Transformers II (3.15) Learning Objectives PDF

Transformers II (3.15) Learning Objectives 3.15.2 Explain the causes of efficiency losses in transformers and methods for minimising losses (Level 2). 3.15.3.1 Describe transformer operation under load (Level 2). 3.15.4.1 Describe power transfer within a transformer (Level 2). 3.15.4.2 Describe tran...

Transformers II (3.15) Learning Objectives 3.15.2 Explain the causes of efficiency losses in transformers and methods for minimising losses (Level 2). 3.15.3.1 Describe transformer operation under load (Level 2). 3.15.4.1 Describe power transfer within a transformer (Level 2). 3.15.4.2 Describe transformer efficiency and how to calculate it (Level 2). 3.15.8 Describe auto-transformers (Level 2). 2024-02-15 B-03b Electrical Fundamentals CASA Part 66 - Training Materials Only Page 195 of 284 Transformer Losses Power Transformers Practical power transformers, although highly efficient, are not perfect devices. Small power transformers used in electrical equipment have an 80% to 90% efficiency range, while large, electrical utility distribution transformers may have efficiencies exceeding 98%. The total power loss in a transformer is a combination of three types of losses. One loss is due to the DC resistance in the primary and secondary windings. This loss is called copper loss or I2R loss. The two other losses are due to eddy currents and to hysteresis in the core of the transformer. Copper loss, eddy-current loss and hysteresis loss result in undesirable conversion of electrical energy into heat energy. Aviation Australia Power utility distribution transformers have very high efficiency Copper Losses Copper loss or I2R loss is the power lost due to the resistance of the wire used in the primary and secondary coils. These losses are usually given off as heat. We can minimise these losses if we lower the resistance of the wire used in the construction of the primary and secondary windings. This is accomplished by increasing the diameter of the wire. Weight and overall physical dimensions must also be considered when reducing copper losses. 2024-02-15 B-03b Electrical Fundamentals CASA Part 66 - Training Materials Only Page 196 of 284 Hysteresis Losses Reversing the magnetism of the transformer's core will release energy in the form of heat. Hysteresis loss is the loss of this heat energy and is proportional to the area and the materials used in the construction of the core, as well as to the frequency of the applied current. Air-core transformers have no hysteresis losses. Hysteresis losses can also be minimised by reducing the area of the core and by using supermalloy and permalloy materials (silicon-iron, alloys) in the construction of the core. Eddy Current Losses The changing magnetic field of a transformer induces an EMF into the core which sets up currents that circulate within the core. These circulating currents are known as eddy currents. If the core were made from solid iron, these eddy currents would combine together to form a very large current, causing very large losses in the form of heat. The induced eddy currents in the core also set up an opposing flux within the core. This results in more current flowing in the primary to try to maintain the core’s magnetic field, further increasing losses. To minimise the loss resulting from eddy currents, transformer cores are laminated. Since the thin, insulated laminations do not provide an easy path for current, eddy-current losses are greatly reduced. Aviation Australia Laminated cores and eddy current losses 2024-02-15 B-03b Electrical Fundamentals CASA Part 66 - Training Materials Only Page 197 of 284 Transformer Efficiency Transformer Efficiency Calculations To compute the efficiency of a transformer, the input power and the output power must be known. The input power is equal to the product of the voltage applied to the primary and the current in the primary. The output power is equal to the product of the voltage across the secondary and the current in the secondary. The difference between the input power and the output power represents a power loss. You can calculate the percentage of efficiency of a transformer by using the standard efficiency formula shown below: Ef f iciency (in%) = P out × 100 P in Where: Pout = total output power delivered to the load Pin = total input power. If input power to a transformer is 650 W and output power is 610 W, what is the efficiency? Solution: Ef f iciency (in%) = P out × 100 P in 610 Ef f iciency (in%) = × 100 650 Ef f iciency (in%) = 93.8% Power loss can also be calculated by subtracting Pout from Pin (VA or W). Power loss in the above transformer is 40 W – wasted due to heat losses. 2024-02-15 B-03b Electrical Fundamentals CASA Part 66 - Training Materials Only Page 198 of 284 Impedance Matching Maximum Power Transfer In order to understand the requirement for impedance matching and how it is accomplished, we must first study power transfer between parts of a circuit. This transfer could be between different stages of an amplifier or from an amplifier to a loudspeaker. Under ideal conditions, 100% of the power produced by a source is used by the load, but the fact that every source of energy has an internal resistance or impedance makes this impossible. The idea is to make the maximum power available at the load. Suppose we have a 24 V source with an internal impedance of 6 Ω, connected to a 2 Ω load. Aviation Australia Max power transfer #1 Now calculate the power in the load and source. Total circuit resistance: 6 + 2 = 8 Ω Circuit current: V I = 24 = R = 3 A 8 Power in the source: P = I 2 R = 3 2 × 6 = 9 × 6 = 54 Power in the load: P = I 2024-02-15 2 R = 3 2 × 2 = 9 × 2 = 18 B-03b Electrical Fundamentals CASA Part 66 - Training Materials Only Page 199 of 284 In this example, the source is consuming three times the power of the load, meaning the source will be glowing with the heat produced and will most likely be damaged. In the diagram below you see the same 24 V 6 Ω source connected to a 6 Ω Load. Aviation Australia Maximum power transfer question #2 With the load increased to 6 Ω, calculate the power in the load and source. Total circuit resistance: 6 + 6=12 Ω Circuit current: V I = 24 = R = 2 A 12 Power in the source: P = I 2 R = 2 2 × 6 = 4 × 6 = 24 Power in the load: P = I 2 R = 2 2 × 6 = 4 × 6 = 24 In this example, the source is consuming the same amount of power as the load, so the source is still producing a large amount of heat. The same applies to a 24 V 6 Ω source connected to an 18 Ω Load. 2024-02-15 B-03b Electrical Fundamentals CASA Part 66 - Training Materials Only Page 200 of 284 Aviation Australia Max power transfer question #3 Again, calculate the power in the load and source if the same 24-V, 6-Ω source is connected to an 18Ω load. Total circuit resistance: 6 + 18 = 24 Ω Circuit current: V 24 I = = R = 1 A 24 Power in the source: P = I 2 R = 1 2 × 6 = 1 × 6 = 6 Power in the load: P = I 2 R = 1 2 × 18 = 1 × 18 = 18 In this example, the load is consuming the same amount of power as in the first example, and is three times that of the source, 6 W. 2024-02-15 B-03b Electrical Fundamentals CASA Part 66 - Training Materials Only Page 201 of 284 Aviation Australia Maximum power transfer graph summary To Summarise If you require minimum heat generated in the source (as in when the source is a battery), then region 3 (Rload should be less than Rsource). If you require maximum power transfer (such as when linking an antenna to a radio), then use region 2 (Rload should be equal to Rsource). Area 1 should never be used as Rload should never be less than Rsource as this would damage the source. Power supplies require a way to connect a source and a load of dissimilar impedance in such a manner that the connection matches both impedances without losing power. Aviation Australia Impedance matching 2024-02-15 B-03b Electrical Fundamentals CASA Part 66 - Training Materials Only Page 202 of 284 The method of connecting dissimilar impedances is to use a transformer; it is just a matter of calculating the turns ratio, which is: ( NP ) 2 = NS ZP ZS This means: T 2 = ZP ZS The proof is shown below. Aviation Australia Max power transfer question #4 If we have a power supply delivering power to a load, and the voltage is 200 V and the load draws 8 A, then the resistance is: V R = 200 = I = 25 Ω 8 Now, if this is an AC power delivered through a transformer and the load connected uses the same power as before, we have the following circuit as an example. Aviation Australia 2024-02-15 Max power transfer question #4 - the power supply is now connected to the load through a transformer B-03b Electrical Fundamentals CASA Part 66 - Training Materials Only Page 203 of 284 The transformer steps the voltage down from 200 V to 50 V, so the current must be stepped up by a factor of 4. T = VP 200 = VS = 4 : 1 50 so 1 T o calculate I Sec T = 1 : 4 = 8 : I Sec ⟹ I Sec = 32 We now a have a voltage of 50 V and a current of 32 A, therefore: V R Sec = 50 = I = 1.5625 Ω 32 From the first circuit, we have a voltage of 200 V and a resistance of 25 Ω, and in the second we have a voltage of 50 V and a rersistance of 1.5626 Ω. This produces a voltage transformation ratio of 4:1. To calculate the impedance transformation ratio the impedance values of each side of the transformer is compared in a ratio. If we assume that the total impedance is equal to the resistance values calculated earlier then R Pri = Z Pri and R Sec = Z Sec. I mpedance Ratio = Z P ri : Z Sec ⟹ 25 : 1.5625 = 16 : 1 As the impedance ratio is 16:1, the impedance transformation ratio (16) is equal to the turns ratio squared (42). Note: We are ignoring transformer losses and assuming 100% efficiency. Transformer Ratings When a transformer is to be used in a circuit, more than just the turn ratio must be considered. The voltage, current and power-handling capabilities of the primary and secondary windings must also be considered. The maximum voltage that can safely be applied to any winding is determined by the type and thickness of the insulation used. When a better (and thicker) insulation is used between the windings, a higher maximum voltage can be applied to the windings. The maximum current that can be carried by a transformer winding is determined by the diameter of the wire used for the winding. If current is excessive in a winding, a higher-than-ordinary amount of power will be dissipated by the winding in the form of heat. This heat may be sufficiently high to cause the insulation around the wire to break down. If this happens, the transformer may be permanently damaged. 2024-02-15 B-03b Electrical Fundamentals CASA Part 66 - Training Materials Only Page 204 of 284 The power-handling capacity of a transformer depends on its ability to dissipate heat. If the heat can safely be removed, the power-handling capacity of the transformer can be increased. This is sometimes accomplished by immersing the transformer in oil or by the use of cooling fins. The power-handling capacity of a transformer is measured in either the volt-amperes or watts. Two common power generator frequencies (60 Hz and 400 Hz) have been mentioned, but the effect of varying frequency has not been discussed. If the frequency applied to a transformer is increased, the inductive reactance of the windings is increased, causing a greater AC voltage drop across the windings and a lesser voltage drop across the load. However, an increase in the frequency applied to a transformer should not damage it. But, if the frequency applied to the transformer is decreased, the reactance of the windings is decreased and the current through the transformer winding is increased. If the decrease in frequency is enough, the resulting increase in current will damage the transformer. For this reason, a transformer may be used at frequencies above its normal operating frequency, but not below that frequency. Aviation Australia Transformer power and frequency ratings 2024-02-15 B-03b Electrical Fundamentals CASA Part 66 - Training Materials Only Page 205 of 284 Auto-Transformers Auto-Transformer Configuration It is not necessary in a transformer for the primary and secondary to be separate and distinct windings. In what is known as an auto-transformer, a single coil of wire is 'tapped' to produce what is electrically a primary and secondary winding. The voltage across the secondary winding has the same relationship to the voltage across the primary that it would have if they were two distinct windings. Aviation Australia Adjustable auto-transformer schematic Auto-transformers may also have an adjustable tap which will provide a variable-output voltage. These auto-transformers are usually referred to by their brand name, Variac. The movable tap in the secondary is used to select a value of output voltage, either higher or lower than Ep, within the range of the transformer. That is, when the tap is at point A, Es is less than Ep; when the tap is at point B, Es is greater than Ep. If a voltage of 200 V is applied between points B and C, then a secondary voltage of 100 V would be available from A and B. 300 V will be available from points A and C. This is because the primary field resulting from the supply of current through B and C creates not only a back EMF of close to 200 V in the turns between B and C, but also an EMF of 100 V in the turns between A and B. These two EMFs are of the same phase because they are induced in one winding by the same field. They can therefore be added together to give a secondary voltage of 300 V between A and C. Alternately the 100 V induced between A and B can be taken as a secondary. 2024-02-15 B-03b Electrical Fundamentals CASA Part 66 - Training Materials Only Page 206 of 284 Aviation Australia Auto-transformer turns ratio 2024-02-15 B-03b Electrical Fundamentals CASA Part 66 - Training Materials Only Page 207 of 284