Commutation, Interpoles and Compensating Windings

Chapter 23 established what has to happen. A coil passing under a brush is momentarily short-circuited, and during that interval its current must reverse completely — from \(+I_a/A\) to \(-I_a/A\) — because it is leaving one parallel path and joining another.

Chapter 30 has since added a complication. The assumption that the commutating coil lies at the magnetic neutral axis, where no EMF is induced in it, holds only on no load. Under load, armature reaction shifts the MNA away from the brushes, so the shorted coil sits in a field of the wrong polarity and generates an EMF that opposes the reversal.

This chapter deals with both problems: the coil's own inductance, and the field it finds itself in.

The coil has self-inductance, and by Chapter 13 any attempt to change its current induces an opposing EMF.

The coil resistance is only a few milliohms, so even a few volts of reactance voltage would drive an enormous circulating current if nothing checked it. What checks it is the brush contact resistance — the subject of Section 31-5.

There are two families of remedy, and they attack the problem in quite different ways.

Small machines rely on resistance commutation alone. Anything above a few kilowatts uses interpoles, and large machines subject to heavy overloads add compensating windings as well.

As the brush moves off one segment and onto the next, the contact area with the leaving segment shrinks while that with the arriving segment grows. Since resistance varies inversely with contact area, the resistance of the outgoing path rises steadily and that of the incoming path falls.

This is the third reason Chapter 22 gave for using carbon brushes despite their poor conductivity — alongside lubrication and their being sacrificial. The 1 V contact drop is the price paid for it.

If an EMF equal and opposite to the reactance voltage could be induced in the commutating coil, the reversal would proceed as though the inductance did not exist. The first method of doing so is simply to move the brushes.

1. Shift the brushes forward, in the direction of rotation, past the new magnetic neutral axis.
2. The commutating coil then lies under the fringe of the next main pole, of opposite polarity to the one it has just left.
3. Moving in that field, it generates a motional EMF which drives the current in the new direction — exactly what is wanted.

Brush shifting is therefore used only on small machines with a steady load, and on machines that must run in one direction only — for a reversible machine the shift would have to be the other way round.

An interpole — also called a commutating pole or compole — is a narrow auxiliary pole fitted midway between the main poles, on the quadrature axis, directly over the commutating zone.

The interpole must do two jobs, and its ampere-turns are the sum of the two requirements.

The flux density required follows from the EMF the commutating coil must generate. A coil of \(T\) turns, its conductors moving at peripheral velocity \(v\) through a field \(B_{ip}\) over an active length \(l\), generates

the factor 2 arising because each turn has two active conductors. Setting \(E_{ip} = E_r\) gives the required density.

Chapter 30 introduced compensating windings. The two devices are complementary, and it is worth being precise about which does what.

• During commutation the coil current must reverse from \(+I_a/A\) to \(-I_a/A\) within \(T_c = (w_b - w_m)/V_c\).

• The coil's self-inductance produces a reactance voltage \(E_r = L\,\Delta I/T_c\) opposing the reversal. It varies as \(I_aN/A\) — the same product as the output power, which is why it caps the machine.

• Under-commutation leaves current to be interrupted at the trailing brush edge, causing arcing and commutator burning. Over-commutation is much less harmful and is deliberately preferred.

• Resistance commutation uses the rising contact resistance of a carbon brush to spread the current transfer over the whole period. It helps, but provides no opposing EMF.

• EMF commutation by brush shift generates a commutating EMF from the main field, but is correct at only one load, demagnetises the machine by \(ZI\theta_m/360\), and cannot serve a reversible machine.

• Interpoles sit on the q-axis and are connected in series with the armature, so their EMF tracks \(I_a\) exactly as the reactance voltage does — and reverses automatically if the machine reverses.

• Interpole polarity: in a generator, the same as the next main pole ahead in the direction of rotation; in a motor, the reverse.

• \(AT_{ip} = ZI/2P + B_{ip}l_g/\mu_0\), and \(B_{ip}\) follows from \(E_{ip} = 2TB_{ip}lv = E_r\).

• The interpole gap is made large so its circuit stays linear and its flux strictly proportional to \(I_a\).

• Interpoles cure sparking; compensating windings cure distortion. Together they cancel the armature mmf all the way round.

• Using \(I_a\) instead of \(2I_a/A\) for the current change. The coil reverses, so the swing is twice the path current.

• Using the commutator velocity in the commutating-EMF formula. \(E_{ip}\) needs the armature velocity \(\pi DN/60\), since that is where the conductors are.

• Omitting the factor 2 in \(E_{ip} = 2TB_{ip}lv\). Each turn has two active conductors.

• Forgetting the armature-cancelling term in \(AT_{ip}\). It is usually the larger of the two.

• Connecting interpoles in shunt. They must be in series, or their strength will not track the load.

• Getting the interpole polarity backwards. Generator: same as the next main pole ahead. Motor: the reverse.

• Making the interpole gap small. A large gap keeps the circuit linear so the flux stays proportional to \(I_a\).

• Expecting resistance commutation to supply an EMF. It only makes the transfer gradual.

• Confusing the roles of interpoles and compensating windings. Interpoles work in the interpolar zone; compensating windings under the pole arc.

• Treating over-commutation as equally bad. It sparks at the leading edge and is much less damaging, which is why designs are biased towards it.

Distinguish carefully between commutator and armature diameters, and between path current and the full reversal.

1. P31.1 A commutator 250 mm in diameter runs at 1000 rev/min with 16 mm brushes and 0.8 mm mica. Find \(T_c\).

   Show answer

   \[V_c = \frac{\pi(0.250)(1000)}{60} = 13.09~\mathrm{m/s}\]

   \[T_c = \frac{15.2\times10^{-3}}{13.09} = 1.161\times10^{-3}~\mathrm{s} = 1.16~\mathrm{ms}\]

2. P31.2 For the machine of P31.1, 4-pole lap-wound with \(I_a = 240\) A and coil inductance 0.05 mH, find \(E_r\).

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   \[\Delta I = \frac{2(240)}{4} = 120~\mathrm{A}\]

   \[E_r = \left(0.05\times10^{-3}\right)\frac{120}{1.161\times10^{-3}} = 5.17~\mathrm{V}\]

   Above the 5 V guideline, so interpoles are required.

3. P31.3 The armature of P31.2 is 350 mm in diameter with an active length of 200 mm, and each coil has 2 turns. Find the required interpole flux density.

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   \[v = \frac{\pi(0.350)(1000)}{60} = 18.33~\mathrm{m/s}\]

   \[B_{ip} = \frac{E_r}{2Tlv} = \frac{5.17}{(2)(2)(0.200)(18.33)} = \frac{5.17}{14.66} = 0.353~\mathrm{T}\]

   Somewhat high; a designer would use a coil of more turns or accept a larger interpole.

4. P31.4 The machine of P31.2 has 400 armature conductors and an interpole gap of 6 mm. Using \(B_{ip} = 0.353\) T, find the interpole ampere-turns and turns.

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   \[I = \frac{240}{4} = 60~\mathrm{A}, \qquad \frac{ZI}{2P} = \frac{(400)(60)}{8} = 3000~\mathrm{AT/pole}\]

   \[\frac{B_{ip}l_g}{\mu_0} = \frac{(0.353)\left(6\times10^{-3}\right)}{4\pi\times10^{-7}} = 1686~\mathrm{AT/pole}\]

   \[AT_{ip} = 3000 + 1686 = 4686, \qquad N_{ip} = \frac{4686}{240} = 19.5 \Rightarrow 20~\text{turns}\]

5. P31.5 A machine's speed is doubled and its armature current halved. What happens to \(E_r\) and to the interpole ampere-turns needed?

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   \[E_r \propto I_aN \quad\Longrightarrow\quad \frac{E_r'}{E_r} = \left(\tfrac{1}{2}\right)(2) = 1\]

   The reactance voltage is unchanged.

   But the interpole ampere-turns are not. The armature-cancelling term \(ZI/2P\) halves with the current, while the gap term must keep \(B_{ip}\) such that \(2TB_{ip}lv = E_r\) — and since \(v\) has doubled while \(E_r\) is unchanged, \(B_{ip}\) must halve too.

   Both terms halve, so \(AT_{ip}\) halves — which is exactly what the series winding delivers, since it carries the halved armature current. The scheme survives the change automatically.

6. P31.6 Why must interpoles be connected in series with the armature rather than the field?

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   Because the reactance voltage they must cancel is proportional to the armature current, and so the commutating EMF must be too.

   A shunt-connected interpole would be excited by the terminal voltage, which barely changes with load. It would be correct at one current and wrong at every other — the same defect as a fixed brush shift.

   The series connection makes the cure automatically proportional to the disease, at every load, with no adjustment. It also reverses the interpole polarity automatically if the machine reverses.

7. P31.7 Why is the interpole air gap made larger than the main air gap?

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   To keep the interpole magnetic circuit dominated by air and therefore linear.

   The whole scheme depends on \(B_{ip} \propto I_a\). If the interpole iron were allowed to saturate — as it would with a small gap and heavy overload — the flux would stop rising in proportion to the current, and the commutating EMF would fall short exactly when the reactance voltage was greatest.

   A large gap costs ampere-turns but guarantees proportionality over the full overload range.

8. P31.8 A DC machine sparks badly at the brushes on load but not on no load. What is happening?

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   On no load there is no armature current, so there is no reactance voltage and no armature reaction; the coil commutates in a genuinely neutral field.

   On load, two things go wrong together: the reactance voltage appears, and armature reaction shifts the MNA away from the brushes so the coil is no longer in a zero-EMF position.

   Likely causes: no interpoles fitted where they are needed; interpoles of insufficient ampere-turns; interpoles connected with wrong polarity; or brushes not on the correct axis. If the sparking gets worse in proportion to load, suspect the interpoles.

9. P31.9 Explain why designers deliberately make interpoles slightly over-strong.

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   An over-strong interpole produces over-commutation — the current reverses slightly early — which causes mild sparking at the leading brush edge.

   An under-strong interpole leaves under-commutation, in which unreversed current is interrupted at the trailing edge. That arc carries the full remaining current and burns the commutator surface, and the damage is progressive.

   The two errors are not symmetric in consequence, so the design is biased to the harmless side — typically a few percent of margin, as in Example 31.4.

10. P31.10 Distinguish the duties of interpoles and compensating windings, and say which machines need both.

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    Interpoles sit on the quadrature axis and act on the interpolar zone. They generate a commutating EMF to cancel the reactance voltage and locally neutralise the armature mmf. They cure sparking.

    Compensating windings sit in slots in the pole faces and act on the region under the pole arc. They cancel the cross-magnetising mmf there, preventing flux distortion. They cure the transient EMF induced by flux shifting when the load changes.

    Machines needing both are those with large, rapidly fluctuating loads — rolling-mill motors, turbo-generators, mine winders. Together the two cancel the armature mmf all the way round the periphery, so the machine behaves almost as though armature reaction did not exist.