By the end of this chapter you should be able to:
State the basic principle of a DC generator and explain the role of flux cutting.
Explain why a conductor moved across the field generates and one moved along it does not.
Explain why the EMF in the coil is alternating for internal as well as external load.
Describe how the commutator converts that alternating EMF into a unidirectional output.
Quantify the ripple and show how it falls as coils are added.
Explain motor action and how the same machine works in reverse.
Use back EMF to explain the starting-current problem and the need for a starter.
Derive \(E_g = \dfrac{\Phi PN}{60}\times\dfrac{Z}{A}\) and apply it to lap- and wave-wound machines.
The Basic Principle
The basic principle of a DC generator is electromagnetic induction — that is, when a conductor cuts across the magnetic field, an EMF is induced in it.
Nothing here is new. This is the dynamically induced EMF of Chapter 12, given by \(e = Blv\sin\theta\), with its direction fixed by Fleming's right-hand rule from Chapter 11. What Part 2 adds is not new physics but a particular arrangement of conductors, iron and sliding contacts that turns the principle into a usable machine.
Demonstrating Induced EMF
A single conductor, a magnet and a galvanometer settle the matter experimentally.
When a conductor is moved vertically upward or downward, the deflection in the galvanometer clearly shows that an EMF is induced in the conductor, since flux is cut by the conductor.
When it is moved horizontally (left or right) there is no deflection, which shows that no EMF is induced, since the flux cut is zero and the conductor moves just parallel to the magnetic lines of force.
The two observations are the two extremes of \(e = Blv\sin\theta\), where \(\theta\) is the angle between the velocity and the field:
Why the Coil EMF Is Alternating
In a generator a coil is rotated at a constant speed of \(\omega\) radians per second in a strong magnetic field of constant magnitude.
An EMF is induced in the coil by the phenomenon of dynamically induced EMF, \(e = Blv\sin\theta\), so that \(e \propto \sin\theta\).
The magnitude and direction of the induced EMF change periodically, depending on \(\sin\theta\).
The wave shape of the induced EMF is AC for internal as well as external load.
AC is converted into DC with the help of the commutator.
"The wave shape of the induced EMF is AC for internal as well as external load" is easy to skim past, and it is the most important statement in the chapter.
There is no DC anywhere inside a DC machine. The armature conductors carry alternating current and experience alternating EMF, always. What the commutator does is present that alternating quantity to the outside world with a constant polarity. A "DC machine" is an AC machine with a mechanical rectifier on its shaft.
This is why the armature core must be laminated (Chapter 7's eddy currents apply in full), and why the machine has iron loss even though it is called a DC machine.
The Commutator Converts AC to DC
Chapter 17 posed the problem and named the two solutions. The DC machine takes the second: switch the connections as the coil rotates, so that the brush touching a given side of the machine always sees the same polarity.
The device is a split ring — the commutator — with the coil ends connected to opposite segments and two fixed brushes bearing on it. Every half revolution, at the instant the coil EMF passes through zero, the segments change places under the brushes. The external connection therefore reverses at exactly the moment the internal EMF reverses, and the two reversals cancel.
The result is a full-wave rectified sine. It never goes negative, which is what "DC" requires, but with a single coil it falls to zero twice per revolution — hardly a useful supply. Its average value is
and its ripple is 100 %. Section 21-5 shows how a real machine fixes this.
Ripple and the Number of Coils
A practical armature carries many coils, evenly spaced around the periphery and connected to many commutator segments. At any instant the brushes tap the coil that is nearest its peak, so the output follows the envelope of a family of shifted sine waves rather than any single one.
With \(n\) coils per pole pair spaced \(180^{\circ}/n\) apart, the output swings between \(E_m\) and \(E_m\cos\left(90^{\circ}/n\right)\), so
For small angles this is approximately \(\tfrac{1}{2}(\pi/2n)^{2}\), so doubling the number of coils cuts the ripple by a factor of four.
| Coils \(n\) | \(90^{\circ}/n\) | \(E_{\min}/E_m\) | Ripple |
|---|---|---|---|
| 1 | 90° | 0 | 100 % |
| 2 | 45° | 0.7071 | 29.3 % |
| 4 | 22.5° | 0.9239 | 7.61 % |
| 8 | 11.25° | 0.9808 | 1.92 % |
| 16 | 5.625° | 0.9952 | 0.48 % |
A real armature has many tens of coils, so the residual ripple is a fraction of a percent — small enough that the output is properly called DC. This is the true reason a DC armature has so many slots: not to increase the EMF, which depends only on the total conductor count, but to make the output smooth.
Motor Action
Everything so far has described generator action. The same machine, supplied with current instead of driven, becomes a motor — and the principle is the companion of the one above.
Faraday's law. A conductor is moved through the field, and an EMF appears:
Direction by Fleming's right-hand rule. Mechanical power in, electrical out.
The force law. A conductor carries current in the field, and a force appears:
Direction by Fleming's left-hand rule. Electrical power in, mechanical out.
From Chapter 17, the torque on a coil of two active conductors at radius \(r\) is
which falls to zero when the coil aligns with the field, and reverses beyond. Left alone the coil would simply oscillate and settle — a compass needle, not a motor.
As a generator it converts the coil's alternating EMF into a unidirectional output.
As a motor it does the reverse: it feeds the coils with current that reverses each half revolution, so the conductors under a given pole always carry current the same way. The torque angle stays near 90°, where \(\sin\theta = 1\), and the torque is both maximum and unidirectional.
One component, one mechanism, two interpretations — which is exactly why a DC machine is reversible without any change of connection.
Back EMF and Starting Current
A motor's armature is rotating in a magnetic field, so by Section 21-1 it must be generating an EMF — whether it is being used as a generator or not. That EMF opposes the supply which produces it, as Lenz's law requires, and is called the back EMF.
The back EMF is what makes a DC motor self-regulating. Load it more, it slows, \(E\) falls, \(I_a\) rises, and the torque increases to meet the load. The motor draws exactly the current its load demands, without any control system whatever.
The armature resistance of a DC machine is deliberately small — a fraction of an ohm — because it is a loss. That is harmless while the machine runs, because \(E\) takes up nearly all of \(V\). At the instant of switching on, however, \(N = 0\), so \(E = 0\) and
which is enormous — commonly twenty to fifty times the rated current, as Example 21.4 shows. Every DC motor above about a kilowatt therefore needs a starter: external resistance in the armature circuit, cut out progressively as the machine speeds up and generates its own back EMF. Chapter 36 designs them.
Problem. A single-coil armature generates a peak EMF of 150 V. Find the average output after commutation and the ripple. How many coils would reduce the ripple below 2 %?
Average output.
Ripple. With one coil the output falls to zero twice per revolution:
Coils needed for 2 %. Require \(1 - \cos(90^{\circ}/n) \lt 0.02\):
So \(n = 8\) coils, giving a ripple of \(1 - \cos 11.25^{\circ} = 1.92\,\%\) \(\checkmark\)
Comment. Eight coils per pole pair is modest — real armatures have far more. Note that adding coils does not raise the peak EMF; it raises the minimum, filling in the troughs. The average rises from 0.637\(E_m\) towards \(E_m\) as a by-product.
Problem. A 230 V DC motor has \(R_a = 0.20~\Omega\) and takes 50 A at rated load. Find the back EMF when running, the current that would flow at the instant of starting, and the external resistance needed to limit the starting current to 1.5 times rated.
Back EMF at rated load.
The back EMF takes up 95.7 % of the supply voltage; only 4.3 % is dropped across the armature resistance.
Starting current with no starter. At standstill \(E = 0\):
Twenty-three times the rated current.
Starter resistance. To limit the current to \((1.5)(50) = 75\) A:
Comment. The starter resistance is fourteen times the armature resistance, and it must be cut out as the motor accelerates or it would waste enormous power and prevent the machine reaching speed. The whole art of DC starting is removing that resistance at the right rate — fast enough to reach full speed, slow enough that the current never exceeds the limit.
The Generated EMF Equation
The EMF of the whole machine now follows from Faraday's law and a count of conductors. Let
- EMF per conductor. By Faraday's law with one turn, the average EMF generated per conductor is \(\mathrm{d}\Phi/\mathrm{d}t\) volts.
- Flux cut in one revolution. A conductor passes every pole once per revolution, so \(\mathrm{d}\Phi = \Phi P\) Wb.
- Time for one revolution. The armature makes \(N/60\) revolutions per second, so \(\mathrm{d}t = 60/N\) seconds.
- EMF per conductor. Dividing, \(\dfrac{\mathrm{d}\Phi}{\mathrm{d}t} = \dfrac{\Phi PN}{60}\) volts.
- Conductors in series per path. The \(Z\) conductors are divided equally among \(A\) parallel paths, giving \(Z/A\) conductors in series in each path.
- EMF per path. Multiplying, since conductors in series add.
For a given DC machine \(Z\), \(P\) and \(A\) are constants, therefore
In terms of angular velocity, substituting \(N = 60\omega/2\pi\) gives the form used in machine analysis:
Problem. A 4-pole lap-wound DC generator has 48 slots with 8 conductors per slot. The flux per pole is 25 mWb and the machine runs at 1000 rev/min. Find the generated EMF.
Conductors.
Parallel paths. Lap-wound, so \(A = P = 4\).
Generated EMF.
Check by the per-conductor route. EMF per conductor \(= \Phi PN/60 = (0.025)(4)(1000)/60 = 1.667\) V. Conductors in series per path \(= Z/A = 384/4 = 96\). Product \(= (1.667)(96) = 160\) V \(\checkmark\)
Comment. The second route is worth doing at least once, because it makes the structure of the equation visible: a small EMF per conductor, multiplied by however many conductors the winding puts in series. The parallel paths do not add EMF — they divide it.
Parallel Paths — Lap and Wave
The factor \(A\) is the one term in the EMF equation that the designer chooses freely, and it changes the machine completely.
As many parallel paths as poles. Each path has few conductors, so the EMF is low, but the total current divides among many paths, so the machine handles high current.
Used for low-voltage, high-current machines.
Only two parallel paths, whatever the number of poles. Each path has many conductors, so the EMF is high, but only two paths share the current.
Used for high-voltage, low-current machines.
For the same armature the ratio of generated EMFs is
so a four-pole machine rewound from lap to wave doubles its voltage — and halves its current capability, because each conductor can still carry only the same current. Chapter 25 treats both windings in detail.
Problem. The armature of Example 21.3 is rewound as a wave winding, all else unchanged. Find the new EMF. If each conductor can safely carry 10 A, find the armature current and the power in each case.
Wave-wound EMF. Now \(A = 2\):
Twice the lap value, as \(P/2 = 2\) predicts.
Armature currents. The armature current is the sum over the parallel paths:
Power.
Comment. The power is identical. This is the essential point about lap and wave: the winding decides how the machine's capability is packaged — as volts or as amperes — but not how much there is of it. The same copper, the same iron and the same flux deliver the same watts either way.
The choice is therefore made by the application. A traction machine on a 600 V line wants wave; a plating rectifier delivering thousands of amperes at a few volts wants lap.
Problem. A DC generator gives 220 V at 1200 rev/min. Find the EMF (a) at 1500 rev/min with the same flux, and (b) at 1200 rev/min with the flux reduced by 10 %. (c) At what speed would the reduced flux still give 220 V?
(a) Speed changed. Since \(E \propto \Phi N\) and \(\Phi\) is unchanged:
(b) Flux changed. Now \(N\) is unchanged:
(c) Compensating speed. To restore 220 V with 90 % flux:
Comment. Part (c) is the whole idea of field weakening, seen from the generator side. Reduce the flux and the machine must turn faster for the same EMF; run a motor on a fixed supply and reduce its flux, and it must speed up until its back EMF again balances the supply. Chapter 37 develops this into a speed-control method — and shows why the torque falls as the speed rises, so that the machine works at constant power above base speed.
Summary and Key Formulas
The basic principle is electromagnetic induction: a conductor cutting across the field has an EMF induced in it.
Moving a conductor across the field gives a deflection; moving it along the field gives none, because the flux cut is zero.
A coil rotating at \(\omega\) in a constant field has \(e \propto \sin\theta\). Its magnitude and direction change periodically, and the wave shape is AC for internal as well as external load.
The commutator converts that AC to DC by reversing the external connection at each zero crossing. The output is a full-wave rectified sine of average \(0.637E_m\) for one coil.
Ripple is \(1 - \cos(90^{\circ}/n)\) for \(n\) coils — 100 % for one coil, under 0.5 % for sixteen. This is why armatures have many slots.
Motor action follows from \(F = BIl\) and Fleming's left-hand rule; the commutator keeps the torque angle near 90° so the torque is maximum and unidirectional.
A running motor generates a back EMF \(E = V - I_aR_a\), which makes it self-regulating. At standstill \(E = 0\) and the current would be \(V/R_a\) — hence the starter.
The EMF equation is \(E_g = \dfrac{\Phi PN}{60}\times\dfrac{Z}{A}\), with \(A = P\) for lap and \(A = 2\) for wave, giving \(E_g \propto \Phi N\).
Lap versus wave changes the voltage-current split but not the power: lap gives low volts and high amps, wave the reverse.
| Quantity | Formula | Notes |
|---|---|---|
| Dynamically induced EMF | \(e = Blv\sin\theta\) | Chapter 12 |
| Force on a conductor | \(F = BIl\) | motor action |
| Total conductors | \(Z = \text{slots} \times \text{conductors/slot}\) | — |
| Flux cut per revolution | \(\mathrm{d}\Phi = \Phi P\) | per conductor |
| Time per revolution | \(\mathrm{d}t = 60/N\) | \(N\) in rev/min |
| EMF per conductor | \(\dfrac{\mathrm{d}\Phi}{\mathrm{d}t} = \dfrac{\Phi PN}{60}\) | — |
| Conductors in series per path | \(Z/A\) | not \(Z/2\) in general |
| Generated EMF | \(E_g = \dfrac{\Phi PN}{60}\times\dfrac{Z}{A}\) | the central equation |
| Compact forms | \(E_g = k\Phi N = k_a\Phi\omega\) | \(k = \dfrac{ZP}{60A}\), \(k_a = \dfrac{ZP}{2\pi A}\) |
| Parallel paths | \(A = P\) lap, \(A = 2\) wave | — |
| Armature current | \(I_a = A \times I_{\text{path}}\) | — |
| Back EMF | \(E = V - I_aR_a\) | motor |
| Starting current | \(I_{a,\text{start}} = V/R_a\) | no back EMF at rest |
| Single-coil average | \(E_{\text{av}} = \dfrac{2}{\pi}E_m\) | full-wave rectified |
| Ripple, \(n\) coils | \(1 - \cos\left(90^{\circ}/n\right)\) | falls as \(1/n^{2}\) |
Common Mistakes
Believing the armature carries DC. It does not, ever. The EMF and current in the conductors are alternating; only the brush output is unidirectional.
Writing \(Z/2\) for the conductors in series per path. It is \(Z/A\). The two agree only for a wave winding, where \(A = 2\).
Confusing \(P\) and \(A\) in the EMF equation. They cancel for a lap winding, giving \(E = \Phi NZ/60\), but not otherwise.
Forgetting to convert flux to webers. A value given in milliwebers must be divided by 1000 before use.
Using \(N\) in rev/s. The 60 in the equation already converts from rev/min.
Thinking more slots gives more EMF. The EMF depends on the total conductor count \(Z\); more slots with the same \(Z\) reduces ripple, not voltage.
Expecting lap and wave to give different power. They give the same watts, packaged differently.
Applying \(E = V + I_aR_a\) to a motor. The back EMF is less than the supply.
Assuming a motor can be switched straight onto the supply. With no back EMF the current would be tens of times rated.
Mixing up Fleming's rules. Right hand for generator action, left hand for motor action.
Chapter Review
Write down \(Z\), \(P\), \(A\) and \(\Phi\) in SI units before substituting — most errors in this topic are bookkeeping, not physics.
P21.1 A 6-pole lap-wound generator has 60 slots with 6 conductors per slot, 20 mWb per pole, running at 1200 rev/min. Find \(E_g\).
Show answer
\[Z = 60 \times 6 = 360, \qquad A = P = 6\]\[E_g = \frac{(0.020)(6)(1200)(360)}{(60)(6)} = \frac{51\,840}{360} = 144~\mathrm{V}\]P21.2 Repeat P21.1 for a wave winding.
Show answer
Three times the lap value, since \(P/2 = 3\).\[E_g = \frac{51\,840}{(60)(2)} = \frac{51\,840}{120} = 432~\mathrm{V}\]P21.3 A 4-pole wave-wound generator must produce 250 V at 900 rev/min with 30 mWb per pole. How many armature conductors are needed?
Show answer
So 278 conductors, rounded up — and in practice to a number compatible with the slot count, such as 280 (40 slots × 7, or 35 slots × 8).\[Z = \frac{60AE_g}{\Phi PN} = \frac{(60)(2)(250)}{(0.030)(4)(900)} = \frac{30\,000}{108} = 277.8\]P21.4 A DC motor on 400 V has \(R_a = 0.25~\Omega\) and runs at rated 60 A. Find the back EMF and the starting current without a starter.
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\[E = 400 - (60)(0.25) = 400 - 15 = 385~\mathrm{V}\]\[I_{a,\text{start}} = \frac{400}{0.25} = 1600~\mathrm{A} = 26.7 \times \text{rated}\]P21.5 For P21.4, find the starter resistance to limit the starting current to twice rated.
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\[R_{\text{total}} = \frac{400}{120} = 3.333~\Omega, \qquad R_{\text{ext}} = 3.333 - 0.25 = 3.08~\Omega\]P21.6 A generator gives 180 V at 1000 rev/min. Find the speed for 240 V at the same flux.
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\[N = 1000\left(\frac{240}{180}\right) = 1333~\mathrm{rev/min}\]P21.7 Find the ripple of a commutated output with 6 coils per pole pair, and the number of coils needed for 1 % ripple.
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For 1 %: \(\cos(90^{\circ}/n) \gt 0.99\), so \(90^{\circ}/n \lt 8.11^{\circ}\) and \(n \gt 11.1\), giving 12 coils.\[\text{ripple} = 1 - \cos\frac{90^{\circ}}{6} = 1 - \cos 15^{\circ} = 1 - 0.9659 = 3.41\,\%\]P21.8 Explain why the armature core of a DC machine must be laminated, even though the machine is called a DC machine.
Show answer
Because nothing inside the armature is DC. As the armature rotates, each part of its iron passes alternately under a north and a south pole, so the flux density in the core alternates at a frequency \(f = PN/120\).By Chapter 7, an alternating flux in solid iron induces eddy currents proportional to the square of the thickness. The core is therefore laminated exactly as a transformer core is, and the machine has hysteresis and eddy-current loss just as an AC machine does. The "DC" in the name describes only what appears at the brushes.
P21.9 A motor's load is suddenly increased. Trace what happens, and explain why no control action is needed.
Show answer
The extra load torque exceeds the motor torque, so the machine slows.
Since \(E \propto \Phi N\), the back EMF falls.
Since \(I_a = (V - E)/R_a\), the armature current rises.
Since \(T \propto \Phi I_a\), the torque rises until it again matches the load.
The machine settles at a slightly lower speed, drawing exactly the current the load demands. The back EMF is a built-in negative feedback loop — which is why a DC motor needs no controller merely to run stably.
P21.10 A machine is to supply 5 kW at 250 V. Each armature conductor path can carry 12 A. Should it be lap- or wave-wound if it has 4 poles?
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The required current is \(I_a = 5000/250 = 20\) A.Both can carry 20 A, so current is not the constraint. Wave is the better choice: it produces twice the EMF from the same conductors, so it reaches 250 V with half as many, saving copper and slot space.\[\text{lap: } I_a = (4)(12) = 48~\mathrm{A}; \qquad \text{wave: } I_a = (2)(12) = 24~\mathrm{A}\]Had the requirement been 5 kW at 50 V — that is, 100 A — only the lap winding could have carried it, and lap would be forced.
MCQ 1. The EMF induced in the armature coils of a DC generator is:
(a) DC (b) AC (c) DC with ripple (d) zeroShow answer
(b) AC — for internal as well as external load. Only the brush output is unidirectional.MCQ 2. A conductor moved parallel to the magnetic lines of force generates:
(a) maximum EMF (b) half the maximum (c) no EMF (d) a reversed EMFShow answer
(c) no EMF, since the flux cut is zero and \(\sin\theta = 0\).MCQ 3. The conductors in series in each parallel path number:
(a) \(Z\) (b) \(Z/2\) (c) \(Z/A\) (d) \(Z/P\)Show answer
(c) \(Z/A\). \(Z/2\) is right only for a wave winding, where \(A = 2\).MCQ 4. For a lap winding the number of parallel paths is:
(a) 2 (b) \(P\) (c) \(P/2\) (d) \(Z\)Show answer
(b) \(P\) — as many paths as poles.MCQ 5. The generated EMF of a DC machine is proportional to:
(a) \(\Phi\) only (b) \(N\) only (c) \(\Phi N\) (d) \(\Phi/N\)Show answer
(c) \(\Phi N\), since \(Z\), \(P\) and \(A\) are fixed for a given machine.MCQ 6. Rewinding a 4-pole machine from lap to wave changes the EMF by a factor of:
(a) 1/2 (b) 2 (c) 4 (d) 1Show answer
(b) 2, since the ratio is \(P/2 = 2\).MCQ 7. The average output of a single-coil commutated generator is:
(a) \(E_m\) (b) \(0.707E_m\) (c) \(0.637E_m\) (d) \(0.5E_m\)Show answer
(c) \(0.637E_m = (2/\pi)E_m\) — the mean of a full-wave rectified sine.MCQ 8. Doubling the number of armature coils reduces the ripple by a factor of about:
(a) 2 (b) 4 (c) 8 (d) unchangedShow answer
(b) 4, since the ripple varies roughly as \(1/n^{2}\).MCQ 9. At the instant of starting, a DC motor's back EMF is:
(a) equal to \(V\) (b) zero (c) maximum (d) negativeShow answer
(b) zero, since \(E \propto N\) and the machine is at rest — which is why the starting current is so large.MCQ 10. Motor action follows from:
(a) Faraday's law and the right-hand rule (b) \(F = BIl\) and the left-hand rule (c) Lenz's law only (d) Ampère's lawShow answer
(b) \(F = BIl\) and Fleming's left-hand rule. The right-hand rule is for generator action.
State the basic principle of a DC generator and describe the experiment that distinguishes motion across the field from motion along it.
Explain why the EMF inside a DC machine is alternating, and what consequences this has for the design of the armature core.
Describe how the commutator converts an alternating coil EMF into a unidirectional output.
Explain quantitatively why a practical armature has many coils rather than one.
Compare generator and motor action, identifying the law and the hand rule that governs each.
Explain what back EMF is, why it makes a DC motor self-regulating, and why it creates a starting problem.
Derive the EMF equation of a DC generator, taking care over the number of conductors in series per parallel path.
Explain why lap and wave windings give different voltages but the same power.
The principle is established and the EMF equation derived. Chapter 22 examines construction in detail — yoke, poles, armature core and its laminations, windings, commutator and brushes — and explains why each part is made as it is.
Chapter 23 returns to the commutator, whose action was described here in outline, and treats it properly: segment count, brush position, and the reason commutation limits the machine's speed and current. Chapters 24 and 25 develop the armature windings and the lap and wave arrangements whose parallel-path counts were simply quoted in Section 21-9.
The EMF equation reappears in Chapter 27 applied to design problems, and its companion — the torque equation \(T = \dfrac{\Phi ZP}{2\pi A}I_a\) — is derived in Chapter 34, where the same constant \(k_a\) found here turns out to relate torque to armature current exactly as it relates EMF to speed.