AMME1705 · Introduction to Electromechanical Systems
DC Motors: Back-EMF, Torque & Speed
This chapter of the University of Sydney AMME1705 Introduction to Electromechanical Systems guide turns a spinning brushed DC motor into the small set of linear equations the unit examines: torque from current (τ = Kt Ia), the speed-dependent back-EMF that pushes against the supply (V_EMF = Kb Ω), and the armature loop balance Va = Ia Ra + V_EMF. From these fall the stall (startup) torque, the no-load speed, and the straight torque-speed line you read to find a motor's operating point.
It also covers how a motor splits electrical power into heat and mechanical work, why open-loop voltage control drifts, and how an encoder measures shaft speed to close the loop — the electromechanical core behind the 33% final exam's motor questions.
What this chapter covers
- 01Torque follows current: τ = Kt·Ia, with correct units (mN·m from mN·m/A)
- 02Back-EMF V_EMF = Kb·Ω and why it opposes the supply voltage
- 03The speed constant Kv (rpm/V) and how Kt and Kv are one constant in SI
- 04The armature loop balance Va = Ia·Ra + V_EMF (Kirchhoff round the motor)
- 05Stall torque and stall current at standstill (Ω = 0, V_EMF = 0)
- 06No-load speed Ω_nl = Kv·Va when the motor draws almost no current
- 07Reading the torque-speed line and how raising Va shifts it up
- 08Power split P_elec = Ia²Ra + τΩ, and rpm-to-rad/s conversion
- 09Encoders: pulses proportional to speed for closed-loop speed control
- 10Exam traps: Ra vs Ra+Rs, rpm vs rad/s, and the Kt-Kv mix-up
Worked example: no-load speed, stall torque and an operating point
- +1(a) No-load speed. With no load the current is almost zero, so the resistive drop vanishes and the back-EMF rises to nearly the full supply: V_EMF ≈ Va = 24 V. Then Ω_nl = Kv·Va = 239 rpm/V × 24 V = 5736 rpm.
- +1(b) Stall current. At standstill Ω = 0, so V_EMF = Kb·0 = 0 and the whole supply sits across Ra. Ia,stall = Va/Ra = 24 V / 2.0 ohm = 12 A (the maximum current).
- +1(b) Stall torque. τ_stall = Kt·Ia,stall = 40 mN.m/A × 12 A = 480 mN.m (= 0.480 N.m) — torque is largest at startup.
- +1(c) Back-EMF at Ia = 3.0 A. From the loop Va = Ia·Ra + V_EMF, so V_EMF = Va − Ia·Ra = 24 − 3.0×2.0 = 24 − 6.0 = 18.0 V.
- +1(c) Torque and speed. τ = Kt·Ia = 40 mN.m/A × 3.0 A = 120 mN.m; Ω = Kv·V_EMF = 239 rpm/V × 18.0 V = 4302 rpm.
Key terms
- Torque constant Kt
- The proportionality between armature current and shaft torque: τ = Kt·Ia. Its SI unit is N·m per amp, though AMME1705 usually quotes it in mN·m/A. Double the current, double the torque.
- Back-EMF (V_EMF)
- The voltage a spinning motor generates against its own supply, proportional to speed: V_EMF = Kb·Ω. It has units of volts and rises with speed, giving the motor its natural speed regulation. At standstill it is zero.
- Speed constant Kv
- Links the back-EMF voltage to speed via Ω = Kv·V_EMF, usually in rpm/V. It is the reciprocal of the back-EMF constant Kb. In consistent SI units Kt [N·m/A] equals 1/Kv [rad/s per V], so Kt [mN·m/A] × Kv [rpm/V] is about 9549.
- Armature loop balance
- Kirchhoff's voltage law round the single armature loop: Va = Ia·Ra + V_EMF. The applied voltage equals the resistive drop across the armature resistance Ra plus the back-EMF. Rearranged, Ia = (Va − V_EMF)/Ra.
- Stall (startup) torque
- The torque at zero speed. With Ω = 0 the back-EMF is zero, so the current is at its maximum Ia = Va/Ra and the torque is largest: τ_stall = Kt·Va/Ra. If a series resistor Rs is present, use Ra + Rs.
- No-load speed
- The top speed a motor reaches when nothing loads the shaft. The current falls to almost zero, the back-EMF rises to nearly the supply, and Ω_nl = Kv·Va.
- Torque-speed line
- For a fixed supply voltage, a straight line from the no-load point (zero torque, top speed) to the stall point (zero speed, maximum torque). More torque always means less speed. Raising Va shifts the whole line up.
- Encoder
- A shaft sensor (optical or Hall) that emits pulses at a rate proportional to speed. Counting the pulses measures the true shaft speed directly, letting a controller close the loop instead of relying on the supply voltage alone.
DC Motors: Back-EMF, Torque & Speed FAQ
Why does a DC motor draw a huge current at start-up?
At the instant of switch-on the shaft is not yet turning, so there is no back-EMF (V_EMF = Kb·0 = 0). The full supply voltage sits across the small armature resistance Ra, giving the maximum current Ia = Va/Ra. That is also why the torque is largest at startup: τ_stall = Kt·Va/Ra. As the motor speeds up the back-EMF grows, the net voltage across Ra falls, and the current drops back.
What is the difference between the torque constant Kt and the speed constant Kv?
Kt links current to torque (τ = Kt·Ia, in N·m/A or mN·m/A); Kv links the back-EMF voltage to speed (Ω = Kv·V_EMF, in rpm/V). They describe the same electromechanical coupling from two sides, and in consistent SI units they are reciprocals: Kt [N·m/A] = 1/Kv [rad/s per V]. A quick check when the exam quotes mixed units is that Kt [mN·m/A] × Kv [rpm/V] is about 9549.
Can AI help me with DC motors and the torque-speed relationship in AMME1705?
Yes, as a study aid. AskSia's Sia tutor can explain the armature loop, back-EMF and the torque-speed line step by step, walk you through why stall means maximum current, and check your reasoning on a practice motor problem so you understand each move. It is built to help you learn the method and build confidence, not to sit your exam or hand you final answers, and it never promises a particular grade — the working, and the marks, are yours.
Studying with AI? Sia — free AI electrical engineering tutor works through AMME1705 step by step.
Exam move
Make the armature loop your reflex for every motor question: write Va = Ia·Ra + V_EMF first, then branch. If the shaft is stalled, set Ω = 0 so V_EMF = 0, giving Ia = Va/Ra and τ_stall = Kt·Va/Ra (maximum current and torque). If the motor is unloaded, set Ia ≈ 0 so V_EMF ≈ Va, giving the no-load speed Ω_nl = Kv·Va (maximum speed). For any operating point in between, go torque → current (Ia = τ/Kt) → back-EMF (V_EMF = Va − Ia·Ra) → speed (Ω = Kv·V_EMF). Watch three things the marks hang on: keep torque in mN·m and only convert speed to rad/s (× 2π/60) when you compute mechanical power τ·Ω; use Ra + Rs if a series resistor sits before the motor; and round only the final answer to the precision the question states. Across the paper the pace is about 1.5 minutes per mark (roughly 80 marks in the 2 hours of writing time), so a six-mark motor question deserves about nine minutes — confirm the exact exam length, mark total and permitted materials on your current Canvas assessment page.
