Components the transformer
- 1. The Transformer - The Transformer Principle When an alternating current is passed through a coil the magnetic field produced will also alternate. If a second coil is placed within this magnetic field the alternating flux will link the two coils and induce an emf in the second coil. This process is called mutual induction and is also known as the transformer effect. AC Supply Alternating magnetic flux AC Supply Alternating Induced emf Primary coil Secondary coil
- 2. The Double Wound Transformer A transformer is commonly used to change (transform) an alternating voltage from one value to another. The basic transformer consists of two coils, called the ‘primary’ and ‘secondary’ each electrically ‘isolated’ from the other, there is no electrical connection between the primary and secondary coils. Primary winding Secondary winding Alternating induced emf Applied AC supply Laminated iron core Magnetic flux If both coils are mounted on an iron core the magnetic flux and coupling between coils will increase, this assembly is called a transformer.
- 3. The Transformer – Turns Ratio An alternating current flowing in one winding will produce an alternating flux in the core and since the turns of both windings cut this flux the emf induced in each turn is the same for each winding. The volts per turn for primary and secondary windings are the same. The ratio of secondary turns (NS) to primary turns (NP) is called the turns ratio. The ratio of secondary voltage (VS) to primary voltage (VP) depends on the turns ratio. VP VSNP NS IP IS Primary Winding Secondary Winding VP VS NP NS == IS IP =
- 4. The Double Wound Transformer Activity – Turns Ratio 1. A transformer is required to provide a 60V output from a 240V ac supply. Calculate (a) the turns ratio required and (b) the number of primary turns, if the secondary is wound with 500 turns. 2. A transformer operating from a 240V ac supply is required to provide two secondary windings one of 12 volts and the other 18 volts. If the primary winding has 2400 turns determine the turns on each secondary winding. 3. A transformer operates from a 240 volt supply and has a secondary voltage of 24 volts, if this secondary winding has 96 turns determine the turns required for the primary winding.
- 5. The Transformer – Power Conservation If we neglect transformer losses since they are very small the efficiency can be taken as 100%, making the secondary power (PS) equal to the primary power (PP) . VS VP NS NP == PS = PP VS IS = VP IP but IP IS = VS VP IP IS = NS NP This provides a more general expression linking voltage and turns to the current in each winding. VP VS NP NS == IS IP =
- 6. Transformer Rating When designing or specifying a transformer two main items of information are needed – Secondary Voltage (V) and Secondary Current (I). The rating of a transformer is defined in terms of the above information in the form ‘VA’ (volt-amp). The ‘VA’ rating is the maximum operating power level a transformer can deliver to a load. The above example assumes an ‘ideal’ transformer with no losses. 120VA Transformer IP 0.5A IS 6.6A VP 240V VS 18V
- 7. The Double Wound Transformer Activity – Transformer Rating 1. A 600VA transformer operating from a 240 volt supply has a 6 volt secondary winding. Determine the ‘full load’ current capability of the secondary winding and the current drawn from the supply when operating at full load. 2. A welding set transformer operating from a 240V mains supply has a rating of 2000 VA (2kVA). If the full load secondary voltage is 3 volts determine the a) the maximum welding current and b) the current drawn from the mains supply under full load. 3. A 60VA transformer has two 12 volt secondary windings, determine the secondary current capability. NOTE: In all cases an ideal transformer is assumed.
- 8. The Transformer - Losses Iron (core) loss Iron losses are made up of ‘hysteresis’ loss and ‘eddy’ current loss both cause heat to be generated. The iron loss is constant and not dependant on load conditions meaning that the no-load iron loss is the same as the full-load iron loss. Copper (I2 R) loss Copper losses are due to the effect of resistance on the currents of primary and secondary windings causing heat to be generated. The copper loss is a power loss and dependant on the square of the load current. Therefore a transformer operating at half load will have only a quarter of the copper loss it has on full-load.
- 9. The Transformer – Efficiency The input power to a transformer provides both the output power and the transformer losses. input power = output power + power losses efficiency = output power input power x 100% = output power input power + power loses x 100%efficiency Primary winding Secondary winding Output power PO Input power PI
- 10. The Transformer – Regulation The resistance and inductance of a transformer provide an impedance to the output current resulting in a volt drop due to the load. On no-load there will be no secondary current and no volt drop. On full-load the output voltage will fall. The difference between the no-load voltage and the full-load voltage, expressed as a percentage of the no-load voltage is called the ‘voltage regulation’. no load output voltage – full load output voltage no load output voltage x 100%Regulation =
- 11. Toroidal Equipment Transformer Common Types of Transformer for 50Hz Operation Package styles vary but all have a laminated iron core to support the magnetic flux linking the windings. Chassis Mounted Equipment Transformer Audio Transformer Circuit Board Mounted Equipment Transformer