Peter's Physics Pages

A Semester of First Year Physics with Peter Eyland

Lecture 17 (Transformers)

In this lecture the following are introduced:
• The Transformer as a Mutual Inductor
• The Uses of Transformers
• The Construction of Transformers
• The Ideal Transformer
• The Complex Mutual Impedance of a Transformer
• The Ideal Transformer as a multiplier for Potential Difference
• The Ideal Transformer as a multiplier for Current
• The Ideal Transformer with a Complex Load
• The Reflected and Shunt Impedances in Equivalent Circuits

Transformers

 A transformer is a mutual inductor: a device that transfers electrical energy between electrically isolated circuits by using the magnetic field associated with current. The Uses of Transformers

Transformers have many uses in power transmission and electronics.

For example, they can be used to
• minimise energy losses in transporting electricity over long distances, and
• match loads with internal resistance so that there is maximum power transfer, and
• couple signals between electronic stages.

The Construction of Transformers

 The basic transformer has two coils of insulated highly conductive copper wire wound around a straight iron core. In practice, the coils are often wound separately around a rectangular iron core. The core traps the magnetic field inside it and so mimimises loss of magnetic flux. The core is not a solid piece of metal but laminated (layers clamped together) to minimise energy loss due to eddy currents that are induced in it. Transformers that operate at many MHz, may be wound on a core of ferrite, a special ceramic magnetic material. Sometimes the coils are wound around a hollow non-magnetic former so that the core material is air. The Ideal Transformer

The ideal transformer has the following properties:
• Zero resistance in the wires.
• No hysteresis (i.e. energy loss) when the core is magnetising and de-magnetising.
• The individual self-inductances and the mutual inductance do not change with current.
• All the magnetic flux from the one coil passes through all the turns of the other coil.
• The capacitance between the turns is zero.

The Complex Mutual Impedance of a Transformer

 In the diagram on the right, the subscripts "p" and "s" stand for primary and secondary. The primary coil is connected to the source and the secondary coil is connected to the load. Let the current in the primary coil be given by:  The e.m.f. across the secondary coil is then: Where is the complex mutual impedance: the ratio of potential produced over initiating current. The negative sign means that the secondary e.m.f. will lag the primary current by 900.

The Ideal Transformer as a multiplier for Potential Difference

 Review: See Mech. Eng. Lecture 7 (Electromagnetism) when link available. In this case the secondary circuit is open circuited, as shown below      Now from the definition of complex mutual impedance, given above: It is seen that the primary e.m.f. has been multiplied by the turns ratio, . The negative sign means that there is a phase reversal (indicated by the current directions).

 The Ideal Transformer as a multiplier for Current In this case the secondary coil is short-circuited, as shown in the diagram below.  The primary current has been divided by the turns ratio, .

As before, the negative sign means that there is a phase reversal and the currents flow as indicated.

The Ideal Transformer with a Complex Load

Four equivalent single loop circuits are developed. A series and a parallel equivalent circuit representing the view from the primary side and a series and a parallel equivalent circuit representing the view from the secondary side.

 A complex load is added to the secondary circuit, as shown in the diagram. From energy conservation, the e.m.f.s supplying energy equal the potential differences produced. For the circuit shown, this gives these equations. • Equation for the primary circuit: The task is to take the loop equations and eliminate the secondary current from them. From the second equation the secondary current is found in terms of the primary current. This is then substituted into the first equation. • Primary Reflected Impedance in Series The equation above, for the primary circuit e.m.f. and current, allows the original two-loop circuit to be re-drafted as a single primary loop with a series impedance, as shown on the right - with This impedance can be thought of as the impedance reflected from the secondary circuit. • Primary Reflected Impedance in Parallel the one-loop circuit with the reflected impedance, can also be re-drafted with a parallel (or shunt) impedance. The equalising condition is that this new parallel circuit must draw the same current from the e.m.f. as the original circuit. The original circuit and the equation for the primary circuit are repeated below.  Equating the impedance from the original primary circuit equation with the impedance of the new parallel combination of primary coil and shunt impedance (without ZP) gives. This gives the shunt impedance needed for equivalence. The equivalent parallel primary circuit is shown on the right, where the shunt impedance is the secondary circuit impedance divided by the square of the turns ratio. The self-inductance of the primary coil Lp will usually be large when compared with the shunt impedance. As with all parallel combinations where one element is large compared with the other, the large element can be neglected in favour of the small element. This means that Lp can be neglected and the circuit will simply be: The equivalent parallel primary circuit has the primary e.m.f., the original primary impedance, and a primary shunt impedance .

 • Equation for the secondary circuit. The task here is to take the circuit loop equations and eliminate the primary current from them.  Re-writing the second equation with the primary current as the subject: Substituting the primary current gives the equation for the secondary circuit. • Secondary Reflected Impedance in Series The equation for the secondary circuit e.m.f. and current, allows the original two-loop circuit to be re-drafted as a single secondary loop with a secondary e.m.f. and a series impedance, as shown on the right: The secondary e.m.f. and reflected series impedance are:  • Secondary Reflected Impedance in Parallel As done before with the primary circuit, the one-loop circuit with the reflected impedance, can also be re-drafted with a parallel (or shunt) impedance. The equalising condition is that the same current is drawn from the e.m.f. as in the equation for the secondary circuit (shown on the right). Equating the impedance from the secondary circuit equation with the impedance of the new parallel combination of secondary coil and shunt impedance (without ZS) gives. This gives the shunt impedance needed for equivalence. The equivalent parallel secondary circuit is shown on the right, where the shunt impedance is the primary circuit impedance multiplied by the square of the turns ratio. The self-inductance of the secondary coil LS will usually be large when compared with the shunt impedance. As with all parallel combinations where one element is large compared with the other, the large element can be neglected in favour of the small element. This means that LS can be neglected and the circuit will simply be: The equivalent parallel secondary circuit has a secondary e.m.f. the original secondary impedance, and a secondary shunt impedance .

The four equivalent single loop circuits are:

 equivalent primary series circuit  equivalent primary parallel circuit equivalent secondary series circuit   equivalent secondary parallel circuit  Example 1

 For the circuit shown, find the power dissipated in the load resistor. (Note that the circuits are not ideally coupled since , but this does not materially affect the situation). Since the load is on the secondary side, choose the equivalent secondary series circuit. The equivalent secondary e.m.f. is: The reflected impedance is given by: The secondary equivalent circuit is then: The secondary current is given by: Substituting for the primary e.m.f. The average power dissipated in the load resistor depends on the r.m.s. current: Example 2

 In the circuit shown, power is supplied at 1000 rad.s-1. The primary circuit has a 400R resistor and a 10μH capacitor. The transformer is ideal with mutual inductance of 500mF and primary self-inductance of 400mF. Find the load required for maximum power transfer. Note: This question requires a knowledge of the maximum power transfer theorem. Since the load is on the secondary side, choose the equivalent secondary series circuit. Where For maximum power transfer the external load impedance must equal the complex conjugate of the internal impedance of the source. The complex conjugate of the internal impedance gives This means that the load has to be a 400R resistor, and a capacitor with reactance 325R, i.e. Summarising:

Transformers are mutual inductors and used in power tranmission, load matching and coupling electronic stages.
Transformers typically have two coils of insulated highly conductive copper wire wound around a iron core.
The ideal transformer has no resistance, no hysteresis loss, constant self and mutual inductance, no flux loss and no capacitance.
The complex mutual impedance is , and The open circuit potentials give The short circuit currents are For a transformer with a complex load:

 equivalent primary series circuit  equivalent primary parallel circuit equivalent secondary series circuit div align="center">   equivalent secondary parallel circuit  Acknowledgement: These notes are based in part on "Alternating Current Circuit Theory" by G.J.Russell and K.Mann NSWUP 1969. email Write me a note if you found this useful

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