Doubly Excited Magnetic Systems

A doubly excited magnetic system has two windings sharing one magnetic circuit, with a movable member between them. Typically one winding is on the stationary member and the other on the rotating one — a field winding and an armature winding, in the language of Part 2.

Chapter 16 ended on a limitation. A singly excited system develops a force proportional to \(i^{2}\), so it can only attract, never repel, and reversing the supply changes nothing. That is enough for a relay but useless for a machine, which must be able to motor and to brake, to run forwards and backwards.

The second winding removes the limitation entirely. As Section 17-3 shows, a new torque term appears which is proportional to the product \(i_1i_2\) rather than to a square. Reverse one current and the torque reverses; reverse both and it does not. Every DC machine, every synchronous machine and every induction machine in this book runs on that term.

With two windings the flux linkage of each depends on both currents:

and the coenergy, obtained by the same integration as in Chapter 15 but now over both currents in turn, is the expression already met in Chapter 14:

Which of the three actually varies with angle depends on the geometry, and this is worth settling before any calculation:

• \(L_1\), the stator self-inductance, varies with \(\theta\) only if the rotor is salient — a shaped rotor presents different reluctance at different angles.

• \(L_2\), the rotor self-inductance, varies with \(\theta\) only if the stator bore is non-uniform, which for a slotted but essentially cylindrical stator it is not. In most machines \(L_2\) is constant.

• \(M\), the mutual inductance, always varies with \(\theta\), being greatest when the two windings' axes coincide and zero when they are perpendicular:

  \[M(\theta) = M_{\max}\cos\theta\]

The torque follows from Chapter 16's rule, differentiating the coenergy with respect to angle while holding both currents constant:

Substituting \(M = M_{\max}\cos\theta\) gives the alignment torque explicitly:

so \(T \propto \sin\theta\) — a result obtained quite differently in Section 17-7, from pure geometry.

The energy method gives the answer but not much physical insight. The classical approach reaches the same result by asking what happens when a magnetisable body is placed in an existing field and allowed to turn.

The following cases are considered to understand the process of torque development by the alignment of two fields:

• Soft iron piece placed in the magnetic field.

• Permanent magnet placed in the magnetic field.

• Electromagnet placed in the magnetic field.

In each case a stationary main field \(F_m\) is produced by permanent magnets, and a rotatable member develops its own field \(F_r\). The angle between the two axes is the torque angle \(\theta\), and the torque acts to bring the two into line.

Consider a soft iron piece, capable of free rotation, placed in the magnetic field of two permanent magnets. The magnetic lines of force are set up in the soft iron piece.

Position 1 — Aligned, \(\theta = 0\)

• The molecular poles get aligned parallel to the magnetic field due to magnetic induction.

• The soft iron piece obtains the polarities as marked.

• This is the stable position of the soft iron piece.

• Torque produced in this case is zero.

Position 2 — Rotated, \(\theta \lt 90^{\circ}\)

• If the soft iron piece is rotated through an angle \(\theta\) (with \(\theta \lt 90^{\circ}\)), then by magnetic induction ends A and B become north and south poles respectively.

• A force of attraction acts on the two ends and the soft iron piece will try to come in line with the main field — that is, the position of least reluctance path.

• This anticlockwise torque tries to decrease the angle \(\theta\) and is considered as negative.

Position 3 — Perpendicular, \(\theta = 90^{\circ}\)

• When the soft iron piece is rotated through an angle \(\theta = 90^{\circ}\), an equal force of attraction and repulsion acts on each end of the soft iron piece, therefore the torque produced is zero.

• This is the unstable position of the soft iron piece, because a slight change in angle \(\theta\) in either direction will create a torque in that direction.

Position 4 — Beyond, \(\theta \gt 90^{\circ}\)

• When the soft iron piece is rotated through an angle \(\theta \gt 90^{\circ}\), then by magnetic induction ends A and B become south and north poles respectively — the polarities have reversed.

• The torque now acts to carry the piece onward to \(\theta = 180^{\circ}\), which is a second stable position identical in every respect to \(\theta = 0\).

Replace the soft iron by a permanent magnet, free to rotate. Now the rotor has a polarity of its own, which travels with it.

Position 1 — In line, \(\theta = 0\)

• There is a force of attraction on the north and south pole of the rotating magnet which, being equal and opposite, cancel each other.

• In this position the torque produced is zero, because the field of the rotating magnet \(F_r\) and the field of the stationary permanent magnet \(F_m\) are in line with each other.

Position 2 — Between 0 and 180°

• When the magnet is rotated through an angle \(\theta\) — less than 90°, equal to 90°, or more than 90° but less than 180° — its north pole will be attracted towards the south pole, and its south pole towards the north pole, of the stationary permanent magnets.

• In other words, the rotor field \(F_r\) tries to come in line with the main field \(F_m\), and torque is developed.

• This anticlockwise torque is considered negative, because it is decreasing the torque angle \(\theta\).

Position 3 — Opposed, \(\theta = 180^{\circ}\)

• When the magnet is rotated through an angle \(\theta = 180^{\circ}\), the two fields \(F_r\) and \(F_m\) are in line with each other but acting in opposite directions.

• Therefore the torque developed is zero, but this is the unstable position, because a slight change in angle \(\theta\) in either direction will create a torque in that direction and the rotor will not regain its original position.

Position 4 — Between 180° and 360°

• When the magnet is rotated through an angle \(\theta\) more than 180° but less than 360° — that is, less than 270°, equal to 270°, or more than 270° — its north and south poles will be attracted towards the south and north poles of the stationary permanent magnets respectively.

• In other words \(F_r\) will again try to come into line with \(F_m\), but now the torque is positive, carrying the rotor onward towards \(360^{\circ}\) — which is the same position as \(0^{\circ}\).

Now replace the permanent magnet by an electromagnet — a coil of current-carrying conductors on a rotatable former. This is the actual armature of a machine, and the analysis gives the classical torque expression.

Let

In the right-angled triangle, the angle \(aob = \theta\), so the distance perpendicular to the force is

The total torque acting on the two conductors is therefore

The force on a current-carrying conductor in a field was established in Chapter 1 as \(F = BIl\), where

Most real machines have both components. A salient-pole synchronous machine has a shaped rotor, so \(L_1\) varies with angle, and a field winding, so \(M\) varies too. The total torque is the sum:

where \(\delta\) is the torque angle — called the load angle in Chapter 76 — and the two coefficients come from the alignment and reluctance terms respectively.

There is a difficulty with everything above, and it is fundamental. By the alignment of two fields torque develops, but the torque produced is not unidirectional.

The reason is visible in the figure of Section 17-8. As the rotor turns, \(\theta\) changes, the torque falls to zero at alignment and then reverses. Left alone, the rotor would oscillate about the aligned position and settle there — which makes an excellent compass needle but a useless motor.

• A doubly excited system has two windings on one magnetic circuit. Its coenergy is \(W' = \tfrac{1}{2}L_1i_1^{2} + \tfrac{1}{2}L_2i_2^{2} + Mi_1i_2\).

• The torque has three components: two reluctance terms and one alignment term, \(T = \tfrac{1}{2}i_1^{2}\mathrm{d}L_1/\mathrm{d}\theta + \tfrac{1}{2}i_2^{2}\mathrm{d}L_2/\mathrm{d}\theta + i_1i_2\,\mathrm{d}M/\mathrm{d}\theta\).

• With \(M = M_{\max}\cos\theta\) the alignment torque is \(-i_1i_2M_{\max}\sin\theta\). It survives without saliency and can be reversed.

• Soft iron in a field has induced poles that reverse past 90°, giving stable positions at 0° and 180° and torque varying as \(\sin2\theta\) — it is reluctance torque.

• A permanent magnet or electromagnet carries fixed polarity, giving one stable position at \(\theta = 0\), an unstable one at 180°, and torque varying as \(\sin\theta\).

• Geometrically, \(T = 2Fr\sin\theta\), and with \(F = BIl\) this gives \(T = 2BIlr\sin\theta = K_L\sin\theta\), so \(T \propto \sin\theta\).

• A salient machine has both components: \(T = T_1\sin\delta + T_2\sin2\delta\). Saliency raises the peak torque and moves it to an angle below 90°.

• Alignment torque is not unidirectional. Continuous torque needs either a rotating main field (AC machines) or current switching in the armature (DC machines, by commutator).

• Writing the mutual term with a factor of one half. The self terms carry \(\tfrac{1}{2}\), the mutual term does not: \(Mi_1i_2\), not \(\tfrac{1}{2}Mi_1i_2\).

• Assuming both self-inductances vary with angle. Usually only \(L_1\) does, because the stator bore is cylindrical and the rotor sees no variation.

• Expecting alignment torque to need saliency. It does not — a perfectly round rotor still produces it, through \(\mathrm{d}M/\mathrm{d}\theta\).

• Confusing the two torque periods. Alignment torque varies as \(\sin\theta\) over 360°; reluctance torque as \(\sin2\theta\) over 180°.

• Treating the soft-iron case as alignment torque. Soft iron has no excitation of its own, so it is singly excited and produces reluctance torque only.

• Placing the stable position at 90°. Torque is greatest at 90°, not zero; the stable position is at \(\theta = 0\) where the torque vanishes and any displacement is opposed.

• Assuming the total torque peaks at 90°. With saliency it peaks earlier — 68.5° in Example 17.4.

• Reversing both currents to reverse a machine. That leaves \(i_1i_2\) unchanged and the torque unchanged. Swap one only.

• Forgetting the factor of two in \(T = 2Fr\sin\theta\). Both poles, or both conductors, contribute.

• Believing alignment torque alone can run a motor. It reverses every half revolution. Continuous rotation requires commutation or a rotating field.

Identify which inductances vary with angle before differentiating — constant terms contribute nothing.

1. P17.1 A doubly excited system has \(M = 0.20\cos\theta\) H with \(i_1 = 12\) A and \(i_2 = 6.0\) A, and a cylindrical rotor. Find the torque at 45° and its maximum.

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   \[T = -i_1i_2M_{\max}\sin\theta = -(12)(6.0)(0.20)\sin\theta = -14.4\sin\theta ~\mathrm{N\,m}\]

   \[T(45^{\circ}) = -(14.4)(0.7071) = -10.2~\mathrm{N\,m}, \qquad T_{\max} = 14.4~\mathrm{N\,m} \text{ at } 90^{\circ}\]

2. P17.2 For the system of P17.1, find the torque if \(i_1\) is reversed, and if both currents are reversed.

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   \(i_1\) reversed: the product \(i_1i_2\) changes sign, so \(T = +14.4\sin\theta\) — the torque reverses.
   Both reversed: the product is unchanged, so \(T = -14.4\sin\theta\) — no change.

3. P17.3 An armature coil has two conductors of length 0.30 m at radius 0.12 m carrying 20 A in a field of 0.75 T. Find \(K_L\) and the torque at 40°.

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   \[K_L = 2BIlr = 2(0.75)(20)(0.30)(0.12) = 1.080~\mathrm{N\,m}\]

   \[T = (1.080)\sin 40^{\circ} = (1.080)(0.6428) = 0.694~\mathrm{N\,m}\]

4. P17.4 A system has \(L_1 = 0.40 + 0.08\cos2\theta\) H, \(L_2\) constant, and \(M = 0.18\cos\theta\) H, with \(i_1 = 8.0\) A and \(i_2 = 4.0\) A. Find both torque components at \(\theta = 60^{\circ}\).

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   \[T_{\text{rel}} = \tfrac{1}{2}(64)\left(-0.16\sin 120^{\circ}\right) = (32)(-0.1386) = -4.44~\mathrm{N\,m}\]

   \[T_{\text{align}} = (32)\left(-0.18\sin 60^{\circ}\right) = (32)(-0.1559) = -4.99~\mathrm{N\,m}\]

   \[T = -9.43~\mathrm{N\,m}\]

5. P17.5 A salient machine has \(T = 300\sin\delta + 80\sin2\delta\) N·m. Find the torque at 45° and 90°, and the maximum torque with its angle.

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   \[T(45^{\circ}) = 300(0.7071) + 80(1) = 212.1 + 80.0 = 292.1~\mathrm{N\,m}\]

   \[T(90^{\circ}) = 300 + 0 = 300~\mathrm{N\,m}\]

   Setting \(300\cos\delta + 160\cos2\delta = 0\) and using \(\cos2\delta = 2\cos^{2}\delta - 1\):

   \[320\cos^{2}\delta + 300\cos\delta - 160 = 0 \quad\Longrightarrow\quad \cos\delta = 0.3796\]

   \[\delta_{\max} = 67.7^{\circ}, \qquad T_{\max} = 300(0.9251) + 80(0.7024) = 333.7~\mathrm{N\,m}\]

   Saliency gives 11.2 % more peak torque than the 300 N·m of a cylindrical machine.

6. P17.6 A soft iron piece is placed in a magnetic field. State its stable and unstable positions and the period of its torque.

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   Stable: \(\theta = 0\) and \(180^{\circ}\) — both aligned with the main field, since the induced polarities simply reverse.
   Unstable: \(\theta = 90^{\circ}\) and \(270^{\circ}\) — the attracting and repelling forces balance, but any displacement grows.
   Period: 180°, with torque varying as \(\sin2\theta\). The piece cannot distinguish the two ends of the main field because it has no polarity of its own.

7. P17.7 Repeat P17.6 for a permanent magnet, and explain the difference.

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   Stable: \(\theta = 0\) only.
   Unstable: \(\theta = 180^{\circ}\).
   Period: 360°, with torque varying as \(\sin\theta\).

   The difference is that a permanent magnet carries a fixed polarity that rotates with it, so it can distinguish the north end of the main field from the south. Soft iron's polarity is induced and flips as it turns, halving the period. This is precisely the distinction between alignment torque and reluctance torque.

8. P17.8 At what torque angle does a cylindrical machine deliver half its maximum torque, and what does this imply for a single-coil armature?

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   \[\sin\theta = 0.5 \quad\Longrightarrow\quad \theta = 30^{\circ} \ \text{or} \ 150^{\circ}\]

   A single-coil armature would therefore spend much of each revolution producing well under its peak torque, and none at all at alignment. Real armatures carry many coils spaced around the periphery, so that as one approaches alignment another is near 90°, keeping the total torque nearly constant. Chapter 23 develops this.

9. P17.9 State the two ways of obtaining unidirectional torque and name the family of machine each produces.

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   1. Rotating the main field, so the rotor is dragged along by its tendency to align. Achieved electrically by a polyphase winding. Gives the synchronous machine (rotor in exact step) and the induction machine (rotor slipping, currents induced).

   2. Switching the armature current, so that conductors under a given main pole always carry current in the same direction. Achieved by the commutator and brushes. Gives the DC machine.

10. P17.10 A DC motor is to be reversed. Explain why swapping both the field and armature connections fails, and which is normally swapped in practice.

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    Torque depends on the product \(i_1i_2\). Reversing both leaves the product unchanged, so the machine continues to turn the same way — the reason a series motor works equally well on AC.

    In practice the armature connections are swapped. The field winding is highly inductive, so interrupting it produces a large voltage transient, and its flux takes considerable time to reverse. Armature reversal is faster and electrically safer.