2. Characteristics and equivalent model
2.1 Constant model
- Expression of the e.m.f
The armature being in rotation, the conductors cut the inductor magnetic flux and carry an alternating induced voltage. The collector rectifies this voltage; the number of slots being large, the e.m.f. between the brushes is almost continuous.
It is given by:
E = N.n.Ф or E = K.Ф.Ω
with: K = N/2π
E : e.m.f (V) ; N : number of active armature conductors ; Ф : flux under inductor pole (Wb) ; n and Ω : rotor speed (n in rev/s, and Ω in rad/s).
It can be noted that:
- The collector is a rotating voltage rectifier.
- If the flux is constant (most common case), E is directly proportional to the speed Ω.
- The current in the armature causes a magnetic field which modifies the e.m.f. This is the armature magnetic reaction, which is attenuated by arranging additional windings on the rotor. This phenomenon is generally neglected.
- Armature model
The armature model is given in Figure 13. It is governed by the equation:
U = E - R.I
E: e.m.f (V) ; U: armature voltage (V) ; I: armature current (A) ; R: armature resistance (Ω) which takes into account the winding, the collector and the brushes.

Figure 13: Armature model
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Moment of the electromagnetic torque
Electromagnetic power is converted into mechanical power.
Pem = E.I = Tem.Ω
With: E = K.Ф.Ω. In this case, we find:
Tem = K.Ф.I
Tem: moment of the electromagnetic torque (N.m) ; Ф: flux under pole inductor (Wb) ; I: armature current (A).
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Expression of the speed
Using the Ohm's law and the expression for the e.m.f. give:

Ω: armature rotation speed (rad/s) ; U: armature voltage (V) ; I: armature current (A) ; R: armature resistance (Ω) ; Ф: flux under pole inductor (Wb).
- No-load characteristic
It is given for a generator with separate excitation, whatever the excitation mode of the machine. This is the magnetization curve of the magnetic circuit. The operating point P is located around the saturation zone in Figure 14.

Figure 14: No-load characteristic
Due to hysteresis, the curve does not pass through the origin. There is a residual e.m.f. ER (as well as a narrow hysteresis cycle, not shown in Figure 14).
2.2 Dynamic model
- Electrical and mechanical equations of the dynamic model
The equations of the dynamic model represented by Figure 15 are given by:

With: J: Moment of inertia; f: coefficient of viscous friction.

Figure 15: Equivalent armature diagram
- Block diagram of the dynamic model
The load and the supply voltage are the main cause of the speed variation, the motor is represented by a controlled system (Figure 16). Using Laplace's formulation, we have:

The block diagram of the dynamic model of the direct current machine is given in Figure 16.

Figure 16 : Bloc diagram of the dynamic model
3. Separately excited motor
- Diagram
This excitation mode requires two separate power sources (Figure 17). The direction of rotation is changed by reversing the terminals of the armature or inductor.

Figure 17: separately excited direct current machines
- Starting conditions
1. The inductor is supplied before the armature by adjusting the excitation current Ie to its rated value.
2. It is necessary to limit the armature current Ist at start-up (Ist < 2.IN in general) by starting at reduced voltage, using a buck DC-DC converter or a controlled rectifier.
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Runaway condition
If the excitation is canceled while the armature is still supplied, the motor races and can destroy the armature. Consequently :
- We must never cut the excitation circuit.
- To stop the motor, we must cut the armature before the inductor (by reducing the armature voltage until the motor stops completely).
4. DC series motor (Series wound DC motor)
- Diagram
The inductor is in series with the armature: a single power source is sufficient. The direction of rotation is changed by reversing the connections of the armature and the inductor (figure 18).
The armature voltage becomes:
U = E + RT.I
with: RT = r + R
R : armature resistance (Ω) ; r : inductor résistance (Ω).

Figure 18: DC series motor
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Starting conditions and runaway conditions
1. The starting current must be limited as in separate excitation (ID < 2.IN in general).
2. We must never start without load under rated voltage, otherwise the motor will race and the armature may be destroyed.
5. Efficiency
- Power tree
Powers and losses of the direct current machine are given in figure 19.

Figure 19: Power tree
The absorbed power is given by:
PA = U.I + Ue.Ie
The copper losses in the armature are:
pJR = R.I²
The copper losses in the inductor are:
pJe = r.Ie² = Ue²/r = Ue.Ie
The useful power:
PU = TU.Ω
with: U: armature voltage (V) ; Ue: inductor voltage (V) ; I: armatre current ; Ie :inductor current (A) ; R: armature resistance (Ω) ; r : inductor resistance (Ω) ; TU : useful torque on the shaft (N.m) ; Ω : rotor speed (rad/s).
It should be noted that:
- At constant speed, mechanical losses (pm) and iron losses (pF) are constant.
- The losses are grouped under the name of collective losses, i.e.: pC = pF + pm, which, to a first approximation, are proportional to the speed.
- Efficiency
The motor efficiency is given by:
