How motors make torque

A wire carrying current $I$ in a magnetic field $B$ feels a force $F = B I L$. Wind the wire into a loop on a shaft and the two sides are pushed in opposite directions: that is torque. Every motor is a way of keeping that force pointing the same way as the rotor turns.

In the course's terms (eqs. 1.4–1.6): a conductor of length $L$ carrying $I$ in a field $B$ feels $F = I L B \sin\theta$, largest when it is perpendicular to the field; the force at radius $r$ gives the torque $\tau = r \times F$; and a coil of $N$ turns and area $A$ gives $\tau = N I A B \sin\varphi$. Torque is proportional to current and field: that is the lever every drive pulls.

The induction motor

This is the motor of Labs I to V, and of most of industry. The three stator windings make a rotating field (lesson A2) at synchronous speed $n_s = 120 f / p$.

The rotor is a cage of bars shorted by end rings. As the field sweeps past the bars it induces currents in them, those currents feel a force, and the rotor follows the field. It can never catch up: at synchronous speed the field would not move relative to the bars, nothing would be induced, and there would be no torque. The fractional lag is the slip:

$$s = \frac{n_s - n_r}{n_s}$$

Typical full-load slip is 2 to 5 %.

FoundationStart here if this is new to you

Imagine running beside a moving conveyor belt, trying to catch a paper blowing on it. If you run exactly as fast as the belt, the paper never moves relative to you. You only feel the wind if the belt runs a little faster than you. The rotor is you, the field is the belt: it only gets pushed while it is slightly slower.

The torque-speed curve

The curve below comes from the motor's equivalent circuit, the same formula as Lab I:

$$T(s) = \frac{3 V_1^2\, (R_2'/s)}{\omega_s\left[(R_1 + R_2'/s)^2 + (X_1 + X_2')^2\right]}$$

Constant torqueFan or pump

Load torque

80 N·m

Supply voltage

100 %

Rotor resistance R₂′

0.40 Ω

The motor starts and settles where the two curves cross.

Starting torque

84.0 N·m

Breakdown torque

197.2 N·m

Breakdown slip

0.194

Operating speed

1,447 rpm

Operating slip

3.56 %

Read it from standstill (left) to synchronous speed (right):

• Starting torque, at $s = 1$. It must beat the load's breakaway torque or the motor never moves.
• Breakdown torque, the peak, at slip $s_\text{bd} = R_2' / \sqrt{R_1^2 + X^2}$. Load the motor beyond it and it stalls.
• The stable region lies between breakdown and synchronous speed. There, a small drop in speed raises torque steeply, so the motor holds speed under load.

Predict first

You raise the rotor resistance R₂′ from 0.4 Ω to 2 Ω. What happens to the breakdown torque?

It risesIt stays the sameIt falls

At the instant of starting there is no induced back-voltage to oppose the supply, so the motor draws 5 to 8 times its full-load current. Every starting method in Chapter 2 exists to deal with that inrush.

Synchronous motors

The rotor is a magnet, DC-excited or permanent, so it locks onto the rotating field and turns at exactly $n_s$. Load makes it fall behind the field by the torque angle δ:

$$T = \frac{3 V_t E_f \sin\delta}{\omega_s X_s}$$

Torque peaks at δ = 90°. Beyond that the rotor slips poles and loses synchronism.

DC motors

Back-EMF $E = k\Phi\omega$, torque $T = k\Phi I_a$ and the armature circuit $V = E + I_a R_a$ give:

$$\omega = \frac{V - I_a R_a}{k\Phi}$$

• Shunt: constant flux, speed drops only a little with load.
• Series: flux follows current, so torque grows as $I_a^2$. With no load the flux vanishes and the speed runs away.
• Compound: a mix of both.

ExplorerGo deeper: derivations and open questions

Where does the breakdown slip come from? Differentiate $T(s)$ with respect to $R_2'/s$ and set it to zero: the maximum occurs when $R_2'/s = \sqrt{R_1^2 + X^2}$, the condition for maximum power transfer into the resistance $R_2'/s$.

Rotor losses. The air-gap power splits into mechanical power $(1-s)P_\text{ag}$ and rotor copper loss $sP_\text{ag}$. A motor running at 30 % slip on a wound-rotor resistance wastes 30 % of its air-gap power as heat. Why does that make rotor-resistance speed control inefficient?