Speed control and DC motor starters

A DC motor's speed obeys one compact relation, $N \propto E_b/\Phi$ , and that single fact is what makes DC motors so easy to command — turn one knob and the motor slows down, turn another and it speeds up, all without changing the load. But the very same machine that is so obliging once running is dangerous the instant you switch it on: a stationary armature has no back-EMF to fight the supply, so raw line voltage sees only the tiny armature resistance and tries to push a huge current through the windings. This article closes the DC-machines story by tying speed control and starting together. We use $N \propto E_b/\Phi$ with $E_b = V - I_aR_a$ to show why varying armature voltage and varying field flux give two complementary speed ranges, then use the same $E_b = V - I_aR_a$ relation with $E_b = 0$ at standstill to size the external resistance a starter must insert to keep the initial current safe.

The core insight

Because $N \propto E_b/\Phi$ with $E_b = V - I_aR_a$ , a DC motor's speed can be lowered below base speed by reducing armature voltage or raised above base speed by weakening the field, while at standstill the same relation (with $E_b=0$ ) shows why a starter must insert external resistance to limit the otherwise huge starting current.

The core idea

The governing relation. From the DC-machine equations already established, the back-EMF is $E_b = k\Phi N$ , so speed rearranges to

N \propto \frac{E_b}{\Phi}, \qquad E_b = V - I_aR_a.

where: $N$ is speed (rpm), $E_b$ is back-EMF (V), $\Phi$ is field flux per pole (Wb), $V$ is supply voltage (V), $I_a$ is armature current (A), $R_a$ is armature resistance ( $\Omega$ ). This one relation is the whole of DC motor speed control — every method changes either $E_b$ or $\Phi$ .

Armature-voltage control (below base speed). Reduce the voltage across the armature — a series rheostat, or a variable supply — while keeping the field at its full rated value. For a fairly constant load torque $I_a$ stays close to constant, so $E_b = V - I_aR_a$ falls almost in step with $V$ . With $\Phi$ held fixed, $N \propto E_b$ falls too: this method takes the motor from base speed down toward zero. It gives smooth, fine control at low speed but wastes power as heat when done with a plain rheostat.

Field-flux control (above base speed). Increase the resistance in the shunt-field circuit instead, which lowers the field current and hence $\Phi$ . With $E_b$ nearly constant (close to $V$ , since the armature drop is small), $N \propto 1/\Phi$ means weakening the flux raises speed — the motor climbs above base speed. Because the field current is small, the $I^2R$ loss in the field rheostat is small too, so this method is efficient — but it cannot push speed down, and it cannot be pushed too far up either, since flux is already near its rated (near-saturation) ceiling and cannot be strengthened much further. Used together, armature-voltage control below base speed and field-flux control above base speed give the DC motor a wide, continuous speed range — one of its standout advantages over machines like the induction motor.

Figure

A speed versus control setting graph with two straight segments meeting at base speed, one rising segment for armature voltage control below base speed and one falling segment for field flux control above base speed

Why a starter is needed. At the instant of switch-on the armature has not yet turned, so no back-EMF exists: $E_b = 0$ . The current equation $I_a = (V - E_b)/R_a$ then collapses to $I_a = V/R_a$ , and since $R_a$ is deliberately made very small (to keep running $I^2R$ loss low), this starting current can be tens of times the rated value — large enough to damage the commutator, brushes and windings in seconds. A starter (3-point or 4-point type) inserts an external resistance in series with the armature at the moment of starting, and cuts it out step by step as the motor accelerates and $E_b$ builds up, keeping the current at each step within a safe band; once the motor reaches running speed the resistance is fully removed and $E_b$ alone limits the current. (The 3-point and 4-point designs differ only in how they wire field-circuit protection — a construction detail, not needed here.)

Derivation ASizing the external starting resistance from E_b = V - I_aR_a

The goal is to turn the single relation $E_b = V - I_aR_a$ , evaluated at the instant of starting where $E_b = 0$ , into a formula for the external starter resistance $R_{st}$ that limits the starting armature current to a chosen safe multiple of the rated current — the calculation every DC-starter sizing problem in this syllabus ultimately reduces to.

Step 1. Write the armature circuit equation with the starter resistance included, $I_a = \dfrac{V - E_b}{R_a + R_{st}}$ , where $R_{st}$ is the external resistance the starter switches in at the first contact position.

Step 2. Apply the starting condition. At the very instant of starting the armature is stationary, so $E_b = 0$ , and the equation reduces to $I_{a,start} = \dfrac{V}{R_a + R_{st}}$ .

Step 3. Decide the safe starting current. A designer chooses $I_{a,start}$ as some multiple $m$ of the rated current, $I_{a,start} = m\,I_{rated}$ , small enough to protect the commutator and windings (commonly $m$ between about $1.5$ and $2$ ).

Step 4. Solve for the total resistance and then $R_{st}$ . Rearranging Step 2 gives $R_a + R_{st} = V/I_{a,start}$ , so

R_{st} = \frac{V}{I_{a,start}} - R_a = \frac{V}{m\,I_{rated}} - R_a.

This is the resistance the first starter contact must insert; as the motor speeds up, $E_b$ grows, so progressively less external resistance is needed to hold the current near $m\,I_{rated}$ , and each subsequent starter step removes some of $R_{st}$ accordingly.

Worked example

Problem. A 220 V shunt motor has armature resistance $R_a = 0.4\ \Omega$ and a rated full-load armature current of 25 A. Part (a): find the starting current if switched directly across the line with no starter. Part (b): find the external starting resistance needed to limit the starting current to 1.5 times rated current.

Step 1. Direct starting, no resistance: with $E_b = 0$ ,

I_{a,start} = \frac{V - E_b}{R_a} = \frac{220 - 0}{0.4} = 550\ \text{A}.

Comparing to rated current, $550/25 = 22$ — the motor would draw 22 times its rated current, confirming why direct starting is unsafe.

Step 2. Choose the safe starting current for part (b): $I_{a,start} = 1.5 \times 25 = 37.5\ \text{A}$ .

Step 3. Apply the derivation's result with $m = 1.5$ :

R_a + R_{st} = \frac{V}{I_{a,start}} = \frac{220}{37.5} = 5.867\ \Omega.

Step 4. Isolate $R_{st}$ :

R_{st} = 5.867 - 0.4 = 5.467\ \Omega \approx 5.47\ \Omega.

Answer: without a starter the motor draws 550 A (22 times rated); an external starting resistance of about $5.47\ \Omega$ limits it to 37.5 A (1.5 times rated), and this resistance is cut out in steps as the motor accelerates and back-EMF builds up.

Competitive stretch

If the same motor's starter is required to limit current to twice rated ( $m=2$ , so $I_{a,start} = 50$ A) instead, the total resistance needed is $220/50 = 4.4\ \Omega$ , so $R_{st} = 4.4 - 0.4 = 4.0\ \Omega$ — a smaller external resistance than for the tighter $1.5\times$ limit, since allowing a larger safe current needs less resistance to hold it back. This illustrates the general trade-off: a stricter current limit (smaller $m$ ) always demands a larger $R_{st}$ .

Common traps

Mixing up the two control zones: field-flux (weakening) control only raises speed above base speed, so trying to use it to go slower than base speed does not work — that needs armature-voltage control instead.
Forgetting $E_b = 0$ at the very instant of starting and mistakenly plugging in a running (nonzero) back-EMF into the starting-current formula, which produces a falsely small and unsafe-seeming answer.
Treating the starter's resistance as permanent — it must be fully cut out once the motor is up to speed; leaving it in wastes power as heat and caps the motor below its rated speed and torque.
Assuming field-rheostat speed control has no downside — it is efficient, but it only works above base speed and cannot be pushed too far, since a very weak field risks overspeed instability.

Ask the tutor: want to try sizing a multi-step starter that removes resistance in several stages as the motor accelerates, or work the reverse problem where a maximum allowed starting current is given and you back out the minimum permissible armature resistance?

Why does weakening the field make the motor go faster, not slower?

Speed follows $N \propto E_b/\Phi$ . While the motor is running, the armature drop $I_aR_a$ is small, so $E_b$ stays close to the supply voltage and barely changes when you adjust the field. So the flux $\Phi$ sits alone in the denominator: shrink $\Phi$ and the ratio $E_b/\Phi$ grows, so $N$ must rise to keep the machine in balance. It feels backwards at first, but the motor is compensating for weaker flux by spinning faster to generate the same back-EMF.

Why is armature resistance made so small in the first place, if it causes the huge starting current problem?

A small $R_a$ keeps the running $I^2R$ loss low, which is essential for efficiency — a motor running continuously cannot afford to waste significant power heating its own windings. The tiny $R_a$ is exactly right for running conditions, where back-EMF does almost all the current-limiting work. The starting problem only appears because, at standstill, $E_b$ is zero and there is nothing except that same small $R_a$ to hold current back — which is precisely why an external, temporary starting resistance is added rather than redesigning the armature with permanently higher resistance.

Does every DC motor need a starter, even a small one?

In principle any DC motor started directly across the line will try to draw the same excessive multiple of rated current, since the ratio $V/(I_{rated}R_a)$ does not depend on the motor's size. In practice, very small motors have low enough absolute current and thermal mass that brief direct starting rarely causes visible damage, so simple designs sometimes skip a formal starter. For any motor of meaningful size, though, a 3-point or 4-point starter (or an equivalent electronic soft-start) is standard practice to protect the commutator and windings.

Speed control and DC motor starters

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads