Function of the Commutator

Consider only one coil \(AB\) placed in a strong magnetic field.

• The two ends of the coil are joined to slip rings.

• Two brushes rest on these slip rings.

• When the coil is rotated in the counter-clockwise direction at \(\omega\) radians per second, \(\phi\) is cut by the coil and an EMF is induced in it.

• The induced EMF is alternating, and the current flowing through the external resistance is also alternating — that is, at the second instant current flows in the external resistance from \(M\) to \(L\), whereas at the fourth instant it flows from \(L\) to \(M\).

Now consider that the two ends of the coil are connected to only one slip ring split into two parts — segments \(A''\) and \(B''\).

• Each part is insulated from the other by a mica layer.

• Two brushes rest on these parts of the ring.

The mechanism is worth stating plainly, because it is the whole of the DC machine in one line. The coil EMF reverses every half revolution, and so does the external connection. Two reversals cancel, and the load sees a constant polarity. The split occurs at the instant the coil EMF passes through zero, so the swap costs nothing.

In an actual machine there are a number of coils connected to a number of segments of the ring, called the commutator, and these coils deliver EMF and current to the external load.

• The actual current flowing in the external load is shown by the firm line, which fluctuates only slightly.

• The number of coils placed on the armature is much greater than shown, and a practically pure direct current is obtained at the output.

Chapter 21 quantified the residual fluctuation: the ripple is \(1 - \cos(90^{\circ}/n)\) for \(n\) coils per pole pair, falling from 100 % with a single coil to under half a percent with sixteen. That is what "fluctuates only slightly" means numerically.

Consider an armature with four coils — 1, 2, 3 and 4 — equally spaced in slots. The number of commutator segments is equal to the number of coils.

When the armature is rotated clockwise, the direction of the induced EMF — and hence the current — is:

The next question is how many brush sets are required and where they are to be placed with respect to the poles. The brushes are to collect or deliver the current, and are placed at such a position that sparking is minimum.

Applying the rule to the four-coil armature:

• At segment 1, current is separated towards conductors 1 and \(4'\) — hence it is a separating point.

• Current is coming towards segment 3 from conductors \(2'\) and 3 — hence it is a meeting point.

• So commutator segment 1 is the position of the negative brush, and segment 3 the position of the positive brush.

Rule (ii) brings out a very important point: in certain positions of the commutator the brushes will actually be short-circuiting the coils connected to the segments with which they are in contact.

• In the case illustrated, two coils are short-circuited, and as such the width of the brushes must be greater than the thickness of the mica insulation between segments.

• The negative brush short-circuits segments 1 and 4, whereas the positive brush short-circuits segments 2 and 3.

• Coils 4 and 2 are short-circuited by the brushes, and hence no EMF should be induced in these coils.

• The armature winding forms a closed circuit and consists of two parallel paths.

A coil stays short-circuited only briefly. The time for which it is so is the commutation period \(T_c\), and it follows directly from the geometry.

The brush must travel its own width, less the mica it started on, relative to the commutator surface:

Two consequences follow immediately, and both matter for Chapter 31:

• \(T_c \propto 1/N\) — the faster the machine runs, the less time there is to reverse the current.

• \(T_c \propto w_b\) — a wider brush gives more time, but short-circuits more coils at once.

The number of coils short-circuited at any instant follows from the segment pitch:

Consider what the short-circuited coil must actually do. Before commutation it belongs to one parallel path and carries \(+I_a/A\); afterwards it belongs to the other path and carries \(-I_a/A\). The total change is

and it must happen within the commutation period \(T_c\).

Collecting the results of this chapter shows why the commutator, and not the magnetics, sets the ceiling on a DC machine.

• A coil connected through slip rings delivers alternating current — the machine is an alternator.

• Replacing them by one ring split into two segments, insulated by mica, leaves the coil EMF unchanged in magnitude but makes the external current unidirectional.

• In a real machine many coils feed many segments, and the output fluctuates only slightly — practically pure DC.

• The function of the commutator is to convert the armature's AC into DC in the external circuit (generator), or the alternating torque into continuous torque (motor).

• Coils are connected so that the EMFs of their two sides add, and the number of segments equals the number of coils.

• Rule (i): place a brush at each meeting point (positive) or separating point (negative) of two EMFs. Rule (ii): brushes are equally spaced, opposite the pole centres, conventionally at the G.N.P.

• Brushes short-circuit the coils they bridge. This is harmless because those coils are at the magnetic neutral axis where no EMF is induced — and the brush must be wider than the mica so that contact transfers make-before-break.

• The commutation period is \(T_c = (w_b - w_m)/V_c\), typically one or two milliseconds, and varies as \(1/N\).

• In that time the coil current must reverse from \(+I_a/A\) to \(-I_a/A\). Coil inductance opposes this, producing a reactance voltage \(E_r = L\,\Delta I/T_c\) which causes under-commutation and sparking.

• \(E_r\) grows with both current and speed, which is why commutation — not the magnetic circuit — caps a DC machine's power.

• Saying the commutator makes the coil EMF direct. It does not. The coil EMF stays alternating and unchanged in magnitude; only the external connection is switched.

• Confusing slip rings with a split ring. Two separate rings give AC; one ring split in two gives DC.

• Thinking the short-circuiting of coils is a fault. It is deliberate and unavoidable, and harmless provided it happens at the neutral axis.

• Making the brush narrower than the mica. That would break the armature circuit under full current and cause a violent arc.

• Taking the current change as \(I_a/A\). It is \(2I_a/A\), because the current reverses rather than merely falling to zero.

• Forgetting to subtract the mica thickness when computing the commutation period.

• Using rev/min directly as a speed. The peripheral speed needs the \(\pi D_cN/60\) conversion.

• Reversing the polarity rule. Meeting point is positive, separating point is negative.

• Assuming a wider brush is always better. It lengthens \(T_c\) but short-circuits more coils, and they must all lie near the neutral zone.

• Believing the commutator only causes wear. Its real cost is a ceiling on the product of current and speed.

Convert speeds to peripheral speed before computing any time, and remember the current reverses rather than merely falling.

1. P23.1 A commutator 200 mm in diameter runs at 1200 rev/min with 15 mm brushes and 1.0 mm mica. Find the commutation period.

   Show answer

   \[V_c = \frac{\pi(0.200)(1200)}{60} = 12.57~\mathrm{m/s}\]

   \[T_c = \frac{(15 - 1.0)\times10^{-3}}{12.57} = 1.114\times10^{-3}~\mathrm{s} = 1.11~\mathrm{ms}\]

2. P23.2 The commutator of P23.1 has 60 segments. How many coils does each brush short-circuit?

   Show answer

   \[\text{pitch} = \frac{\pi(0.200)}{60} = 10.47~\mathrm{mm}, \qquad \frac{15}{10.47} = 1.43\]

   The brush spans between one and two pitches, touching 2 segments, so it short-circuits 1 coil.

3. P23.3 For the machine of P23.1, lap-wound with 4 poles, carrying 160 A with coil inductance 0.05 mH, find the reactance voltage.

   Show answer

   \[\Delta I = \frac{2(160)}{4} = 80~\mathrm{A}\]

   \[E_r = \left(0.05\times10^{-3}\right)\frac{80}{1.114\times10^{-3}} = 3.59~\mathrm{V}\]

   Below the 5 V guideline, so the machine should commutate acceptably without interpoles.

4. P23.4 For the machine of P23.3, what armature current would bring the reactance voltage to 5 V?

   Show answer

   Since \(E_r \propto I_a\):

   \[I_a = 160\left(\frac{5.00}{3.59}\right) = 223~\mathrm{A}\]

   An overload of only 39 % would reach the limit — which is why DC machines of any size are fitted with interpoles.

5. P23.5 An 8-pole machine carries 480 A. Find the number of brush sets and current per set for lap and wave windings.

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   \[\text{lap: } A = 8, \quad 8 \text{ sets}, \quad I_{\text{set}} = \frac{480}{8} = 60~\mathrm{A}\]

   \[\text{wave: } A = 2, \quad 2 \text{ sets}, \quad I_{\text{set}} = \frac{480}{2} = 240~\mathrm{A}\]

6. P23.6 Compare the reactance voltage of the lap and wave versions in P23.5, all else equal.

   Show answer

   Since \(E_r \propto 1/A\):

   \[\frac{E_{r,\text{wave}}}{E_{r,\text{lap}}} = \frac{A_{\text{lap}}}{A_{\text{wave}}} = \frac{8}{2} = 4\]

   The wave winding has four times the reactance voltage, because each of its two paths carries four times the current and so each coil must reverse four times as much.

   This is a real disadvantage of wave windings on high-current machines, and a further reason why lap is preferred for low-voltage, high-current duty.

7. P23.7 Why must a brush be wider than the mica between segments?

   Show answer

   If it were narrower, there would be an instant when the brush touched neither segment, breaking the armature circuit while it carried full current. The winding's inductance would then force a violent arc, exactly as when an inductive circuit is switched off.

   A wider brush ensures the transfer is make-before-break: the brush touches the next segment before leaving the previous one. The resulting short circuit is deliberate, and is harmless because the coil concerned is at the neutral axis where its EMF is zero.

8. P23.8 Explain the difference between under- and over-commutation and say which is the more serious.

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   Under-commutation — the reversal is delayed by the coil's inductance, so at the end of \(T_c\) the current has not finished reversing. The remainder is interrupted as the brush leaves the segment, producing an arc at the trailing edge.

   Over-commutation — the reversal happens too early, which can cause sparking at the leading edge but is far less damaging.

   Under-commutation is the serious one, because it is what inductance naturally produces and because burning at the trailing edge rapidly ruins the commutator surface. Interpoles are deliberately made slightly over-strong to push the machine towards over-commutation.

9. P23.9 A machine's speed is doubled and its current halved. What happens to the reactance voltage?

   Show answer

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

   It is unchanged. Note that the output power is also unchanged, since \(P \propto EI_a \propto NI_a\) — which is the point of Section 23-9: the reactance voltage tracks the power, not the speed or current separately.

10. P23.10 Chapter 18's sizing equation says a machine gives more power if it runs faster. Why does this argument fail for DC machines?

    Show answer

    The sizing equation \(P = \omega T = \omega\sigma\pi D^{2}l/2\) assumes the only limits are magnetic saturation and heat removal. For a DC machine there is a third limit: commutation.

    Since \(E_r \propto N I_a\) and the output is also proportional to \(N I_a\), raising the power by any means raises the reactance voltage in exact proportion. Once \(E_r\) reaches the limit that interpoles can correct, the machine cannot deliver more power however good its magnetics or cooling.

    This is the fundamental reason DC machines were displaced by inverter-fed AC drives, which have no commutator and therefore no such ceiling.