Set 12 — Speed Control of DC Motors
A dc motor's speed is set by the ratio of back emf to flux, and both terms are open to interference. Weaken the field and the speed rises; insert resistance in the armature and it falls; change the applied voltage and it follows. This set works one problem on each, and a fourth in which the load itself dictates how the current must move.
The recurring trap is to change one quantity and forget its consequences. Weakening the field at constant torque raises the armature current, which lowers the back emf, which pulls the speed back below the ideal ratio. Every problem here is arranged so that the second-order effect matters.
One relation underlies every method of speed control:
\[ N = \frac{E_a}{k\phi} = \frac{V_t - I_a(R_a + R_{ext})}{k\phi} \]Three quantities on the right can be altered — the flux, the armature-circuit resistance and the applied voltage — and they give the three classical methods.
Always compare two operating points rather than computing absolute speeds. The machine constant cancels:
\[ \frac{N_2}{N_1} = \frac{E_{a2}}{E_{a1}}\times\frac{\phi_1}{\phi_2} \]The load fixes the torque, and the torque fixes the current. Since \(T \propto \phi I_a\), a constant-torque load subjected to field weakening forces the armature current up in exact inverse proportion to the flux:
\[ \phi_1I_{a1} = \phi_2I_{a2} \]Never change the flux without recomputing \(E_a\). The new armature current changes the armature drop, so \(E_{a2} \ne E_{a1}\) and the naive \(N \propto 1/\phi\) is wrong.
At constant flux the relation collapses to \(N \propto E_a\), and \(T \propto I_a\). This is armature-resistance control: speeds below base speed, at the price of \(I_a^2R_{ext}\) wasted in the added resistor.
In a series motor the flux follows the armature current. Unsaturated, \(\phi \propto I_a\), so \(T \propto I_a^2\) and, with the resistance neglected, \(N \propto V_t/I_a\).
A 250 V dc shunt motor has a shunt field resistance of 250 Ω and an armature resistance of 0.25 Ω. With no additional resistance in the field circuit it runs at 1500 rpm and draws an armature current of 20 A. A resistance of 250 Ω is now inserted in series with the field, the load torque remaining the same. Find the new speed and the new armature current.
Set out the two relations the problem turns on. Speed is governed by back emf and flux; torque is governed by current and flux. The magnetic circuit is unsaturated, so the flux tracks the field current and the ratio \(\phi_1/\phi_2\) may be replaced by \(I_{f1}/I_{f2}\):
The added resistance halves the field current. The field circuit is a simple series path across the 250 V supply:
So \(\phi_1/\phi_2 = 2\): the flux is halved.
Constant load torque doubles the armature current. With \(T_a \propto \phi I_a\) unchanged,
This is the price of field weakening: the same torque now costs twice the armature current, and four times the armature copper loss.
The two back emfs must be computed separately, because the armature drop is no longer the same:
Combine flux and emf:
Just short of double. Halving the flux would double the speed exactly if the back emf held constant; the extra armature drop caused by the doubled current pulls it back by about 2%.
A 220 V shunt motor has an armature resistance of 0.5 Ω and takes a current of 40 A on full load. By how much must the main flux be reduced to raise the speed by 50%, the developed torque remaining constant?
Name the unknown as a ratio. Let \(x = \phi_1/\phi_2\), the factor by which the flux is weakened. Everything else can be written in terms of it.
Constant torque fixes the new current in terms of \(x\):
Both back emfs follow:
The second emf now depends on the unknown, which is what makes this problem quadratic rather than linear.
Impose the 50% speed rise:
Solve, and reject the unphysical root:
The larger root demands \(I_{a2} = 40 \times 9.405 = 376\ \text{A}\) in a machine rated at 40 A, with a back emf of only 32 V. It satisfies the algebra and nothing else, so \(x = 1.595\).
Convert the ratio into a reduction. The question asks how much the flux falls, not what the ratio is:
Rounding \(x\) to 1.6 before this step gives the frequently quoted 37.5%, but 1.6 does not satisfy the quadratic; carrying the exact root through gives 37.3%.
Verify. With \(x = 1.595\): \(I_{a2} = 63.8\ \text{A}\), \(E_{a2} = 220 - 31.9 = 188.1\ \text{V}\), and
A 220 V shunt motor with an armature resistance of 0.5 Ω is excited to give a constant main field. At full load it runs at 500 rpm and takes an armature current of 30 A. A resistance of 1 Ω is now placed in the armature circuit. Find the speed at
- full-load torque
- double full-load torque
Constant flux simplifies both relations. With \(\phi\) fixed the speed ratio loses its flux term and the torque becomes proportional to current alone:
Establish the reference condition before any resistance is added:
Part (a): full-load torque. Torque unchanged means current unchanged, so \(I_{a2} = 30\ \text{A}\) still — but it now flows through 1.5 Ω:
A 15% speed reduction bought with 900 W dissipated in the added resistor — roughly 14% of the armature input.
Part (b): double full-load torque. At constant flux the current must double:
Read the two answers together. The same added resistor gives 427 rpm at one torque and 317 rpm at another: the speed now depends heavily on the load, whereas without the resistor it would have fallen only from 500 rpm to about 463 rpm over the same range.
A dc series motor drives a load whose torque varies as the square of the speed. The magnetic circuit may be assumed unsaturated, the motor resistance negligible and the stray losses ignored. Estimate the percentage reduction in the motor terminal voltage that will halve the speed, and the percentage fall in the motor current that accompanies it.
Start with the machine, not the load. In an unsaturated series motor the field carries the armature current, so the flux is proportional to it and the torque becomes a square law:
Now impose the load. The load demands \(T \propto N^2\), and in steady state the two torques are equal:
Current and speed are locked together for this particular load, whatever the applied voltage. This is the whole content of the second half of the question.
Halving the speed therefore halves the current:
With the resistance neglected the back emf equals the terminal voltage, so \(E_{a1} = V_1\) and \(E_{a2} = V_2\). The flux ratio is the current ratio, so \(\phi_1/\phi_2 = 2\), and the speed relation becomes
Express as a reduction:
Consistent with the power view: torque falls as \(N^2\) and speed as \(N\), so the output power falls as \(N^3\) — to one-eighth. With the current halved, the voltage must fall to a quarter to make \(VI\) one-eighth.
Key Formulas
| Quantity | Relation | Notes |
|---|---|---|
| Speed of a dc motor | \(N = \dfrac{V_t - I_a(R_a+R_{ext})}{k\phi}\) | The master relation — all problems |
| Two-point speed ratio | \(\dfrac{N_2}{N_1} = \dfrac{E_{a2}}{E_{a1}}\times\dfrac{\phi_1}{\phi_2}\) | Machine constant cancels — all problems |
| Back emf | \(E_a = V_t - I_a(R_a+R_{ext})\) | Problems 1–3 |
| Torque | \(T_a \propto \phi I_a\) | Problems 1–3 |
| Constant-torque condition | \(\phi_1I_{a1} = \phi_2I_{a2}\) | Problems 1, 2 |
| Shunt field current | \(I_f = \dfrac{V_t}{R_{sh}+R_{ext}}\) | Problem 1 |
| Unsaturated flux | \(\phi \propto I_f\), so \(\dfrac{\phi_1}{\phi_2} = \dfrac{I_{f1}}{I_{f2}}\) | Problem 1 |
| Flux reduction | \(\dfrac{\phi_1-\phi_2}{\phi_1} = 1 - \dfrac{1}{x},\ x = \dfrac{\phi_1}{\phi_2}\) | Ratio is not the reduction — Problem 2 |
| Constant flux | \(N \propto E_a\) and \(T \propto I_a\) | Armature control — Problem 3 |
| Rheostat loss | \(P_{ext} = I_a^2R_{ext}\) | The cost of armature control — Problem 3 |
| Series motor, unsaturated | \(\phi \propto I_a\), \(T_a \propto I_a^2\) | Problem 4 |
| Series motor speed | \(N \propto \dfrac{V_t - I_a(R_a+R_{se})}{I_a}\) | Problem 4 |
| Fan-type load | \(T_L \propto N^2\), \(P \propto N^3\) | Problem 4 |
| Voltage reduction | \(\dfrac{V_1-V_2}{V_1}\times100\%\) | Problem 4 |
Common Mistakes
Holding the armature current fixed while weakening the field. A constant-torque load forces \(I_a\) up in inverse proportion to the flux — from 20 A to 40 A here — Problems 1 and 2.
Writing \(N \propto 1/\phi\) and stopping there. The back emf changes too, because the new current changes the armature drop; ignoring it gives 3000 rpm instead of 2939 rpm — Problem 1.
Confusing the flux ratio with the flux reduction. \(\phi_1/\phi_2 = 1.595\) corresponds to a 37.3% reduction, not a 59.5% one — Problem 2.
Accepting both roots of the quadratic. One of them demands 376 A in a 40 A machine; a root has to be physically checked, not merely computed — Problem 2.
Rounding an intermediate root before using it. Taking \(x = 1.6\) in place of 1.595 changes the answer by 0.2 percentage points and does not satisfy the equation it came from — Problem 2.
Forgetting to add the external resistance to \(R_a\). The armature drop is \(I_a(R_a+R_{ext})\), so 45 V rather than 15 V at full-load torque — Problem 3.
Assuming the added resistor changes the torque. It does not: at constant flux the torque still depends only on \(I_a\), and the load decides that — Problem 3.
Using \(T \propto I_a\) for a series motor. Its flux rises with current, so unsaturated torque goes as \(I_a^2\) — Problem 4.
Halving the voltage to halve the speed of a series motor. The flux falls with the current as well, so a quarter of the voltage is required on a fan load — Problem 4.
Neglecting the armature drop when the problem does not permit it. Problem 4 explicitly calls the resistance negligible; Problems 1 to 3 do not, and the drop is what makes their answers fall short of the ideal ratios.
Three handles, three characters. Field weakening raises the speed above base speed at roughly constant power, and charges for it in armature current. Armature-resistance control lowers the speed below base speed, and charges for it in wasted heat and in a speed that sags with load. Voltage control does the same job without either penalty, which is why it displaced the other two as soon as controlled rectifiers became cheap.
Every calculation on this page assumed the motor was already turning. At standstill the back emf is zero, the armature drop is the only thing limiting the current, and a 220 V machine with 0.5 Ω would draw 440 A — more than ten times full load. Something must stand in for the missing back emf during run-up, and removing it again as the machine accelerates.
Next: Set 13 — Starting, Starters and Braking, where the sections of a multi-step starter are designed from the permitted current swing, and the same three handles are used in reverse to bring a loaded machine to rest.