Set 10 — Armature Reaction and Commutation
Every emf computed so far has assumed the field winding has the air gap to itself. It does not. The armature carries hundreds of amperes in conductors sitting in that same gap, and its magnetomotive force distorts the main flux, shifts the neutral axis, and — once the brushes are moved to follow that shift — subtracts directly from the field's ampere-turns. This set puts numbers on all three.
The second half turns to the other consequence. A coil short-circuited by a brush must reverse its current in under a millisecond, and its own inductance resists. The reactance voltage, the interpole flux density that cancels it, the ampere-turns that produce it and the brush current density that betrays a badly commutated machine are all worked out for one 100 kW generator.
The armature is a winding too, and it has its own mmf. With \(Z\) conductors each carrying \(I_c = I_a/A\), the armature's total ampere-turns are \(ZI_c/2\), shared between \(P\) poles:
\[ \mathcal{F}_a\ \text{per pole} = \frac{ZI_c}{2P}\ \ \text{A-turns} \]With the brushes on the geometrical neutral axis that mmf is purely cross-magnetising — it acts at right angles to the main field, distorting the flux towards one pole tip and away from the other without changing the total. Shift the brushes by \(\theta_m\) mechanical degrees and part of it turns round and opposes the field directly:
\[ \mathcal{F}_d = \frac{ZI_c\,\theta_m}{360}, \qquad \mathcal{F}_c = \frac{ZI_c}{2P} - \frac{ZI_c\,\theta_m}{360} \qquad\text{per pole} \]Mechanical and electrical degrees differ by the pole-pair factor: \(\theta_e = \theta_m\times P/2\).
Demagnetising ampere-turns are paid for in field current. With \(N_{sh}\) shunt turns per pole the field must be raised by \(\Delta I_f = \mathcal{F}_d/N_{sh}\) to restore the flux, or the emf read from the OCC at the reduced effective current \(I_f - \Delta I_f\).
A compensating winding cancels the armature mmf under the pole face, conductor for conductor. It sits in slots in the pole shoes and is in series with the armature, so it tracks \(I_a\) automatically:
\[ Z_{cw}\ \text{per pole} = \frac{Z}{PA}\times\frac{\text{pole arc}}{\text{pole pitch}} \]Commutation is a current reversal in a shorted coil, and the time available for it is set by the commutator geometry. With a brush spanning one segment,
\[ T_c = \frac{w_b - t_{\text{mica}}}{v_c}, \qquad v_c = \frac{\pi D_cN}{60} \]The coil's own inductance fights the reversal. The average emf opposing it — the reactance voltage — is the whole difficulty of commutation:
\[ E_r = L\frac{di}{dt} = \frac{2LI_c}{T_c} \]Interpoles supply an equal and opposite emf. A small pole on the neutral axis, wound in series with the armature, gives the commutating coil a rotational emf that cancels \(E_r\). Its ampere-turns must first neutralise the armature mmf and then drive its own air gap:
\[ \mathcal{F}_{ip} = \frac{ZI_c}{2P} + \frac{B_{ip}\,l_g}{\mu_0} \]
A 4-pole, wave-wound dc generator has 720 armature conductors and delivers an armature current of 100 A. The brushes are given a lead of 8 mechanical degrees from the geometrical neutral axis. Find
- the current in each conductor and the total armature ampere-turns per pole
- the demagnetising and cross-magnetising ampere-turns per pole
- the brush lead in electrical degrees
- the extra shunt-field current needed to make good the demagnetisation, the shunt field having 1200 turns per pole
Conductor current first — everything scales from it. A simplex wave winding has two paths whatever the pole count:
Total armature ampere-turns per pole. There are \(Z\) conductors carrying \(I_c\), and two conductors make one turn, so the armature's mmf is \(ZI_c/2\) ampere-turns spread over \(P\) poles:
Compare this with a shunt field of, say, 1200 turns at 2 A — 2400 A-turns. The armature outweighs the field almost two to one, which is why armature reaction is not a small correction in a dc machine.
Shifting the brushes redirects part of that mmf. A brush lead of \(\theta_m\) puts the conductors lying in a band of width \(2\theta_m\) at each of the \(P\) brush axes into the demagnetising zone — their mmf now lies along the pole axis, opposing the field. That is a fraction \(P(2\theta_m)/360\) of the whole armature, so
What is left is cross-magnetising. The armature's mmf has not changed in size, only in direction:
A check on the formula: if the brushes were shifted a full pole pitch, \(\theta_m = 360/2P = 45^\circ\), and \(\mathcal{F}_d\) would come to \(720\times50\times45/360 = 4500\) — the entire armature mmf, with nothing cross-magnetising left. The formula behaves correctly at its limit.
Convert the lead to electrical degrees. One mechanical revolution carries the winding past \(P/2\) electrical cycles:
Pay for the demagnetisation in field current:
Two thirds of an ampere of extra excitation, needed at full load and not at no load — which is one reason a shunt generator's terminal voltage droops with load by more than its armature resistance can explain.
A 6-pole dc machine has a simplex lap-wound armature of 720 conductors carrying a total armature current of 600 A. The ratio of pole arc to pole pitch is 0.75. A compensating winding, embedded in the pole faces and connected in series with the armature, is to neutralise the armature reaction under the pole arc. Find
- the armature ampere-turns per pole, and the part of them lying under the pole arc
- the number of compensating conductors required per pole, and in total
- the ampere-turns left for the interpoles to deal with
Conductor current, then armature mmf. A lap winding gives one path per pole:
Only part of the armature lies under a pole face. The pole arc covers 75% of the pole pitch, so 75% of the armature conductors — and 75% of the mmf — are the compensating winding's business:
Count the compensating conductors. Work in ampere-conductors, which avoids the factor of two entirely. The armature has \(ZI_c/P\) ampere-conductors per pole, of which the fraction \(\psi\) lies under the arc; the compensating conductors carry the full armature current \(I_a\), not \(I_c\):
Verify in ampere-turns. Fifteen conductors per pole are \(15/2 = 7.5\) turns, each carrying 600 A:
Exactly the figure computed in step 2. In practice 15 is rounded to a convenient even number of bars per pole and the small mismatch is accepted.
What is left over. The remaining quarter of the armature mmf lies in the interpolar space, where there is no pole face to bury a winding in:
That is precisely the region where commutation happens, and it is the interpole's job. Compensating windings and interpoles are not alternatives; they divide the air gap between them.
Why the winding must be in series with the armature. Armature reaction is proportional to \(I_a\) and reverses when \(I_a\) reverses. A compensating winding carrying the same current cancels it at every load and in both directions of power flow, with no adjustment. Excited from anywhere else it would cancel correctly at one current only — and would over-compensate at light load, which is worse than not compensating at all.
A 250 V, 400 A, 4-pole dc generator has a simplex lap-wound armature of 200 conductors and runs at 600 rpm. Its commutator is 0.32 m in diameter; each brush is 10 mm wide, and the mica between segments is 1 mm thick. Each armature coil has an inductance of 0.02 mH. Find
- the number of commutator segments and the segment pitch
- the peripheral speed of the commutator and the time available for commutation
- the average reactance voltage of the coil undergoing commutation
- the emf the interpoles must induce in that coil
Segments and coils. Every coil has two ends and every segment receives two ends, so segments equal coils, and coils are half the conductors:
The 10 mm brush is just under one segment pitch wide, so it short-circuits one coil at a time — the simplest case, and the one the standard formula assumes.
Commutator peripheral speed:
The commutation period is the time the brush takes to cross the coil's two segments — brush width less the mica, which the brush must ride over before the next segment is engaged:
Under a millisecond, and it is the same at every point of the commutator. A useful check: with a brush one segment wide, \(T_c \approx 60/(CN) = 60/(100\times600) = 1\) ms, independent of the commutator diameter — a bigger commutator runs faster in exactly the proportion that its segments are wider.
The current in that coil must reverse completely in that time. A lap winding gives four paths, so
A rate of change of 223 kA/s in a coil of twenty microhenries. Four and a half volts sounds trivial against 250 V, but it appears across a coil that is short-circuited by the brush, so the only thing limiting the resulting current is the contact resistance.
What the interpoles must do. By Lenz's law \(E_r\) opposes the reversal: it tries to keep the old current flowing, so at the end of the period the coil current has not reached \(-I_c\) and the shortfall must be broken by the brush leaving the segment — as an arc. Cancelling \(E_r\) therefore means giving the coil a rotational emf of the same size and the opposite sign:
That is what an interpole is: a small pole placed exactly where the coil is when it is short-circuited, of the polarity needed to drive the current towards its new value rather than away from it.
The machine of Problem 3 — 250 V, 400 A, 4-pole, lap-wound, 200 conductors, 600 rpm — has an armature 0.70 m in diameter with a core 0.40 m long. Its interpoles have an air gap of 6 mm, and each armature coil is a single turn. Design the interpoles: find
- the armature peripheral speed
- the flux density the interpole must produce in its air gap
- the ampere-turns per interpole, split into the part that neutralises the armature and the part that drives the gap
- the number of turns on each interpole
Leakage and the reluctance of the iron may be neglected.
Armature peripheral speed — not the commutator speed, because the coil sides move with the armature, not with the commutator:
Work backwards from the emf required. The commutating coil is a single turn, so it has two active conductors, each of length 0.40 m, each cutting the interpole flux. Their emfs add round the coil because the two sides lie under interpoles of opposite polarity:
A quarter of a tesla, against roughly 0.8 T under the main poles. Interpoles are small and lightly worked — but they must never saturate, or their contribution would stop growing just when the load current needs it most.
The interpole has two jobs, and its ampere-turns are the sum. First it must cancel the armature mmf that occupies the same interpolar space:
Then it must drive its own air gap, using the magnetic-circuit method of Part 1:
Turns follow from the current the winding carries. The interpole winding is in series with the armature, so it carries the full \(I_a = 400\) A:
Round up, never down: ten turns give 4000 A-turns, a 7.7% margin that is trimmed on test by a small divertor across the interpole winding, or by adjusting the interpole air gap with shims. Rounding down would leave the machine under-commutated at full load.
Notice which term dominates. Two thirds of the interpole's ampere-turns are spent simply undoing the armature, and only a third produce the useful commutating flux:
| Component | A-turns | Share | Scales with |
|---|---|---|---|
| Neutralising the armature mmf | 2500 | 67% | \(I_a\) |
| Driving the interpole gap | 1213 | 33% | \(I_a\) |
| Total per interpole | 3713 | 100% | \(I_a\) |
Both terms are proportional to the armature current — the first obviously, the second because \(E_r \propto I_c\) and therefore \(B_{ip} \propto I_c\). That is the whole reason the interpole winding is put in series with the armature: the correction it provides then tracks the load automatically, at every current and in both directions.
A 4-pole shunt generator has a lap-wound armature of 480 conductors and runs at 1000 rpm with a field current of 1.8 A in a shunt winding of 1000 turns per pole. Its open-circuit characteristic at that speed is the one used in Set 9:
| \(I_f\) (A) | 0 | 0.2 | 0.4 | 0.6 | 0.8 | 1.0 | 1.2 | 1.4 | 1.6 | 1.8 | 2.0 | 2.2 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| \(E_a\) (V) | 6 | 30 | 60 | 88 | 112 | 132 | 148 | 161 | 172 | 180 | 186 | 190 |
On load the armature carries 200 A and the brushes are shifted 5 mechanical degrees. Find
- the demagnetising ampere-turns per pole and the equivalent loss of field current
- the generated emf on load, and the drop caused by armature reaction, in volts and per cent
- the field current that would restore the no-load emf, and the field-circuit resistance needed for it
Conductor current and demagnetising ampere-turns. Lap winding, so \(A = P = 4\):
For reference the total armature mmf is \(ZI_c/2P = 3000\) A-turns per pole, so the 5° shift has turned about one ninth of it into direct opposition; the other 2667 A-turns remain cross-magnetising.
Convert ampere-turns into field current. The demagnetising mmf and the shunt field act on the same magnetic circuit, so they may simply be subtracted once both are expressed in the same units:
This is the whole method: armature reaction is turned into an equivalent field current, and the OCC — which knows nothing about armature current — is then read at the reduced value.
Read the OCC at the effective current. Between 1.4 A and 1.6 A the curve rises 11 V over 0.2 A, a slope of 55 V/A:
The drop caused by armature reaction:
Fifteen volts, before a single volt of \(I_aR_a\) has been counted. This is the reason a shunt generator's external characteristic falls away faster than its armature resistance alone predicts — and note that the demagnetisation costs more volts than it would on the straight part of the curve, because 1.467 A sits on a steeper stretch of the OCC than 1.8 A does.
Restoring the emf. To generate 180 V the effective current must be 1.8 A, so the actual field current must be raised by the whole of \(\Delta I_f\):
The rheostat must be cut from 100 Ω down to 84.4 Ω — an 18.5% increase in field current to make good a 5° brush shift at full load. It also means the field copper loss rises with it, from \(180\times1.8 = 324\) W to \(180\times2.133 = 384\) W.
In the machine of Problems 3 and 4 the conductor current is 100 A, so each brush arm collects 200 A. A brush arm is made up of six carbon blocks side by side, each 10 mm wide and 40 mm long, giving a total contact area of 24 cm². Taking the contact area on each of the two segments to be proportional to the fraction of the segment the brush covers, find
- the current density at each contact half way through the commutation period, under ideal linear commutation
- the same two densities if the reversal is retarded, so that at half period the coil current is still \(+50\) A instead of zero
- the current that must be broken at the trailing edge if, at the end of the period, the coil current has reached only \(-80\) A
Set up the current sharing. While the brush bridges two segments, it collects \(2I_c = 200\) A in total, but not equally: the coil being commutated carries a current \(i\) that runs from \(+I_c\) to \(-I_c\), and it adds to one contact while subtracting from the other. Calling the segment the brush is leaving "outgoing" and the one it is meeting "incoming",
Meanwhile the areas change: after a fraction \(t/T_c\) of the period the brush still covers a fraction \((1 - t/T_c)\) of the outgoing segment and \(t/T_c\) of the incoming one.
Ideal linear commutation makes the two vary together. If the current falls at a constant rate, \(i = I_c\left(1 - 2t/T_c\right)\), then
The current and the area shrink in exactly the same proportion, so the density is constant — and equal at both contacts. That is the definition of ideal commutation, and the reason it is the design target: no part of the brush is worked harder than any other at any instant.
At half period, ideally: the coil current is zero and each contact has half the area:
Now retard the reversal. The coil's inductance holds the current up, so at \(t = T_c/2\) it is still \(+50\) A. The areas are unchanged — they depend only on where the brush is — but the currents are not:
Half the brush is running at 150% of its design density while the other half loafs at 50%. The overloaded half is the trailing edge, which is exactly where burnt commutators are found.
The end of the period is where the damage is done. At \(t = T_c\) the brush leaves the outgoing segment altogether, so the contact area there is zero. Whatever current is still flowing in it must stop instantly:
Twenty amperes interrupted in a coil of 0.02 mH: the energy \(\tfrac12Li^2 = 4\ \text{mJ}\) has nowhere to go except into an arc between the brush and the departing segment. Repeated a hundred times a revolution, ten times a second, it pits the copper and blackens the brush face.
Read the three cases side by side:
| Condition | \(i\) at \(T_c/2\) | \(J_{\text{out}}\) | \(J_{\text{in}}\) | Current broken at \(T_c\) |
|---|---|---|---|---|
| Linear (ideal) | 0 | 8.33 A/cm² | 8.33 A/cm² | 0 |
| Retarded (under-commutation) | +50 A | 12.5 A/cm² | 4.17 A/cm² | 20 A |
| Accelerated (over-commutation) | −50 A | 4.17 A/cm² | 12.5 A/cm² | 0 |
Over-commutation — too strong an interpole — overloads the leading edge instead, but it does not leave a current to break, which is why designers deliberately err on that side. Sparkless commutation is not achieved by making the interpole exactly right; it is achieved by making it slightly too strong.
Key Formulas
| Quantity | Relation | Notes |
|---|---|---|
| Total armature mmf | \(\mathcal{F}_a = \dfrac{ZI_c}{2P}\) per pole | Problems 1, 2, 4, 5 |
| Demagnetising mmf | \(\mathcal{F}_d = \dfrac{ZI_c\,\theta_m}{360}\) per pole | \(\theta_m\) in mechanical degrees — Problems 1, 5 |
| Cross-magnetising mmf | \(\mathcal{F}_c = \mathcal{F}_a - \mathcal{F}_d\) | Problems 1, 5 |
| Mechanical to electrical | \(\theta_e = \theta_m\times P/2\) | Problem 1 |
| Equivalent field current | \(\Delta I_f = \mathcal{F}_d/N_{sh}\) | Read the OCC at \(I_f-\Delta I_f\) — Problems 1, 5 |
| Compensating conductors | \(Z_{cw} = \dfrac{Z}{PA}\times\dfrac{\text{pole arc}}{\text{pole pitch}}\) | Per pole, in series with armature — Problem 2 |
| Commutator speed | \(v_c = \pi D_cN/60\) | Problem 3 |
| Commutation period | \(T_c = \dfrac{w_b - t_{\text{mica}}}{v_c}\) | Brush spanning one segment — Problem 3 |
| Segments and pitch | \(C = Z/2\), pitch \(= \pi D_c/C\) | Problem 3 |
| Reactance voltage | \(E_r = \dfrac{2LI_c}{T_c}\) | Linear commutation — Problems 3, 6 |
| Interpole emf | \(E_{ip} = 2N_{\text{coil}}B_{ip}\,l\,v_a\) | Set equal to \(E_r\) — Problems 3, 4 |
| Interpole ampere-turns | \(\mathcal{F}_{ip} = \dfrac{ZI_c}{2P} + \dfrac{B_{ip}l_g}{\mu_0}\) | Problem 4 |
| Interpole turns | \(N_{ip} = \mathcal{F}_{ip}/I_a\), rounded up | Series winding — Problem 4 |
| Coil current, linear | \(i(t) = I_c\left(1 - 2t/T_c\right)\) | Problem 6 |
| Contact currents | \(i_{\text{out}} = I_c + i,\ i_{\text{in}} = I_c - i\) | Problem 6 |
Common Mistakes
Using the armature current in place of the conductor current. Every armature-reaction formula counts \(I_c = I_a/A\), because that is what a single conductor carries. Using 100 A instead of 50 A doubles every ampere-turn on the page — Problems 1 and 5.
Putting the brush shift into the formula in electrical degrees. \(\mathcal{F}_d = ZI_c\theta/360\) wants mechanical degrees. Sixteen electrical degrees on a 4-pole machine is eight mechanical, and the difference is a factor of two — Problem 1.
Forgetting the factor of two between conductors and turns. Ampere-turns are \(ZI_c/2\), ampere-conductors are \(ZI_c\). Compensating windings are most safely counted in ampere-conductors, where the factor never appears — Problem 2.
Sizing the compensating winding for the whole pole pitch. It can only cancel what lies under the pole face, a fraction 0.75 here; the rest belongs to the interpoles — Problem 2.
Exciting the compensating winding from anywhere but the armature circuit. Armature reaction follows \(I_a\) instant by instant, and only a series connection follows it back — Problem 2.
Using the armature diameter for the commutation period, or the commutator diameter for the interpole emf. The brush rides on the commutator; the coil sides sweep past the interpole on the armature. Here they are 0.32 m and 0.70 m — Problems 3 and 4.
Writing \(E_r = LI_c/T_c\). The current swings from \(+I_c\) to \(-I_c\), so the change is \(2I_c\) and the reactance voltage doubles — Problem 3.
Giving the interpole only enough ampere-turns for its own air gap. It must first cancel the armature mmf occupying the same space — two thirds of the total here — before any flux appears at all — Problem 4.
Rounding the interpole turns down. Nine turns instead of ten leaves the machine under-commutated at full load, which is the sparking condition; the excess of a rounded-up winding is trimmed with a divertor — Problem 4.
Blaming sparking on brush current density. The density is comfortable throughout; what causes the arc is the residual current in a coil whose segment has just left the brush — Problem 6.
The armature turns out to be a second magnetising winding that nobody designed and nobody can switch off. Its mmf per pole is fixed by the load current, its direction by the brushes, and its consequences run in two directions at once: along the pole axis, where it costs volts that Set 9's magnetisation curve prices exactly; and across the interpolar gap, where it opposes the very flux the commutating coil needs. Compensating windings answer the first, interpoles the second, and both are wired in series with the armature so that the cure grows with the disease.
With that, the dc generator is complete. Every quantity it produces has been accounted for — the emf from the winding and the flux, the flux from the field circuit and its own magnetisation curve, and the corrections the armature current forces on both. Nothing in any of it depended on the direction of the current. Reverse it, feed the machine from a supply instead of a prime mover, and the same emf now opposes the source rather than driving the load.
Next: Set 11 — DC Motors: Back EMF, Torque and Speed, where the same equations are read with one sign changed, and the difference between supply voltage and back emf is what sets the current, the torque and the speed together.