Set 8 — Armature Windings, Lap and Wave
Set 7 asked for the number of parallel paths and was handed an answer: \(A = P\) for lap, \(A = 2\) for wave. This set earns it. The winding is laid out conductor by conductor — back pitch, front pitch, resultant pitch, commutator pitch — and the parallel paths, the brush arms and the current in each strand of copper all fall out of the geometry.
The drill is arithmetic on integers: pitches must be odd, the winding must close on itself, and a simplex wave winding refuses outright to exist for certain coil counts. The last two problems put money on it, comparing the voltage, current, resistance and rating of one armature wound both ways.
A winding is fully described by four pitches. Number the conductors so that odd numbers lie in the top layer and even numbers in the bottom. The back pitch \(Y_b\) is the jump across the back of the armature from one side of a coil to the other; the front pitch \(Y_f\) is the jump at the commutator end into the next coil; the resultant pitch \(Y\) is the advance per coil; and the commutator pitch \(Y_c\) is the number of segments spanned by one coil. Both \(Y_b\) and \(Y_f\) must be odd, so that every coil joins a top conductor to a bottom one.
Lap and wave differ in one sign. A lap coil comes back to the segment next to the one it left; a wave coil travels on to the next pole pair:
\[ \text{lap: } Y = Y_b - Y_f = \pm2,\ \ Y_c = \pm1 \qquad \text{wave: } Y = Y_b + Y_f = 2Y_A,\ \ Y_c = Y_A \]The plus sign is progressive, the minus retrogressive. Everything else on this page follows from that one difference.
The average pitch fixes the coil span. A full-pitched coil spans one pole pitch, so
\[ Y_A = \frac{Y_b+Y_f}{2} \approx \frac{Z}{P}\ \text{conductors} \qquad\text{and for a wave winding}\qquad Y_A = \frac{Z \pm 2}{P} \]A simplex wave winding is only possible for certain coil counts. The winding must close on itself after passing once round the armature, which forces
\[ Y_c = \frac{C \pm 1}{P/2} \quad\text{to be a whole number,}\qquad C = \frac{Z}{2} = \text{commutator segments} \]If it is not, one coil is left unconnected as a dummy and the count is reduced by one.
The winding fixes the number of parallel paths, and the paths fix the current. Lap gives \(A = P\) and needs \(P\) brush arms; simplex wave gives \(A = 2\) whatever the pole count and needs only two. Each conductor then carries
\[ I_c = \frac{I_a}{A}, \qquad \text{conductors in series per path} = \frac{Z}{A} \]Armature resistance follows the same rule twice over. A path contains \(Z/A\) conductors and there are \(A\) paths in parallel, so with \(r\) ohms per conductor
\[ R_a = \frac{Zr}{A^2} \]Equalisers belong to lap windings only. Points \(C/(P/2)\) segments apart lie under like poles and should be at the same potential; joining them lets the circulating current caused by unequal pole fluxes bypass the brushes. A wave winding needs none, because each of its two paths already samples every pole.
A 4-pole dc machine has a double-layer armature winding in 24 slots, one coil per slot, so that each slot carries two coil sides. The winding is to be simplex lap and progressive. Determine
- the number of conductors, coils and commutator segments
- the pole pitch, in conductors and in slots
- the back pitch, front pitch, resultant pitch and commutator pitch for a full-pitched coil
- the number of parallel paths and the number of brush arms
Count the copper first. Two coil sides per slot means two conductors per slot, and every coil has two sides, so coils and slots are equal in number. Each coil brings two ends to the commutator and each segment receives two ends, so segments equal coils:
Odd conductors 1, 3, 5, … lie in the top layer, even conductors 2, 4, 6, … in the bottom. Slot \(k\) therefore holds conductors \(2k-1\) and \(2k\).
The pole pitch is the distance between adjacent pole centres, measured in whatever units are convenient:
Choose the back pitch. A full-pitched coil spans one pole pitch, so \(Y_b\) should be near 12 — but it must be odd, otherwise the coil would join two top conductors and never reach the bottom layer. The two candidates are 11 and 13, and taking
sends conductor 1 (top of slot 1) to conductor 14 (bottom of slot 7) — a span of exactly 6 slots, which is the full pole pitch.
The front pitch follows from the lap condition. For a simplex progressive lap winding the coil must return to the segment next to the one it started from, so the resultant advance is two conductors:
Trace the winding to confirm it closes. Start at conductor 1, go back \(+13\), come front \(-11\), and repeat, reducing modulo 48:
| Coil | Start (top) | Back, \(+13\) | Front, \(-11\) |
|---|---|---|---|
| 1 | 1 — slot 1 | 14 — slot 7 | 3 |
| 2 | 3 — slot 2 | 16 — slot 8 | 5 |
| 3 | 5 — slot 3 | 18 — slot 9 | 7 |
| … | … | … | … |
| 24 | 47 — slot 24 | 60 − 48 = 12 — slot 6 | 1 — closed |
Every odd conductor is used once and the twenty-fourth coil lands back on conductor 1. The winding is closed and simplex.
Paths and brushes. Each brush arm short-circuits the coil under it and taps the winding once per pole, so a lap winding produces as many paths as poles and needs as many brush arms:
If a chorded coil is wanted — shorter than full pitch, which improves commutation — drop both pitches by two: \(Y_b = 11\), \(Y_f = 9\). The resultant pitch is still \(+2\) and the commutator pitch still \(+1\), so the winding is unchanged electrically; only the coil now spans 5 slots instead of 6.
The armature of a 4-pole dc generator has 21 slots with two coil sides per slot, wound double layer with one coil per slot. It is to carry a simplex wave winding. Find
- the average pitch, back pitch and front pitch
- the commutator pitch, for both the progressive and the retrogressive winding
- the number of parallel paths, the conductors in series per path, and the minimum number of brush arms
Copper count. As before, two coil sides per slot:
The pole pitch is \(Z/P = 42/4 = 10.5\) conductors — not a whole number, which is the first hint that a wave winding on an odd slot count cannot be exactly full-pitched.
The wave condition sets the average pitch. In a wave winding the coil does not come back; it goes on, and after \(P/2\) coils it has travelled once round the armature and must land two conductors away from where it started — not on top of it, or the winding would close after one lap and leave the rest of the copper unused. Hence \(P Y_A = Z \pm 2\) and
Split the average pitch into back and front pitches, both odd, summing to \(2Y_A\). For the progressive winding the symmetric choice is available:
Conductor 1 (top of slot 1) joins conductor 12 (bottom of slot 6): a span of 5 slots against a pole pitch of 5.25 slots, so the coil is very slightly chorded — the closest to full pitch that 21 slots and 4 poles allow.
The commutator pitch is the average pitch expressed in segments. There are half as many segments as conductors, so
Both are whole numbers here, so either winding may be executed on this armature.
Check that the progressive winding closes. Each coil advances the starting conductor by \(Y = Y_b + Y_f = 22\), reduced modulo 42:
| Coil | Start (top) | Back, \(+11\) | Front, \(+11\) |
|---|---|---|---|
| 1 | 1 — slot 1 | 12 — slot 6 | 23 |
| 2 | 23 — slot 12 | 34 — slot 17 | 45 − 42 = 3 |
| 3 | 3 — slot 2 | 14 — slot 7 | 25 |
| 4 | 25 — slot 13 | 36 — slot 18 | 47 − 42 = 5 |
After every \(P/2 = 2\) coils the winding has been once round the armature and moved forward by two conductors. Since \(\gcd(22,42) = 2\), the sequence visits all 21 odd conductors before returning to 1, so all 21 coils are included and the winding is closed.
Paths and brushes. Only two points of the winding are ever at the extreme potentials, so
Two brush arms are sufficient — that is the practical advantage of the wave winding on a machine with many poles. In practice all four arms are often fitted anyway, purely to spread the current over more brush area; it is permissible because in a wave winding the segments under like-polarity brushes are already at the same potential.
A simplex wave winding is proposed for each of the three armatures below. Decide in each case whether it can be executed, give the commutator pitch where it can, and state the remedy where it cannot.
- 4 poles, 20 coils
- 6 poles, 36 slots with one coil per slot
- 8 poles, 201 coils
Then state the general rule for the coil counts a given pole number will accept.
The test is whether the commutator pitch is a whole number. A wave coil advances \(Y_c\) segments, and after \(P/2\) coils it has been once round the commutator and moved on by one segment. That is the condition:
A fractional \(Y_c\) is meaningless — a coil cannot end on half a segment — so the winding simply cannot be made.
Apply it to all three:
| Case | \(P/2\) | \(C\) | \((C+1)/(P/2)\) | \((C-1)/(P/2)\) | Verdict |
|---|---|---|---|---|---|
| (a) | 2 | 20 | 10.5 | 9.5 | Impossible |
| (b) | 3 | 36 | 12.33 | 11.67 | Impossible |
| (c) | 4 | 201 | 50.5 | 50 | Retrogressive only |
Case (a): 4 poles, 20 coils. With \(P/2 = 2\) the numerator \(C\pm1\) must be even, so \(C\) must be odd. Twenty coils will not do. Leave one coil unconnected — a dummy coil, wound and wedged into its slot for mechanical and magnetic balance but with both ends insulated and no commutator connection — and the winding proceeds on 19 active coils:
The commutator then has 19 segments. The dummy coil generates an emf but delivers no current and reaches no commutator segment, so the armature produces \(19/20\) of the emf that twenty active coils would have given.
Case (b): 6 poles, 36 coils. Now \(P/2 = 3\), so \(C\pm1\) must be divisible by 3, which fails for any \(C\) that is itself a multiple of 3. Since \(36 = 3\times12\), both numerators fail. One dummy coil gives 35 active coils:
So only the progressive winding is available, with \(Y_c = 12\) and 35 commutator segments. Since \(\gcd(12,35) = 1\), the winding passes through every segment before closing.
Case (c): 8 poles, 201 coils. Here \(P/2 = 4\) and \(200/4 = 50\) exactly, so the retrogressive winding exists with \(Y_c = 50\). The progressive one does not, because \(202/4 = 50.5\). A wave winding needing only two brush arms on an 8-pole machine is exactly why this arrangement is chosen for high-voltage, moderate-current machines.
The general rule. Write \(p = P/2\) for the number of pole pairs. A simplex wave winding exists if and only if
For 4 poles (\(p=2\)) every odd \(C\) works. For 6 poles (\(p=3\)) any \(C\) not divisible by 3 works. For 8 poles (\(p=4\)) only \(C = 4m\pm1\) works — three coil counts in four are refused. A lap winding, by contrast, accepts every \(C\) without exception, because \(Y_c = \pm1\) can never be fractional.
An 8-pole shunt generator delivers 396 A at 500 V. Its armature has 1200 conductors and a resistance of 0.03 Ω when lap-wound; the shunt field resistance is 125 Ω. Find
- the armature current, and the current in each path and each conductor for the lap winding
- the same two currents if the identical armature were rewound simplex wave and made to carry the same armature current
- the conductor cross-section each winding would need at a current density of 5 A/mm²
- the armature resistance of the wave-wound version
The armature supplies the load and its own field. This is a generator, so the two currents add:
Lap winding: eight paths. The armature current divides equally between them, and a conductor belongs to exactly one path, so the path current is the conductor current:
Each path contains \(Z/A = 1200/8 = 150\) conductors in series.
Wave winding: two paths. The same 400 A now has only two routes:
Size the copper. At 5 A/mm² the two windings need very different conductors:
Ten square millimetres is an ordinary strip; forty is a bar four times the area, and 600 of them would never fit in the same slots. This is the real reason large-current machines are lap-wound: the wave winding at the same armature current is not merely inconvenient, it is unbuildable.
Armature resistance scales as \(1/A^2\). Each path has \(Z/A\) conductors in series and there are \(A\) such paths in parallel, so with \(r\) ohms per conductor \(R_a = Zr/A^2\). From the lap figure,
Halving the number of paths quadruples the conductors in series and quarters the parallel count — sixteen times the resistance, from the same copper.
What this machine actually generates. With the lap winding the emf is only a little above the terminals:
A wave-wound armature would generate four times as much — 2048 V — because the emf goes inversely with the number of paths and \(A\) falls from 8 to 2, at the same flux and speed, which is a different machine altogether. Problem 6 works that comparison out properly.
A 6-pole, 250 V dc generator has a simplex lap-wound armature of 72 coils and delivers 600 A. The resistance of each armature path is 0.09 Ω. Determine
- the current in each path, and the total armature resistance
- the commutator spacing between equipotential points, the number of points joined by one equaliser ring, and the largest number of rings the winding will take
- the number of rings and of tappings if the winding is equalised to 33⅓%
- the circulating current that would flow through the brushes, in the absence of equalisers, if one pole's flux were 2% low
Paths and path current. A lap winding gives one path per pole:
Six paths of 0.09 Ω in parallel; the full-load armature drop is \(600\times0.015 = 9\) V, or 3.6% of 250 V.
Why a lap winding needs equalisers at all. Each of the six paths lies under one pole and takes its emf from that pole alone. If the six pole fluxes are not identical — and they never are, because air gaps differ by fractions of a millimetre — the six paths generate different emfs, and current circulates between them. That current has nowhere to go except through the brushes, which are asked to carry it on top of their share of the load current. A wave winding is immune, because each of its two paths passes under every pole and so averages the fluxes.
Find the equipotential points. Points on the winding one pole pair apart occupy the same magnetic position and should be at the same potential. There are \(P/2 = 3\) pole pairs, so equipotential points lie
Segments 1, 25 and 49 form one such set; 2, 26 and 50 the next, and so on. Each equaliser ring therefore joins \(P/2 = 3\) points.
The maximum number of rings is reached when every segment is tapped:
Equalising to 33⅓% means using one third of that maximum:
The eight rings are spread evenly by tapping every third segment: ring 1 at segments 1, 25, 49; ring 2 at 4, 28, 52; ring 3 at 7, 31, 55; and so on to ring 8 at 22, 46, 70. Between 25% and 50% equalisation is normal practice; complete equalisation is rarely worth the copper.
Estimate the circulating current. Take the machine at its nominal 250 V and let one pole be 2% weak, so that one path generates 245 V while its neighbours generate 250 V. The weak path and a strong path form a closed loop of two path resistances:
Twenty-eight amperes of pure circulation, from a 2% flux error — 28% of the 100 A each path is meant to carry. Without equalisers it passes through the brushes and appears as unequal brush heating and sparking; with them it flows harmlessly round the equaliser ring, which is why the rings are made of comparatively light copper: they carry only this difference current, never load current.
A 4-pole armature carries 480 conductors, each of resistance 0.004 Ω and each able to carry 25 A continuously. The machine runs at 1500 rpm with a useful flux of 20 mWb per pole. For a simplex lap winding and for a simplex wave winding, find
- the generated emf
- the permissible armature current
- the armature resistance and the full-load terminal voltage
- the armature copper loss and the output
Everything in the emf equation is fixed except \(A\). Evaluate the part that the winding cannot touch:
The current rating is set by the conductor, not by the machine. Each path is a chain of conductors carrying 25 A, and the paths are in parallel:
Armature resistance. With \(Zr = 480\times0.004 = 1.92\ \Omega\) of copper in total,
Terminal voltage at full load:
The wave machine's drop is twice as many volts but the same percentage — 5% in both cases, since \(I_aR_a/E_a\) is independent of \(A\).
Losses and output. The armature copper loss is identical, as it must be: the same conductors carry the same 25 A in both machines.
| Quantity | Lap, \(A = 4\) | Wave, \(A = 2\) |
|---|---|---|
| Conductors in series per path | 120 | 240 |
| Generated emf \(E_a\) | 240 V | 480 V |
| Armature current \(I_a\) | 100 A | 50 A |
| Power developed \(E_aI_a\) | 24 kW | 24 kW |
| Armature resistance \(R_a\) | 0.12 Ω | 0.48 Ω |
| Copper loss \(I_a^2R_a\) | 1200 W | 1200 W |
| Terminal voltage \(V_t\) | 228 V | 456 V |
| Output \(V_tI_a\) | 22.8 kW | 22.8 kW |
| Brush arms | 4 | 2 |
| Equalisers | required | not required |
Read the last two rows of the table. Halving \(A\) doubles the voltage and halves the current; the product, the power, is untouched, and so is every loss. The same iron, the same copper and the same speed give the same 22.8 kW either way. What the winder is choosing is not how much power the machine delivers but the voltage at which it delivers it — and, with that, whether the machine needs four brush arms and a set of equalisers or two brush arms and none.
Key Formulas
| Quantity | Relation | Notes |
|---|---|---|
| Conductors, coils, segments | \(Z = 2C = S\,n\) | \(S\) slots, \(n\) conductors per slot — Problems 1, 2 |
| Pole pitch | \(Z/P\) conductors \(= S/P\) slots | Problems 1, 2 |
| Average pitch | \(Y_A = \dfrac{Y_b+Y_f}{2}\) | Equals the pole pitch when full-pitched — Problem 1 |
| Lap resultant pitch | \(Y = Y_b - Y_f = \pm2,\ Y_c = \pm1\) | \(+\) progressive, \(-\) retrogressive — Problem 1 |
| Wave resultant pitch | \(Y = Y_b + Y_f = 2Y_A\) | Both pitches odd — Problem 2 |
| Wave average pitch | \(Y_A = \dfrac{Z \pm 2}{P}\) | Must be a whole number — Problems 2, 3 |
| Wave commutator pitch | \(Y_c = \dfrac{C \pm 1}{P/2}\) | The feasibility test — Problems 2, 3 |
| Parallel paths | \(A = P\) lap, \(A = 2\) wave | Simplex windings — Problems 1, 2, 4, 6 |
| Brush arms | \(P\) for lap, 2 for wave | Minimum needed — Problems 1, 2 |
| Conductor current | \(I_c = I_a/A\) | Also the path current — Problems 4, 5, 6 |
| Conductors per path | \(Z/A\) | Sets the emf — Problems 4, 6 |
| Armature resistance | \(R_a = \dfrac{Zr}{A^2}\) | \(r\) per conductor — Problems 4, 6 |
| Generated emf | \(E_a = \dfrac{\phi PN}{60}\times\dfrac{Z}{A}\) | Set 7 — Problems 4, 6 |
| Equaliser spacing | \(C/(P/2)\) segments, \(P/2\) points per ring | Lap only — Problem 5 |
| Circulating current | \(I_{\text{circ}} = \dfrac{\Delta E}{2R_{\text{path}}}\) | Between two unequal paths — Problem 5 |
Common Mistakes
Choosing an even back pitch. An even \(Y_b\) joins a top conductor to another top conductor, so the coil never reaches the bottom layer and the winding cannot be made. Round the pole pitch to the nearest odd number — 13 or 11, never 12 — Problems 1 and 2.
Using \(Y = Y_b - Y_f\) for a wave winding. A lap coil doubles back, so its pitches subtract; a wave coil goes on, so they add. Getting this the wrong way round turns a resultant pitch of 22 into one of 0 — Problem 2.
Assuming any coil count will take a wave winding. It will not: \((C\pm1)\) must divide by \(P/2\). Thirty-six coils on six poles is impossible however the pitches are juggled — Problem 3.
Counting the dummy coil as active. Its ends go nowhere, so it contributes no emf and no commutator segment; a 20-slot armature wave-wound on 19 coils has 19 segments, not 20 — Problem 3.
Dividing the load current instead of the armature current. In a generator the armature also feeds the shunt field, so the 396 A delivered is not the 400 A that flows in the paths — Problem 4.
Scaling armature resistance as \(1/A\). Halving the paths both doubles the conductors in series and halves the parallel branches, so the resistance rises as \(1/A^2\) — a factor of 16 here, not 2 — Problems 4 and 6.
Believing a wave winding "gives more power". It gives more volts and proportionately fewer amperes; the product is unchanged, and so is the copper loss — Problem 6.
Fitting equalisers to a wave winding. They would be pointless: each of the two paths already passes under every pole and so sees the average flux. Equalisers are a lap-winding remedy for a lap-winding problem — Problem 5.
Sizing equaliser rings for load current. They carry only the difference current between paths — here 27.8 A against a path current of 100 A — and are made light accordingly — Problem 5.
Confusing the number of brush arms with the number of brushes. An arm may hold several carbon blocks side by side for contact area; the winding fixes the number of arms, the current density fixes the number of blocks on each — Problems 1 and 2.
The number of parallel paths is no longer a rule to be quoted. It is a consequence of the commutator pitch, which is a consequence of how the winder chooses to bring each coil back to the commutator, and the whole of Set 7's \(A = P\) or \(A = 2\) now rests on a winding table that can be traced coil by coil. With it come the practical consequences: brush arms, conductor size, armature resistance as \(Zr/A^2\), and the equalisers that a lap winding cannot do without.
Every emf calculated so far has taken the flux per pole as given. For a separately excited machine that is fair, but a shunt generator has no external supply for its field — it must raise its own excitation from the few volts its residual magnetism provides, and whether it succeeds depends on the field resistance, on the speed, and on the shape of its own magnetisation curve. The flux, in other words, is not an input at all; it is the solution of a self-consistency problem.
Next: Set 9 — Generator Characteristics and Voltage Build-Up, where the open-circuit characteristic and the field-resistance line are made to intersect, the critical resistance and critical speed are read off, and the machines that refuse to excite are diagnosed.