

If v1 goes beyond this range, however, H is above Hsat, and the effective inductance seen at the primary becomes small. The flux is also not well-confined to the core.

ABvsatpeak

,1

Because of the reduced inductance, the current in the primary becomes large in these peak parts of the cycle.

L

1i t

CORE SATURATION

46

If v1 goes beyond this range, however, H is above Hsat, and the effective inductance seen at the primary becomes small. The flux is also not well-confined to the core.

ABvsatpeak

,1

The voltage in the secondary also ceases to be a clean sine wave, and acquires harmonics.

L

2v t

CORE SATURATION

47

CORE SATURATION

HysteresisIn Lecture 21 it was stated that the area of the hysteresis loop represents the energy dissipated as heat.

49

Ferromagnetic materialsSOFT and HARD Ferromagnetic materials

Soft ::: transformer cores, solenoids, ….Hard :: permanent magnets

Area of hysteresis loop is equivalent to energy lost in one cycle.

The hysteresis power loss for one cycle can be expressed as:

fWPhh

where f is the cycle frequency and Wh is the energy dissipated in each cycle.

We want to find an expression for Wh.

At a sufficiently high H the core saturates and B is essentially constant.

Recall

The saturation problem is also the reason for not using very low frequencies. The impedance in the primary becomes small as the frequency becomes small, so that the current becomes large and the core saturates.

LjZL

48

ELEC 3105 BASIC EM AND POWER ENGINEERING

Transformer and the power grid

49

TRANSFORMER AND THE POWER GRID

Typical power requirement for a small city: 1000MW

Maximum voltage provided by typical generator: 30 KV

Thus need a current of 3.3*104 A

Suppose generator is 100 km from city.

Copper transmission line 2.5 cm in radius has resistivity

cma 5.2

km100

cm 7.1

50

TRANSFORMER AND THE POWER GRID

Copper transmission line 2.5 cm in radius has resistivity

cma 5.2

km100

cm 7.1

86.02aA

R

Resistance over length of transmission line length

Power loss in wire conduit

Need very large core to handle this current without saturation.

MWRI 9502 Leaves about 50 MW for city.

51

TRANSFORMER AND THE POWER GRID

Solution is to step up the voltage and thus reduce the current and power loss in copper conduit.

cma 5.2

km100

Typical AC high voltage line: 765 kV

Current required: I = 1000 A

Power loss in wire conduit MWRI 86.02 Leaves about 999.14MW for city.

52

GEOMAGNETIC STORMS AND CORE SATURATION

1989 http://www.geolab.emr.ca/geomag/e_gic_history.html - year1989

A great magnetic storm occurred on 13 March 1989, that caused a nine-hour blackout of the 21,000 MV Hydro Québec, power system. A vivid description of that failure has been provided by G. Blais and P. Metsa (1993) of Hydro Québec:

"Telluric currents induced by the storm created harmonic voltages and currents of considerable intensity on the La Grande network. Voltage asymmetry on the 735-kV network reached 15%. Within less than a minute, the seven La Grande network static var compensators on line tripped one after the other....With the loss of the last static var compensator, voltage dropped so drastically on the La Grande network (0.2 p.u.) that all five lines to Montréal tripped through loss of synchronism (virtual fault), and the entire network separated. The loss of 9,450 MV of generation provoked a very rapid drop in frequency at load-centre substations. Automatic underfrequency load-shedding controls functioned properly, but they are not designed for recovery from a generation loss equivalent to about half system load. The rest of the grid collapsed piece by piece in 25 seconds."

Many other power utilities in North America experienced problems ranging from minor voltage fluctuations to tripping out of lines and capacitors. A summary of these effects and the times of their occurrence is given by Cucchi and Ponder (1991).

53

GEOMAGNETIC STORMS AND CORE SATURATION

Auroral electrojet 106 A

54

GEOMAGNETIC STORMS AND CORE SATURATION

Auroral electrojet 106 A

High voltage line

B

km100Generating station

Consuming stationkm1000

Auroral electrojet fluctuates by 105 A in one minute

Tm

A

r

IB oo 6

3

5

106.6101502

10

2

Change in flux through circuit

mmT 336 10100010100106.6

Vt

emf 110

55

GEOMAGNETIC STORMS AND CORE SATURATION

Auroral electrojet 106 A

High voltage line

B

km100Generating station

Consuming stationkm1000

Auroral electrojet fluctuates by 105 A in one minute

At this frequency transformer coils look like short circuitsVt

emf 110

1R

A DC current of

is produced A100

56

GEOMAGNETIC STORMS AND CORE SATURATION

Geomagnetic Effects on Power Systems - English

Geomagnetic Storms

Sun

Sun article

57

97.315 BASIC EM AND POWER ENGINEERING

Lecture 27

END Transformer core saturation

START Transformer application

58

TRANSFORMER 120V AC TO 5V DC

LR

GR

Gv

21 NN

1 : 02

DCV 5

In order to convert AC line power to low voltage DC, a step down transformer is used first, followed by a rectifier bridge, followed by coarse filtering, followed by limiting.

5 9

A Thin Film Piezoelectric Transformer for Microelectronic Applications

With the onset of miniaturization, many applications in the electronics industry now require small, low profile components with a high efficiency of operation. Electromagnetic transformers, having thousands of wire turns around a ferrite core, have become an obstacle to the progress of miniaturization, as they are among the most bulky devices on a circuit board. Piezoelectric transformers have recently received some attention as a possible alternative. A piezoelectric transformer essentially consists of two mechanically coupled and electrically insulated piezoelectric resonators (Fig. 1). When an electrical signal near the frequency of mechanical resonance is applied to the input section of the transformer, strong mechanical vibration occurs due to the piezoelectric converse effect. This vibration is transferred to the output section, inducing a voltage on its electrodes due to the piezoelectric direct effect, with a consequent voltage gain.

It is now possible to manufacture piezoelectric films of sufficient thickness and quality to make thin film piezoelectric transformers through the use of a composite film technology developed here at Queen's University. A ring transformer with vibrational motion in the radial direction has been established as the best candidate for a transformer through use of Mason's model for piezoelectric transformers, predicting a operating range of 1-10 MHz with voltage gains of 0.1-100 and a maximum efficiency of operation of 90% (depending on the dimensions and quality of the film). Current work has been focused on optimizing the material parameters and the poling procedure of the piezoelectric composite films, as well as developing a method of characterization for the piezoelectric parameters.

60

Slides not used after this one

6 2

TRANSFORMER APPLICATIONS

• The transformer is usually the most efficient way to convert AC voltages.

• This can be done by impedance matching.

• Impedance matching is the term used for achieving maximum power transfer to a given load from a generator.

• Quite often the generator is set as is the load.

• For maximum power transfer we can use a transformer to convert the power from the generator to the load as shown below.

Consider:

1v

2v

1i

2i

LR

GR

Gv

21 NN

63

TRANSFORMER APPLICATIONS

Consider:

1v

2v

1i

2i

LR

GR

Gv

21 NN

We have

1

1

2

2v

N

Nv and

LLR

v

N

N

R

vi 1

1

22

2

Assuming large self inductance,

we can write

LR

v

N

Ni

N

Ni 1

2

1

2

2

1

2

1

We also know that

from Ohm’s law

G

G

R

vvi 1

1

equate

64

TRANSFORMER APPLICATIONS

Consider:

1v

2v

1i

2i

LR

GR

Gv

21 NN

G

G

LR

vv

R

v

N

N11

2

1

2

Thus

G

G

GLR

vv

RRN

N

1

2

1

211

LG

LG

RRNN

Rvv

2

1

2

1

In terms of this is ……

2v

65

TRANSFORMER APPLICATIONS

Consider:

1v

2v

1i

2i

LR

GR

Gv

21 NN

LG

LG

RNN

RNN

Rvv

2

1

1

2

2

In terms of this is ……

2v

1

1

2

2v

N

Nv

We want to design a transformer to have a maximum power transfer to the load.

Thus:

maximum2

2

2 L

L R

vP

and …….

66

TRANSFORMER APPLICATIONS

Consider:

1v

2v

1i

2i

LR

GR

Gv

21 NN

2

2

1

1

22

1

LG

LG

L

L

RNN

RNN

Rv

RP

maximum2

2

2 L

L R

vP

…….

222

2

22

1

LLGG

GL

L RRkRRk

vkRP

2

1

2

N

Nkwith

67

TRANSFORMER APPLICATIONS

Consider:

1v

2v

1i

2i

LR

GR

Gv

21 NN

…….

4

22222

2

2200

LG

LGGGLLLGG

GL

L

RkR

RRkRvkRRRkRRkvR

dk

dP

2

1

2

N

Nkwith

Differentiating with respect to k and setting to zero will give the condition for maximum transfer to the load.

68

TRANSFORMER APPLICATIONS

Consider:

1v

2v

1i

2i

LR

GR

Gv

21 NN

GGLLG

GL RvkRRkRvR 2

2

20

2

1

2

N

Nk

with

Simplifying

22

222

GGLGvRvRkR

L

G

L

R

Rk gives

G

L

R

R

N

N 1

2

This is the transformer turns ratio which gives maximum power transfer to the load.

69

TRANSFORMER APPLICATIONSConsider:

1v

2v

1i

2i

LR

GR

Gv

21 NN

If we convert the load resistance to the apparent generator side resistance, we get:

3 4

P o w e r t r a n s f o r m a t i o nI n p u t a d m i t t a n c e

E q u i v a l e n t c i r c u i t f o r t h e p r i m a r y

1Lj LLL

RRN

NR

L

L 2

2

2

1

2

1

L

rms

R

vP

2

LR

vP

2

2

1

LR

NN

vNN

P

2

2

2

1

2

22

2

2

1

2

PR

vP

L

2

2

2

Lecture 25 Slide 34

LLRR

N

N 2

2

2

1 Gives

GL

L

G RRR

R

2 Maximum power transfer condition to load

GR

Gv GL

RR