Doubly Excited Magnetic System | Electrical Machines - Electrical Engineering (EE) PDF Download

Introduction

A doubly-excited magnetic system is a magnetic circuit or electromechanical transducer in which two independent windings are used to produce magnetic fields. Each winding is supplied from its own excitation source so that both currents (or voltages) can be controlled independently. Common examples are synchronous machines, separately excited DC machines, certain types of tachometers, and some loudspeaker drive arrangements.

Consider a simple model of a doubly-excited system with one coil on the stator and one coil on the rotor. The stator coil has resistance R₁ and the rotor coil has resistance R₂. Each coil is excited from a separate electrical source so that the currents i₁(t) and i₂(t) are independent.

Assumptions for analysis

• The relationship between flux linkage and current is linear at any rotor position (i.e. magnetic linearity is assumed).
• Hysteresis and eddy-current losses are neglected.
• Leakage fluxes of the windings are negligible (mutual and self inductances represent the dominant coupling).
• Electric fields are neglected; magnetic fields are predominant and quasi-static magnetic analysis applies.

Flux linkage relations

The flux linkages of the two windings are expressed in terms of self and mutual inductances as

ψ₁ = L₁ i₁ + M i₂

ψ₂ = L₂ i₂ + M i₁

Voltage equations (KVL)

Applying Kirchhoff's voltage law to each winding gives the instantaneous voltage equations (including resistive drops and time rate of change of flux linkage).

Substituting the flux linkages into the KVL equations leads to coupled differential equations containing self- and mutual-inductance terms.

Time-varying inductances and notation

In rotating machines and many transducers the inductances L₁, L₂ and mutual inductance M depend on rotor position θ_m (which itself is a function of time). We therefore treat inductances as functions of θ_m(t) while currents are functions of time. Rewriting the voltage equations to make these dependencies explicit gives

Instantaneous power in windings

Multiplying the first voltage equation by i₁ and the second by i₂ gives expressions for instantaneous electrical power supplied to each winding.

These two expressions represent the instantaneous electrical input powers to the two coils respectively: the sum of resistive losses and the rate of change of magnetic energy (and mechanical work when the structure moves).

Energy equation for the doubly-excited system

Integrating the instantaneous power expressions with respect to time and adding them yields the energy balance for the magnetic system:

This energy equation shows that the total electrical energy input W_e to the system is partitioned into resistive (electrical) losses and useful energy terms consisting of energy stored in the magnetic field and mechanical work (or change of co-energy) associated with motion.

The useful field energy term may be written as:

Energy stored in the magnetic field

The instantaneous energy stored in the magnetic field for given currents and position is

W_f = 1/2 L₁ i₁² + 1/2 L₂ i₂² + M i₁ i₂

When the rotor is held stationary (no mechanical output), all useful electrical input goes into field energy. For a fixed rotor position, differentials dL₁, dL₂, dM are zero, so the above expression directly gives the stored magnetic energy at that position.

Electromagnetic torque

When the rotor is free to move, a portion of the input electrical power can be converted into mechanical power. The electromagnetic torque arises from the dependence of inductances on rotor angle. The rate of change of field energy with time includes a contribution due to change of inductances with position:

Integrating with respect to time and separating the mechanical contribution leads to a general expression for a moving transducer in which L₁, L₂, M, i₁ and i₂ may vary with time and position:

Comparing the mechanical power term with torque times angular velocity gives the electromagnetic torque expression:

Differentiating with respect to rotor angle θ_m produces the torque in terms of derivatives of inductances with angle:

Interpretation of terms:

• The first two terms (those proportional to dL₁/dθ_m and dL₂/dθ_m) are known as reluctance torque. They arise because the magnetic circuit's reluctance (hence inductance) changes with rotor position; currents flowing produce a torque that tends to position the rotor to reduce the system's magnetic reluctance.
• The last term (proportional to dM/dθ_m) is the co-alignment torque (sometimes called interaction torque). It is produced by the interaction of the two magnetic fields and tends to align the fields (i.e. minimise magnetic co-energy).

Note: For machines with a uniform air gap the inductances do not vary with rotor position and therefore reluctance torque is zero; only co-alignment torque (if M varies) can be present.

Numerical example

For a doubly excited system, the inductances are given as follows -

L₁ = 10 + 2 cos 2θ; L₂ = 5 + 2 cos 2θ; M = 20 cos θ

The coils are excited by current i₁ = 0.5 A and i₂ = 0.6 A.

• Find the torque as a function of θ.
• Find the energy stored in the system as a function of θ.

Solution (stepwise calculation)

The torque in a doubly-excited system for fixed currents is given by

τ(θ) = 1/2 i₁² dL₁/dθ + 1/2 i₂² dL₂/dθ + i₁ i₂ dM/dθ

Differentiate the given inductances with respect to θ.

dL₁/dθ = d(10 + 2 cos 2θ)/dθ = -4 sin 2θ

dL₂/dθ = d(5 + 2 cos 2θ)/dθ = -4 sin 2θ

dM/dθ = d(20 cos θ)/dθ = -20 sin θ

Substitute currents and derivatives into the torque expression.

Compute the first term: 1/2 × (0.5)² × (-4 sin 2θ) = -0.5 sin 2θ × 0.25 = -0.5 × 0.25 sin 2θ = -0.125 × 4 sin 2θ; simplifying gives -0.5 sin 2θ.

Compute the second term: 1/2 × (0.6)² × (-4 sin 2θ) = 0.5 × 0.36 × (-4 sin 2θ) = -0.72 sin 2θ.

Compute the third term: (0.5)(0.6)(-20 sin θ) = -6 sin θ.

Add the three contributions.

τ(θ) = (-0.5 - 0.72) sin 2θ - 6 sin θ

τ(θ) = -1.22 sin 2θ - 6 sin θ N·m

Energy stored in the magnetic field W_f(θ)

Use the standard expression for field energy:

W_f = 1/2 L₁ i₁² + 1/2 L₂ i₂² + M i₁ i₂

Substitute the given inductances and currents.

Compute 1/2 L₁ i₁² = 1/2 × (10 + 2 cos 2θ) × (0.5)² = 0.5 × (10 + 2 cos 2θ) × 0.25 = 0.125 × (10 + 2 cos 2θ) = 1.25 + 0.25 cos 2θ.

Compute 1/2 L₂ i₂² = 1/2 × (5 + 2 cos 2θ) × (0.6)² = 0.5 × (5 + 2 cos 2θ) × 0.36 = 0.18 × (5 + 2 cos 2θ) = 0.9 + 0.36 cos 2θ.

Compute M i₁ i₂ = (20 cos θ) × 0.5 × 0.6 = 6 cos θ.

Add the three parts.

W_f(θ) = (1.25 + 0.9) + (0.25 + 0.36) cos 2θ + 6 cos θ

W_f(θ) = 2.15 + 0.61 cos 2θ + 6 cos θ (joules)

Difference between singly excited and doubly excited systems

In electromechanical energy conversion, the excitation system provides the magnetic field necessary for interaction with armature circuits to produce torque or force. Depending on how many separate excitation inputs are present, systems are classified as singly or doubly excited.

What is a singly excited system?

A singly excited system has only one electrical input (one excitation winding). The magnetic field is produced by a single coil and the other part of the machine (e.g. rotor) is not independently excited. A typical schematic shows a single coil on a magnetic core.

• Torque in such systems is often of the reluctance or saliency type: the rotor tends to move to a position of minimum reluctance for the magnetic flux.
• Examples include variable reluctance devices and simple electromagnetic actuators.

What is a doubly excited system?

A doubly excited system has two independent excitation inputs (two windings supplied from separate sources). One common example is the synchronous machine where the stator winding and the rotor field winding are excited independently.

• Both windings contribute to the resultant magnetic field and can interact to produce co-alignment torque as well as reluctance torque if inductances vary with position.
• Because excitation is independent, the control and characteristics (such as torque production and power flow) are richer than in singly excited systems.

The following table highlights the major differences between singly excited system and doubly excited system -

Conclusion

• A singly excited system uses one coil to produce the working magnetic flux and typically produces reluctance torque when inductances vary with position.
• A doubly excited system uses two independent coils and therefore supports interaction (co-alignment) torque in addition to possible reluctance torque; this enables independent control of fluxes and richer operating behaviour in machines and transducers.
• Energy and torque in doubly-excited systems are derived from the same basic relations: flux linkages expressed by self and mutual inductances, KVL for each winding, and the positional dependence of inductances which leads to mechanical torque.