DC Machines: Generator EMF, Back EMF, Torque and Speed Calculations
A DC machine can work as a generator or as a motor without any change in construction. Large DC machines have largely been replaced by induction motors with variable frequency drives, but the DC machine remains the clearest way to understand electromechanical energy conversion, and its equations carry straight over to the small permanent-magnet DC motors used in robots, toys, wipers and cordless tools.
This article explains construction, the generator EMF equation, the motor back EMF and torque equations, the types of excitation, and works through numerical examples including starting resistance.
Construction in Brief
| Part | Function |
|---|---|
| Yoke | Outer frame that carries magnetic flux and supports the machine |
| Field poles and field winding | Produce the main magnetic flux Φ per pole |
| Armature core | Laminated slotted cylinder on the shaft that carries the armature conductors |
| Armature winding | Conductors in which EMF is induced; lap or wave connected |
| Commutator | Copper segments that convert alternating EMF in the coils into DC at the brushes |
| Brushes | Carbon blocks that connect the rotating commutator to the external circuit |
In a lap winding, the number of parallel paths A equals the number of poles P, suiting low-voltage, high-current machines. In a wave winding, A = 2 regardless of poles, suiting high-voltage, low-current machines.
Generator EMF Equation
Each conductor cuts the flux of all P poles in one revolution. Total flux cut per revolution is PΦ, and the time for one revolution is 60/N seconds. The EMF per conductor is therefore PΦN/60. With Z conductors divided into A parallel paths, Z/A conductors are in series in each path.
Eg = P Φ Z N / (60 A) volts P = number of poles Φ = flux per pole (Wb) Z = total conductors N = speed (rpm) A = parallel paths (lap: A = P, wave: A = 2) Terminal voltage (generator): V = Eg − Ia Ra
Worked example
A 4-pole DC generator has 720 armature conductors, flux per pole 0.03 Wb, and runs at 1000 rpm. Find the generated EMF for lap and for wave winding.
- Lap winding, A = 4: Eg = 4 × 0.03 × 720 × 1000 / (60 × 4) = 86400/240 = 360 V.
- Wave winding, A = 2: Eg = 86400/120 = 720 V.
- The same machine gives double the voltage in wave connection, but it has half as many parallel paths, so for the same conductor current rating it delivers half the total current. Power capacity is the same.
Motor Operation: Back EMF, Speed and Torque
When the same machine runs as a motor, its rotating armature still generates an EMF, called back EMF Eb, which opposes the applied voltage. The armature current is set by the difference between supply voltage and back EMF. This self-regulating behaviour is why a DC motor draws more current when loaded: speed drops slightly, Eb falls, and Ia rises to produce more torque.
V = Eb + Ia Ra Eb = P Φ Z N / (60 A) => N ∝ Eb / Φ Gross torque: T = 0.159 × (P Z / A) × Φ × Ia (N·m) T = 9.55 × Eb × Ia / N (N·m, N in rpm) Mechanical power developed = Eb × Ia
Worked example: shunt motor under load
A 220 V DC shunt motor has armature resistance Ra = 0.5 Ω. On light load it takes Ia = 20 A at 1000 rpm. When load increases, Ia rises to 40 A. Assume constant flux and find the new speed and both torques.
- Eb1 = 220 − 20 × 0.5 = 210 V.
- Eb2 = 220 − 40 × 0.5 = 200 V.
- N2 = N1 × Eb2/Eb1 = 1000 × 200/210 = 952 rpm.
- T1 = 9.55 × 210 × 20/1000 = 40.1 N·m.
- T2 = 9.55 × 200 × 40/952 = 80.3 N·m.
Doubling armature current doubles the torque at constant flux, while speed falls by under 5%. This nearly constant speed is the defining feature of the shunt motor.
Types of Excitation and Why a Starter Is Needed
| Type | Field connection | Behaviour | Typical use |
|---|---|---|---|
| Separately excited | Field fed from separate supply | Precise speed control | Drives, test rigs |
| Shunt | Field in parallel with armature | Nearly constant speed | Lathes, fans, conveyors |
| Series | Field in series with armature | Very high starting torque, speed rises sharply at light load | Traction, cranes, hoists, universal motors |
| Compound | Both shunt and series fields | Between shunt and series | Presses, shears, elevators |
| Permanent magnet | Magnets instead of field winding | Like shunt, compact | Small motors, robots, automotive |
At standstill, N = 0, so Eb = 0, and the only limit on current is Ra. For the motor above, starting current would be 220/0.5 = 440 A, twenty-two times the light-load value. This damages the commutator and brushes and causes a large supply dip. A starter inserts external resistance and removes it in steps as speed and back EMF build up.
Limit starting current to 40 A: R_total = V / I_start = 220 / 40 = 5.5 Ω R_external = 5.5 − 0.5 = 5.0 Ω
Speed Control and Common Mistakes
From N ∝ (V − Ia Ra)/Φ, speed can be controlled in three ways: varying armature voltage (modern method, using a controlled rectifier or chopper), adding armature resistance (wasteful), or weakening the field to go above base speed.
- Opening the field circuit of a running shunt motor causes Φ to collapse and speed to rise dangerously; starters include a no-volt coil to prevent this.
- Using P instead of A in the EMF equation for a wave winding.
- Forgetting brush drop (often taken as about 1 V per brush in textbook problems) when the problem specifies it.
- Confusing gross torque with shaft torque; shaft torque is lower by friction, windage and iron losses.
Summary
The generated or back EMF of a DC machine is PΦZN/(60A). In a motor, V = Eb + IaRa, speed is proportional to Eb/Φ, and torque is proportional to ΦIa. Because back EMF is zero at standstill, a starting resistance is always needed for anything beyond small motors.