The quarter-turn problem, and the switch that solves it
In rung 1 you saw the fundamental motor force: a wire carrying current I, of length L, sitting in a magnetic field B, feels a sideways push F = B·I·L. Loop that wire into a coil between a north and a south pole and the two sides feel opposite pushes — one up, one down — so the coil twists. Beautiful. But chase the motion through half a turn and the dream collapses. As the coil swings past vertical, the side that was being pushed up is now on the other side, where the field would push it *down*. The torque reverses. Left alone, the coil would just rock back and forth and settle, like a compass needle snapping to north.
The fix is almost embarrassingly clever: at the exact instant the coil passes vertical — the instant the torque is about to reverse — flip the direction of the current in the coil. Then the force on each conductor flips too, and the torque keeps pointing the same way around. The device that performs this flip, automatically, twice per revolution, is the commutator: a copper ring split into segments, riding on the shaft, with two stationary carbon brushes pressed against it. As the shaft turns, the brushes slide from one segment to the next, reversing which way current threads the coil. It is a mechanical rotary switch, synchronised to the rotor by being bolted to it.
Position of coil Without commutator With commutator
---------------- ------------------ ----------------
0° (horizontal) torque + (forward) torque + (forward)
90° (vertical) torque 0 -> reverses current FLIPS here
180° (horizontal) torque - (backward!) torque + (forward)
270° (vertical) torque 0 current FLIPS again
360° (back to start) torque + torque +
Result: rocking + stall vs. steady one-way spin
N pole brush
+--------+ |
| ==== | <- coil sides [== seg A ==] split-ring
| |||| | (force up/down) [== seg B ==] commutator
+--------+ |
S pole brushBack-EMF: the motor that is also, secretly, a generator
Here is the deepest idea in the whole machine, and it is a kind of cosmic fairness. The same conductors that the field *pushes* on to make torque are also moving *through* that field. And Faraday's law says a conductor moving through a magnetic field generates a voltage. So the moment the rotor starts spinning, the motor begins acting as a generator, producing a voltage of its own. By Lenz's law this generated voltage opposes the supply that is driving the current — it pushes back. We call it the [[ee-back-emf|back-EMF]], written E.
How big is the back-EMF? It scales with how fast the rotor spins and how strong the field is: E = k·Φ·ω, where ω is the rotor's angular speed (rad/s), Φ is the magnetic flux per pole, and k is a constant baked in by the machine's geometry — number of conductors, poles, windings. Hold the field fixed (as a magnet-based or separately-excited motor does) and back-EMF is simply proportional to speed: E = kₑ·ω. Spin twice as fast, push back twice as hard. At standstill, ω = 0, so the motor generates *nothing* — it pushes back with zero volts. That single fact, as we'll see, explains the violent gulp of current a motor takes the instant you switch it on.
One equation runs the whole show: V = E + I·R
Now put the electrical side under a microscope. The armature — the spinning winding — is just a coil of copper wire, so it has a small resistance R (the armature resistance, often a fraction of an ohm). When you apply a supply voltage V across the brushes, that voltage has to do two jobs: overcome the back-EMF E the motor is generating, and drive the current I through the winding's resistance R. Walk around the armature loop with Ohm's law and Kirchhoff's voltage law and you get the most important equation in the chapter:
Armature circuit (separately-excited / PM, field held constant)
V ----[ brush ]----+
|
( ) E = k_e * w <- back-EMF (a generator)
| (the spinning coil)
[R] armature resistance
|
V ----[ brush ]----+
KVL: V = E + I*R
Solve for current: I = (V - E) / R = (V - k_e*w) / R
Torque: T = k_T * I (k_T = k_e in SI units)
The self-regulating loop:
start: w = 0 -> E = 0 -> I = V/R (HUGE) -> big torque -> speeds up
then : w rises -> E rises -> I = (V-E)/R drops -> torque drops -> ...
settle: until torque produced = torque the load demands.Read that little loop at the bottom slowly, because it *is* the motor's whole behaviour. The instant you flip the switch the rotor is still: E = 0, so the only thing holding back the current is the tiny armature resistance, and I = V/R is enormous — this is the inrush current that dims your lights and the stall current that defines maximum torque. The huge torque flings the rotor into motion; as ω climbs, E = kₑ·ω climbs with it; the *difference* V − E shrinks; so current and therefore torque fall. The motor keeps accelerating only until the torque it makes exactly matches the torque the load demands. It is a self-balancing system, and it balances itself with no controller at all — just physics in a copper loop.
The speed–torque line, drawn from scratch
Two equations, T = k·I and V = k·ω + I·R, are all we need to predict everything. Eliminate the current and you get the motor's signature characteristic — a straight line linking speed and torque. Substitute I = T/k into the voltage equation, rearrange for ω, and out drops:
Start from: V = k*w + I*R and T = k*I (so I = T/k)
V = k*w + (T/k)*R
k*w = V - (R/k)*T
----------------------------------------------------
w = V/k - (R / k^2) * T <-- the speed-torque LINE
----------------------------------------------------
Two anchor points pin the line down:
* No-load (T = 0): w_0 = V / k <- top speed, free-spinning
* Stall (w = 0): T_stall = k*V / R <- max torque, zero speed
w
w_0 |*
| *.
| *. slope = -R/k^2 (small R => nearly flat,
| *. a 'stiff' constant-speed motor)
| *.
+---------*----------- T
0 T_stallStand back and admire how much physics is compressed into that one line. The vertical intercept ω₀ = V/k is the no-load speed: with no load, the motor speeds up until back-EMF almost equals the supply (E ≈ V), the current dwindles to nearly nothing, and the shaft coasts at a top speed set only by voltage. The horizontal intercept T_stall = kV/R is the stall torque: the most twist the motor can ever produce, at the instant of switch-on. And the slope −R/k² tells you how 'droopy' the motor is — how much speed it gives up as you load it. A motor with tiny armature resistance has a nearly horizontal line: load it heavily and it barely slows. Crank V up or down and the whole line slides vertically — which is exactly how a speed controller works.
A worked example: finding the operating point
Numbers turn the abstract line into a real machine. Take a small permanent-magnet DC motor on a 24 V supply, with armature resistance R = 0.5 Ω and a combined constant k = kₑ = k_T = 0.1 (in SI units: 0.1 V·s/rad, equivalently 0.1 N·m/A). Let's find its stall torque, no-load speed, and where it settles when driving a load that demands 1.2 N·m.
GIVEN: V = 24 V, R = 0.5 ohm, k = 0.1 (V.s/rad = N.m/A)
1) Stall torque (w = 0, so E = 0, I = V/R):
I_stall = V/R = 24 / 0.5 = 48 A
T_stall = k*I = 0.1 * 48 = 4.8 N.m
2) No-load speed (T = 0, so I ~ 0, E ~ V):
w_0 = V/k = 24 / 0.1 = 240 rad/s
= 240 * 60 / (2*pi) ~ 2292 rpm
3) Operating point driving a load needing T = 1.2 N.m:
I = T/k = 1.2 / 0.1 = 12 A (current it draws)
E = V - I*R = 24 - 12*0.5 = 18 V (back-EMF generated)
w = E/k = 18 / 0.1 = 180 rad/s ~ 1719 rpm
Check it lies on the line: w = V/k - (R/k^2)*T
= 240 - (0.5/0.01)*1.2
= 240 - 60 = 180 rad/s [OK]
4) Where did the power go, at this operating point?
P_in = V*I = 24 * 12 = 288 W
P_loss = I^2 * R = 12^2 * 0.5 = 72 W (heat in copper)
P_mech = E*I = T*w = 18 * 12 = 216 W (useful shaft power)
Efficiency = 216 / 288 = 75 %That last block hides the soul of the machine. The product E·I — back-EMF times current — equals T·ω, electrical power converted into mechanical power, to the watt. Back-EMF isn't just a current-limiter; it is the exact bookkeeping entry for energy crossing from the electrical world into the spinning one. Whatever the supply delivers (V·I) that *doesn't* get converted is lost as I²R heat in the winding. Notice how efficiency sinks near stall — at stall, *all* 1152 W (24 V × 48 A) becomes heat and the shaft delivers zero useful work — and peaks somewhere in the mid-load range where back-EMF is high and current still modest.
Where the field comes from: shunt, series, separately-excited
Everything above quietly assumed the field flux Φ was *fixed* — true for a permanent-magnet motor or a 'separately-excited' one whose field coil has its own constant supply. But classic DC machines build the field from an electromagnet, and *how you wire that field winding* relative to the armature completely rewrites the motor's personality. There are three families, and they behave like three different animals.
SEPARATELY-EXCITED / PM field independent of armature current
Phi = constant => straight, predictable speed-torque line
T = k*I, w = V/k - (R/k^2)T (the equations we used all along)
Use: precise speed control, servos, anything with a drive.
SHUNT (field // armature) field sees the full supply voltage V
Phi ~ constant (V fixed) => behaves much like separately-excited:
nearly flat line, 'stiff' speed. Speed barely droops with load.
Use: machine tools, fans, conveyors -> wants steady speed.
SERIES (field in series w/ armature) field current = armature current!
Phi grows with I => T = k*Phi*I ~ I^2 (torque rockets at low speed)
w = V/(k*Phi) - ... -> as load (I) drops, Phi drops, speed SOARS.
*** A series motor on no load can RUN AWAY and destroy itself. ***
Use: starter motors, traction (trains, cranes, old EVs) -> huge
starting torque, self-tapering as it gets up to speed.
T | series (T ~ I^2, steep, enormous at standstill)
| \
| \___
| shunt \______ (nearly flat, steady speed)
+------------------- speedThe series motor's quirk is worth savouring because it's why it ruled railways for a century. Its field current *is* its armature current, so flux climbs with load and torque grows roughly as I² — colossal twist exactly when a train needs to heave away from a platform. But run the same equation backwards: remove the load, current falls, flux collapses, and ω = V/(kΦ) heads for the sky. A truly unloaded series motor will accelerate until it flies apart, which is why you *never* belt-drive one (a snapped belt unloads it) and why tram drivers respected them. The induction motor and electronic drives of later rungs eventually inherited these jobs — but they were chasing the speed–torque curves the humble DC motor drew first.