Set 45 — Universal and AC Series Motors
A series motor makes its flux from its own armature current. Reverse the supply and both the flux and the current reverse, so their product does not — the torque holds its direction through every half cycle. That is the whole reason a single machine can run on direct current and on alternating current alike, and it is why these motors start unaided where a single-phase induction motor cannot.
What alternating current does add is reactance. The series winding now carries a quadrature voltage drop that takes no power but leaves less of the supply available along the current, so the back e.m.f. and the speed both fall and the power factor drops below unity. Every problem here is that one right-angled triangle: first to compare the same machine on the two supplies, then to price a compensating winding, and finally to show why the same square-law torque that makes these motors indispensable in traction and hand tools also makes them destroy themselves if the load is removed.
A series machine does not care about the sign of the supply. Torque comes from the product of flux and armature current, and in a series motor the same current makes both. Reverse the supply and both reverse, so the product — and the torque — does not:
\[ T \propto \phi\,i_a \propto i_a^2 \qquad\Longrightarrow\qquad T_{av} = K I_{rms}^2 \]This is why the same machine runs on a.c. and on d.c., and why it is called universal.
On d.c. the circuit is resistive:
\[ V = E_a + I\left(R_a+R_{se}\right) \]On a.c. the same winding has reactance, and the rotational e.m.f. is in phase with the current — because the flux that generates it is itself made by that current. The three terms therefore form a right-angled triangle:
\[ V^2 = \left(E_a + IR\right)^2 + \left(IX\right)^2, \qquad \cos\phi = \frac{E_a+IR}{V} \]The reactance drop is in quadrature and steals nothing from the e.m.f. directly; it simply leaves less of \(V\) available along the current, so \(E_a\) and the speed both fall.
The e.m.f. and torque equations are the d.c. ones, unchanged:
\[ E_a = \frac{\phi ZNP}{60A}, \qquad T = \frac{\phi ZPI_a}{2\pi A} \]Below saturation \(\phi \propto I_a\), so \(T \propto I_a^2\) and \(N \propto E_a/I_a\).
The speed–torque law follows from those two, and it is the defining feature of the machine:
\[ N = \frac{K_1(V-IR)}{I} \quad\text{with}\quad T = K_2I^2 \;\Longrightarrow\; N \approx \frac{K_3}{\sqrt{T}} - K_4 \]As the torque falls towards zero the speed rises without limit. A series motor must never be run uncoupled.
A compensating winding removes the armature's cross reactance. Wound in the stator and carrying the armature current in opposition, it cancels the armature m.m.f., reduces the total reactance, raises the power factor and improves commutation.
Everything that makes the a.c. performance worse scales with frequency. \(X = 2\pi fL\), so a machine acceptable at 50 Hz may be unusable at 400 Hz; universal motors are accordingly built with few field turns, a laminated yoke and a low inductance.
A 230 V universal motor has a total armature-plus-field resistance of 20 Ω and a total series reactance of 30 Ω at 50 Hz. Operating on 230 V d.c. it draws 2 A and runs at 8000 rev/min. It is now connected to a 230 V, 50 Hz a.c. supply and the load is adjusted until the current is again 2 A r.m.s. Determine
- the back e.m.f. and speed on a.c.;
- the a.c. power factor and the input power on each supply;
- the torque on each supply, and hence why the a.c. speed is lower.
On d.c. the reactance does nothing, because nothing is changing:
On a.c. the three voltages form a right-angled triangle. The rotational e.m.f. is in phase with the current, because the flux producing it is made by that same current; the reactance drop is at 90° to it:
Speed is proportional to \(E_a/\phi\), and the flux is set by the current. The current is the same on both supplies, so the flux is the same and the speeds are in the ratio of the e.m.f.s alone:
Power factor and input power:
Torque is identical on the two supplies. Gross mechanical power divided by angular speed:
They must be equal, and this is the check on the whole calculation: \(T \propto I^2\), and the current was made equal by construction.
So why is the a.c. speed lower? Not because the torque is different — it is not. The supply voltage has to cover a reactance drop of \(IX = 60\) V that simply does not exist on d.c., and because that drop is in quadrature it removes
from what is available along the current. Every one of those 7.96 volts comes out of the back e.m.f., and the speed falls in proportion. Two further effects, not included here, push the a.c. speed lower still: hysteresis and eddy-current loss in the field iron, and the fact that torque depends on the mean square while the flux saturates on the peaks.
A 250 V, 50 Hz single-phase a.c. series motor takes 20 A. The total effective resistance of the armature, field and brushes is 1.2 Ω and the total reactance is 3.0 Ω. At this load it runs at 1500 rev/min and its iron, friction and windage losses amount to 200 W. Determine the back e.m.f., the power factor, the input power, the gross torque and the efficiency.
The voltage triangle, with the two drops known:
The power factor is the ratio of the in-phase side to the hypotenuse:
Notice how the back e.m.f. does the work of raising the power factor. At standstill \(E_a = 0\) and the factor collapses to \(R/\sqrt{R^2+X^2} = 1.2/3.23 = 0.37\) — which is why an a.c. series motor draws a heavy and badly lagging starting current.
Input power, and the gross mechanical power:
The reactance consumes no power — it only bends the phasor — so the input less the copper loss is exactly the gross mechanical power.
Gross torque at 1500 rev/min:
Output and efficiency, after the rotational losses:
Where the losses sit. Of the 4854 W drawn, 480 W goes in copper and 200 W in iron and friction. The reactance, despite carrying a 60 V drop — a quarter of the supply voltage — wastes nothing at all. Its damage is done through the power factor: to deliver 4854 W the supply has to provide \(250\times20 = 5000\) VA, and the extra 146 VA is carried by the cable and the meter without earning anything.
Of the 3.0 Ω total reactance of the motor in Problem 2, 1.8 Ω is due to the cross-magnetising armature reaction. A compensating winding is added in series with the armature to cancel that reaction; it removes the 1.8 Ω but adds 0.15 Ω of resistance. At the same 20 A, determine the new back e.m.f., power factor, input power, speed and torque, and set the two designs side by side.
The new circuit constants:
The voltage triangle again:
Input and gross mechanical power:
Speed rises with the back e.m.f., the current and hence the flux being unchanged:
Torque is unchanged, as it must be at the same current:
The two designs compared, both at 20 A and 250 V:
| Quantity | Uncompensated | Compensated | Change |
|---|---|---|---|
| Resistance \(R\) | 1.20 Ω | 1.35 Ω | +12.5 % |
| Reactance \(X\) | 3.00 Ω | 1.20 Ω | −60 % |
| Back e.m.f. \(E_a\) | 218.7 V | 221.8 V | +1.4 % |
| Power factor | 0.971 | 0.995 | +2.5 % |
| Input power | 4854 W | 4977 W | +2.5 % |
| Copper loss | 480 W | 540 W | +12.5 % |
| Gross mechanical power | 4374 W | 4437 W | +1.4 % |
| Speed | 1500 rev/min | 1522 rev/min | +1.4 % |
| Torque | 27.84 N·m | 27.84 N·m | none |
Reading the table honestly. The dramatic 60 % cut in reactance buys a 2.5 % improvement in power factor and 1.4 % more output at the same current, and costs 60 W of extra copper loss. On these figures alone the compensating winding hardly pays. Its real justification is at the other end of the speed range: at standstill and at low speed the back e.m.f. is small, the reactance dominates the impedance, and the uncompensated machine's power factor falls towards 0.37 while the compensated machine holds about 0.75. Starting current and starting power factor are where compensation earns its keep.
The other reason for compensating. Cross-magnetising armature reaction distorts the flux under the pole tips and shifts the magnetic neutral with load, which on an a.c. machine causes severe sparking at the brushes as the coil undergoing commutation is short-circuited across a transformer e.m.f. The compensating winding removes the distortion at source. Large a.c. series traction motors are always compensated; small universal motors for hand tools rarely are, because at 8000 rev/min the back e.m.f. is large enough to keep the power factor respectable on its own.
A 220 V series motor has 4 poles, a wave-connected armature of 480 conductors and a total resistance of 0.5 Ω. At an armature current of 20 A the flux per pole is 8 mWb, and the magnetic circuit may be taken as unsaturated so that the flux is proportional to the current. Find the speed and the torque at 20 A and at 40 A, and verify the two proportionalities they illustrate.
Fix the winding constants. A wave winding has two parallel paths whatever the pole number:
At 20 A. Back e.m.f. from the circuit, then speed from the e.m.f. equation:
Check by power: \(E_aI = 210\times20 = 4200\) W and \(\omega = 2\pi\times1641/60 = 171.8\) rad/s, giving \(T = 4200/171.8 = 24.45\) N·m. ✓
At 40 A the flux doubles, the magnetic circuit being unsaturated:
The torque proportionality. Both the flux and the current have doubled, and the torque depends on their product:
This square law is the whole reason series motors are used for traction, cranes and hand tools: a modest current overload gives an enormous torque overload.
The speed relation. Since \(N \propto E_a/\phi \propto (V-IR)/I\):
Doubling the current has not quite halved the speed — the resistance drop takes a slightly larger bite at the higher current, so the ratio is 0.476 rather than 0.500.
Where the square law fails. Once the poles begin to saturate the flux stops following the current, and \(T \propto \phi I_a\) tends towards \(T \propto I_a\). A real machine's torque curve therefore starts as a parabola and straightens into a line at high current. Every calculation in this set assumes the parabolic region, and the assumption should be stated whenever a series-motor answer is quoted.
For the 220 V series motor of Problem 4, derive the speed and the torque as explicit functions of the armature current, tabulate them from 1 A to 50 A, and obtain the speed–torque relation. If the armature is mechanically safe only up to 3000 rev/min, find the smallest torque and current at which the motor may be run.
Write the flux as a constant times the current. From Problem 4, 8 mWb at 20 A:
Substitute into the speed equation:
And into the torque equation:
Both formulas reproduce Problem 4 exactly: at 20 A, \(N = 1718.8-78.1 = 1640.7\) and \(T = 0.06112\times400 = 24.45\). ✓
Tabulating across the working range:
| \(I\) (A) | 1 | 2 | 5 | 10 | 20 | 30 | 40 | 50 |
|---|---|---|---|---|---|---|---|---|
| \(N\) (rev/min) | 34297 | 17109 | 6797 | 3359 | 1641 | 1068 | 781 | 609 |
| \(T\) (N·m) | 0.06 | 0.24 | 1.53 | 6.11 | 24.45 | 55.00 | 97.79 | 152.79 |
The first two columns are not operating points; they are the reason the machine must be coupled to something.
Eliminate the current to get the speed–torque law. From the torque equation \(I = \sqrt{T/0.06112}\), so
A rectangular-hyperbola-like curve: steeply falling at small torque, flattening at large torque, and with no finite speed at zero torque.
The safe minimum load. Setting \(N = 3000\) rev/min:
So this motor must never be asked to deliver less than 7.6 N·m — 31 % of its 20 A torque — and its current must never be allowed below 11.2 A.
Why the speed has no upper bound. At no load the only torque required is friction. The current falls until \(0.06112I^2\) matches it, the flux falls with the current, and the speed needed to generate a back e.m.f. of nearly 220 V from that vanishing flux rises without limit. Mathematically \(N \to \infty\) as \(T \to 0\); physically the armature bursts. A series motor is therefore always directly coupled to its load — gears or a rigid shaft, never a belt that can slip off or a chain that can break.
Return to the 230 V universal motor of Problem 1, with \(R = 20\ \Omega\) and \(X = 30\ \Omega\) at 50 Hz. Tabulate its back e.m.f., speed and power factor on d.c. and on 50 Hz a.c. at currents of 1, 2, 3 and 4 A, taking the d.c. speed at 2 A as 8000 rev/min. Then find what happens if the same motor is connected to a 230 V, 400 Hz aircraft supply.
Establish the speed constant from the one given point. With \(N = K E_a/I\) and \(E_{dc} = 190\) V at 2 A giving 8000 rev/min:
The two e.m.f. expressions:
Working one line, at \(I = 3\) A: \(E_{ac} = \sqrt{52900-8100}-60 = 211.66-60 = 151.66\) V, against 170 V on d.c.
The full comparison:
| \(I\) (A) | \(T\) (N·m) | \(E_{dc}\) (V) | \(N_{dc}\) (rev/min) | \(E_{ac}\) (V) | \(N_{ac}\) (rev/min) | \(N_{ac}/N_{dc}\) | a.c. p.f. |
|---|---|---|---|---|---|---|---|
| 1 | 0.113 | 210.0 | 17684 | 208.04 | 17519 | 0.991 | 0.991 |
| 2 | 0.454 | 190.0 | 8000 | 182.04 | 7665 | 0.958 | 0.965 |
| 3 | 1.021 | 170.0 | 4772 | 151.66 | 4257 | 0.892 | 0.920 |
| 4 | 1.814 | 150.0 | 3158 | 116.21 | 2447 | 0.775 | 0.853 |
The torque column is common to both supplies, since \(T = 0.1134I^2\) depends only on the current.
The gap widens with load, and steeply. At 1 A the a.c. speed is within 1 % of the d.c. speed; at 4 A it is 22 % below. The reason is that the reactance drop \(IX\) grows with the current while the supply voltage does not, so it eats an ever larger share of the 230 V:
The lost voltage \(230-\sqrt{230^2-(IX)^2}\) is 1.96, 7.96, 18.34 and 33.79 V — and every volt of it comes off the back e.m.f.
Now the 400 Hz supply. The inductance is unchanged, so the reactance scales with frequency:
Impossible. The reactance drop alone would exceed the supply, so 2 A cannot be reached at all.
Find the largest current the machine can draw at 400 Hz, which is the stalled value with \(E_a = 0\):
The motor is useless: it can develop at best a twentieth of its 50 Hz torque, at a power factor of \(20/240.8 = 0.08\).
What the designer must do about it. Reactance is proportional to the square of the field turns, so a 400 Hz universal motor is built with far fewer field turns and a correspondingly larger current, together with a thinner-laminated and lower-permeability magnetic circuit. That is exactly the direction in which universal motors have always been designed — a weak field, a heavily loaded armature, and a high speed to recover the power. It is also why a universal motor makes a poor 50 Hz machine at low speed and an excellent one at 10 000 rev/min.
Key Formulas
| Quantity | Relation | Notes |
|---|---|---|
| D.C. circuit equation | \(V = E_a + IR\), \(R = R_a+R_{se}\) | No reactance — Problems 1, 4 |
| A.C. voltage triangle | \(V^2 = (E_a+IR)^2+(IX)^2\) | \(E_a\) in phase with \(I\) — Problems 1, 2, 3, 6 |
| A.C. power factor | \(\cos\phi = (E_a+IR)/V\) | Improves as the speed rises |
| Stalled power factor | \(\cos\phi = R/\sqrt{R^2+X^2}\) | \(E_a = 0\) — Problems 2, 6 |
| E.m.f. equation | \(E_a = \phi ZNP/60A\) | Same as any d.c. machine — Problem 4 |
| Torque equation | \(T = \phi ZPI_a/2\pi A\) | Equivalently \(T = E_aI_a/\omega\) |
| Series flux law | \(\phi = kI_a\) below saturation | Fails once the poles saturate — Problem 4 |
| Torque law | \(T \propto I_a^2\) | Same on a.c. with \(I_{rms}\) — Problems 1, 4, 5 |
| Speed law | \(N \propto (V-IR)/I\) | Hyperbolic in current — Problem 5 |
| Speed–torque | \(N = K_3/\sqrt{T}-K_4\) | Unbounded as \(T \to 0\) — Problem 5 |
| Speed ratio a.c. to d.c. | \(N_{ac}/N_{dc} = E_{ac}/E_{dc}\) at equal \(I\) | Equal current means equal flux — Problems 1, 6 |
| Gross mechanical power | \(P_{mech} = E_aI = P_{in}-I^2R\) | Reactance dissipates nothing — Problems 2, 3 |
| Efficiency | \(\eta = (E_aI-P_{rot})/VI\cos\phi\) | Problem 2 |
| Frequency scaling | \(X = 2\pi fL\) | Eight times worse at 400 Hz — Problem 6 |
| Maximum stalled current | \(I_{max} = V/\sqrt{R^2+X^2}\) | The ceiling when \(IX\) approaches \(V\) |
Common Mistakes
Writing \(V = E_a + IR + IX\) on a.c. The reactance drop is in quadrature; the three voltages form a right-angled triangle, not a sum — Problems 1, 2 and 3.
Putting the back e.m.f. in quadrature with the current. In a series machine the flux is made by the armature current, so the rotational e.m.f. is in phase with it — Problem 1.
Expecting less torque on a.c. than on d.c. at the same current. The torque is identical; only the speed differs — Problem 1.
Attributing the lower a.c. speed to the reactance "consuming power". A reactance consumes none; it removes voltage that would otherwise have appeared as back e.m.f. — Problems 1 and 2.
Assuming a compensating winding raises the torque. At the same current the torque is unchanged; what improves is the power factor, and with it the speed and the available output — Problem 3.
Forgetting that a wave winding always has \(A = 2\). Using \(A = P\) puts the speed and the torque out by the pole number — Problem 4.
Taking \(T \propto I^2\) beyond saturation. Once the flux stops following the current the law degenerates towards \(T \propto I\) — Problem 4.
Halving the speed when the current is doubled. The resistance drop makes the ratio 0.476, not 0.500 — Problem 4.
Treating the no-load speed as merely "high". It is unbounded: the equation has no finite value at zero torque, and the armature will burst — Problem 5.
Reusing a 50 Hz reactance at another frequency. \(X\) scales with \(f\), and at 400 Hz it can exceed the supply voltage entirely — Problem 6.
The series motor achieves on a single-phase supply what Set 44's split-phase windings had to work so hard for: it starts by itself, with the largest torque per ampere of any machine in this book, and it does so because its torque depends on the square of the current rather than on its sign. The price is a commutator, brushes that wear, a power factor spoiled by the series reactance, and a speed that runs away if the load is ever removed.
Every machine so far — d.c., transformer, induction, synchronous, universal — has been designed to be connected directly to a supply and left to find its own operating point. The last set of the book looks at machines built the other way round: they cannot run at all without a controller, and in exchange the controller is given complete authority over position, speed and torque. Some of them have no rotor winding whatsoever, and one of them makes torque out of nothing but the rotor's preference for a shorter magnetic path.
Next: Set 46 — Stepper, BLDC and Reluctance Motors, where the shaft position is counted in pulses, the back-e.m.f. constant and the torque constant turn out to be the same number, and torque comes from \(\tfrac12i^2\,dL/d\theta\).