ELEC 3105 Basic EM and Power Engineering Losses in Transformers Ideal / Real Transformer Some common applications Some common issues with transformers

Transformer Losses There are two dominant loss mechanisms in transformers Eddy currents Hysteresis

Eddy Currents 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. 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.

Induction heating

Eddy Currents B increasing on sheet I 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. Depending upon the amount of silicon this material is called 'transformer iron', 'electrical iron' or 'armature iron'. The alternative name 'silicon steel' is a misnomer because steel is iron alloyed with carbon; and carbon does 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. 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.

Eddy Currents Note on expression for loss When alternating field is present at frequency f Then As the frequency increases so do the power losses. Can fix this using ferrites

Eddy Currents Ferrites

Eddy Currents Ferrites

Eddy Currents Ferrites

Eddy Currents Ferrites

Eddy 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. F = IBL

Eddy Currents Braking action due to Eddy currents in a pendulum.

Eddy Currents MHD = magnetohydrodynamic Opposing force due to induced currents in MHD generator. MHD = magnetohydrodynamic Like Hall Effect

Eddy Currents Induced currents in plasma engine.

Plasma accelerator

Eddy Currents Currents due to back emf in a motor

Eddy Currents Eddy Currents Currents produced by generator.

Eddy Currents Eddy Currents A magnet falls more slowly through a metallic tube than it does through a nonmetallic tube. 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 Currents What's going on 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. What's going on 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.

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. )

ELEC 3105 Basic EM and Power Engineering Hysteresis Losses

The hysteresis power loss for one cycle can be expressed as: In a previous lecture it was stated that the area of the hysteresis loop represents the energy dissipated as heat. The hysteresis power loss for one cycle can be expressed as: where f is the cycle frequency and Wh is the energy dissipated in each cycle. We want to find an expression for Wh.

H y s t e r e s i s 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. H We shall use the model device a toroidal coil with an iron core. Increase the current in time interval dt

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

H y s t e r e s i s 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. where the volume of the toroid is:

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 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 0abc In 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 bcd When 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.

H y s t e r e s i s Area of Hysteresis loop Volume of core Power loss

Ferromagnetic materials H y s t e r e s i s

ELEC 3105 Basic EM and Power Engineering START Losses in Transformers again !!!!

There are two dominant loss mechanisms in transformer Transformer Losses There are two dominant loss mechanisms in transformer Eddy currents Hysteresis Total core loss k constant that depends on geometry

ELEC 3105 Basic EM and Power Engineering Ideal / Real Transformer

Ideal Transformer What would the ideal transformer look like? NO LOSS PERFECT IMPEDANCE TRANSFORMATION AT ALL FREQUENCIES NO LOSS PERFECT FLUX COUPLING INFINITE CORE PERMEABILITY TRANSFORMER DRAWS NO CURRENT WHEN SECONDARY IS OPEN Sorry: We must deal with real transformers.

Equivalent circuit reproduces basic transformer equation Real Transformer Consider: Ideal transformer Magnetization inductance Equivalent circuit reproduces basic transformer equation Leakage inductance

Real Transformer Add losses: Resistance of secondary Ideal transformer Eddy currents and Hysteresis losses in core Resistance of primary

Transformer efficiency Ideal transformer 𝑃 𝑜𝑢𝑡 = 𝑃 𝑖𝑛 Efficiency = 100 % 𝜂=100% Real transformer 𝑃 𝑜𝑢𝑡 ≠ 𝑃 𝑖𝑛 Efficiency < 100 % 𝜂<100% 𝑃 𝑖𝑛 = 𝑃 𝑜𝑢𝑡 + 𝑃 𝑐 + 𝑃 𝑒 Copper loss in winding resistance 𝜂= 𝑃 𝑜𝑢𝑡 𝑃 𝑜𝑢𝑡 + 𝑃 𝑐 + 𝑃 𝑒 100% Eddy current & core loss Real transformer

VOLTAGE REGULATION Observed voltage drop when we draw a current from the secondary % 𝑅𝑒𝑔𝑢𝑙𝑎𝑡𝑖𝑜𝑛= 𝑉 𝑠 𝑛𝑜 𝑙𝑜𝑎𝑑 − 𝑉 𝑠 𝑙𝑜𝑎𝑑𝑒𝑑 𝑉 𝑠 𝑙𝑜𝑎𝑑𝑒𝑑 100% Real transformer

ELEC 3105 Basic EM and Power Engineering Transformer core saturation

Core saturation At a sufficiently high H the core saturates and B is essentially constant. 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) Recall the Hysteresis curve B versus H

Core saturation Area of the core Frequency Then at maximum Recall for the primary that: Then at maximum Area of the core Frequency

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

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

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

ELEC 3105 Basic EM and Power Engineering Transformer and the power grid

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

Transformer and the power grid Copper transmission line 2.5 cm in radius has resistivity Resistance over length of transmission line length Power loss in wire conduit Leaves about 50 MW for city. Need very large core to handle this current without saturation.

Transformer and the power grid Solution is to step up the voltage and thus reduce the current and power loss in copper conduit. Typical AC high voltage line: 765 kV Current required: I = 1000 A Power loss in wire conduit Leaves about 999.14MW for city.

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).

Geomagnetic storms and core saturation Auroral electrojet 106 A

Geomagnetic storms and core saturation Auroral electrojet fluctuates by 105 A in one minute Change in flux through circuit Auroral electrojet 106 A High voltage line Generating station Consuming station

Geomagnetic storms and core saturation Auroral electrojet fluctuates by 105 A in one minute At this frequency transformer coils look like short circuits A DC current of is produced Auroral electrojet 106 A High voltage line Generating station Consuming station

Geomagnetic storms and core saturation Geomagnetic Effects on Power Systems - English Geomagnetic Storms Sun article Sun

97.315 Basic EM and Power Engineering Lecture 27 END Transformer core saturation START Transformer application

Transformer 120V AC to 5V DC 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.

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.

Slides not used after this one

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:

Transformer Applications Consider: We have and Assuming large self inductance, we can write We also know that from Ohm’s law equate

Transformer Applications Consider: Thus In terms of this is ……

Transformer Applications Consider: In terms of this is …… We want to design a transformer to have a maximum power transfer to the load. Thus: and …….

Transformer Applications Consider: with …….

Transformer Applications Consider: Differentiating with respect to k and setting to zero will give the condition for maximum transfer to the load. with …….

Transformer Applications Consider: Simplifying This is the transformer turns ratio which gives maximum power transfer to the load. with gives

Transformer Applications Consider: If we convert the load resistance to the apparent generator side resistance, we get: Lecture 25 Slide 34 Gives Maximum power transfer condition to load