Chapter 9: Ideal Transformer 10/9/003 Electromechanical Dynamics 1
Introduction Transformers are one of the most useful electrical devices provides a change in voltage and current levels provides galvanic isolation between different electrical circuits changes the apparent magnitude value of an impedance 10/9/003 Electromechanical Dynamics
Voltage Induction For a coil consisting of N turns placed in a time-varying sinusoidal flux, the flux induces a sinusoidal ac voltage dφ( t) e( t) N dt the rms value of the voltage E where π f 4.44 f N N Φ Φ max max f is the sinusoidal frequency Φ max is the peak flux as defined by Φ Φ max sin ( π f +φ ) the peak flux is useful for working with iron cores and assessing the impact of losses and saturation Φ max B max A core 10/9/003 Electromechanical Dynamics 3
Applied Voltage Consider a coil connected across an AC voltage source the coil and source resistances are negligible the induced voltage E must equal the source voltage; KVL a sinusoidal AC flux Φ must exist to generate the induced voltage on the N turns of the coil Φ max E g 4.44 f N Φ max varies in proportion to E g placing an iron core in the coil will not change the flux Φ magnetization current I m drives the AC flux the current is 90 out-of-phase and lagging with respect to the voltage with an iron core, less current is needed to drive the AC flux 10/9/003 Electromechanical Dynamics 4
Induced Voltages Example a coil, having 4000 turns, links an AC flux with a peak value of mwb at a frequency of 60 Hz calculate the rms value of the induced voltage what is the frequency of the induced voltage? Example a coil, having 90 turns, is connected to a 10 V, 60 Hz source the rms magnetization current is 4 A calculate the peak value of the flux and the mmf find the inductive reactance and the inductance of the coil 10/9/003 Electromechanical Dynamics 5
Elementary Transformer Consider an air-core coil excited by an AC source E g draws a magnetization current I m produces a total flux Φ A second coil is brought close to the first a portion Φ m1 of the flux couples the second coil, the mutual flux an AC voltage E is induced the flux linking only the first coil is called the leakage flux, Φ f1 Improved flux coupling concentric windings, iron core weak coupling causes small E the magnetization current I m produces both fluxes Φ m1 and Φ f1 the fluxes are in-phase the voltages E g and E are inphase terminal orientation such that the coil voltages are in-phase are said to possess the same polarity 10/9/003 Electromechanical Dynamics 6
Ideal Transformer An ideal transformer transformer has no losses core is infinitely permeable all fluxes link all coils there are no leakage fluxes Voltage relationship consider a transformer with two coils of N 1 and N turns a magnetizing current I m creates a flux Φ m the flux varies sinusoidally and has a peak value of Φ max the induced voltages are from these equations, it can be deduced that the ratio of the primary and secondary voltages is equal to the ratio of the number of turns E 1 and E are in-phase polarity marks show the terminal on each coil that have a peak positive voltage simultaneously 10/9/003 Electromechanical Dynamics 7 E E 1 E E 4.44 4.44 1 f f N N 1 N N 1 Φ a Φ max max
Ideal Transformer Current relationship let a load be connected across the secondary of an ideal transformer current I will immediately flow I E Z coil voltages E 1 and E cannot change when connected to a fixed voltage source and hence flux Φ m cannot change current I produces an mmf mmf NI if mmf acts along, it would profoundly change Φ m Φ m can only remain fixed if the primary circuit develops a mmf which exactly counterbalances mmf current I 1 must flow such that I1 N 1 I N a I 1 and I must be in-phase when I 1 flows into the positive polarity marking of the primary, I flows out of the positive polarity marking of the secondary 10/9/003 Electromechanical Dynamics 8 1
Ideal Transformer Ideal transformer model let a N 1 /N then E E 1 / a and I 1 I / a + I 1 Φ m I E 1 E + 10/9/003 Electromechanical Dynamics 9
Ideal Transformer Example a not so ideal transformer has 00 turns in the primary coil and 10 turns in the secondary coil the mutual coupling is perfect, but the magnetization current is 1 A the primary coil is connected to a 480 V, 60 Hz source calculate the secondary rms voltage, peak voltage Example for the transformer above, a load is connected to the secondary coil that draws 80 A of current at a 0.8 lagging pf calculate the primary rms current and draw the phasor diagram 10/9/003 Electromechanical Dynamics 10
Impedance Ratio Transformers can also be used to transform an impedance the source sees the effective impedance Z X E 1 I 1 on the other side, the secondary winding of the transformer sees the actual impedance Z E I the effective impedance is related to the actual impedance by E1 a E a E Z X a Z I I a I 1 10/9/003 Electromechanical Dynamics 11
Shifting Impedances Impedances located on the secondary side of a transformer can be relocated to the primary side the circuit configuration remains the same (series or shunt connected) but the shifted impedance values are multiplied by the turns ratio squared Impedance on the primary side can be moved to the secondary side in reverse manner the impedance values are divided by the turns ratio squared 10/9/003 Electromechanical Dynamics 1
Shifting Impedances In general, as an impedance is shifted across the transformer the real voltage across the impedance increases by the turns ratio the actual current through the impedance decreases by the turns ratio the required equivalent impedance increases by the square of the turns ratio Example using the shifting of impedances calculate the voltage E and current I in the circuit, knowing that the turns ratio is 1:100 10/9/003 Electromechanical Dynamics 13
Ideal Transformer Homework Problems: 9-4, 9-6, 9-10* Note: problem 9-10 is a design problem 10/9/003 Electromechanical Dynamics 14