Equation Linking Speed, Back Emf and Flux in Dc Motor

The speed equation is very essential to analyze the speed of rotation of the DC motor. As the electrical input given to the DC motor produces mechanical output. The output depends on several parameters like input voltage, field current, armature resistance, etc.

Let us find the parameters involved in determining the speed of a DC motor.

We know, the voltage equation of a DC motor, it is given by,

$E_b = V - I_aR_a$

Similarly, the back emf of a DC motor is given by,

$E_b = \frac{\phi Z N P}{60 A}$

Equating the above two equations,

$\frac{\phi Z N P}{60 A} = V - I_aR_a$

$N = \frac{V-I_aR_a}{\phi} * \frac{60A}{ZP} \hspace{0.1cm} rpm --->(1)$

$N = \frac{E_b}{\phi} * \frac{60A}{ZP} \hspace{0.1cm} rpm$

$\boxed { N = K \hspace{0.1cm} \frac{E_b}{\phi} \hspace{0.1cm} rpm } --->(2)$

$where \hspace{0.2cm} K \hspace{0.1cm} is \hspace{0.1cm} the \hspace{0.1cm} proportionality \hspace{0.1cm} constant = \frac{60A}{ZP}$

The above equation implies that speed of a DC motor is directly proportional to back emf and is inversely proportional to the flux.

Speed relation for series and shunt motor

Let the initial flux for the DC motor be φ ₁. The armature current(I_a1) in the series motor is proportional to the flux produced, Hence φ ₁ α I_a1. The speed thus produced be N₁ rpm.

So the speed equation(2) becomes,

$N_1 \hspace{0.1cm} \propto \hspace{0.1cm} \frac{E_{b1}}{\phi_1} --->(3)$

Now if the flux is reduced to φ ₂. The armature current(I_a2) and speed(N₂) vary corresponding to the magnetic flux in the field winding.

The speed equation changes as follows,

$N_2 \hspace{0.1cm} \propto \hspace{0.1cm} \frac{E_{b2}}{\phi_2} --->(4)$

Dividing both equations (3) and (4), we get,

$\boxed { \frac{N_2}{N_1} = \frac{E_{b2}}{ E_{b1}} * \frac{\phi_1}{\phi_2}}$

For shunt motor, the flux is practically constant, i.e., φ ₁ = φ ₂

Therefore, the speed relation becomes,

$\boxed { \frac{N_2}{N_1} = \frac{E_{b2}}{ E_{b1}} }$

Torque and speed equation of DC motor

We have derived the speed equation as,

$N \hspace{0.1cm} \propto \hspace{0.1cm} \frac{E_{b}}{\phi}$

The torque equation of a DC motor has been derived as,

$T \hspace{0.1cm} \propto \hspace{0.1cm} \phi I_a$

From the above equations, it can be observed that, increase in flux would increse the torque but decreases the speed of the motor. But it is not possible, because torque is the rotation force, which tend to rotate the motor. Hence increase in torque always increase the speed of the motor. This inconsistency can be reconciled in the following way.

Let us assume that the motor is operated at a steady speed. Let the flux(φ) of the motor is decreased by decreasing the field current.

The decrease in flux will drop the back emf(E_b = Nφ/K) instantly, causing the armature current to increase because the voltage equation of DC motor is given by I_a = (V – E_b)/R_a. Thus, a small reduction in flux will cause the armature current to increase heavily.

Hence, in the torque equation, T α φI_a, the decrease in flux is counterbalanced by a huge increase in armature current. This will increase the net torque T, which increases the speed of the motor.

Speed Regulation of DC motor

When the load in the motor is reduced from its rated value to zero, the speed of rotation of the motor changes. This change in speed is known as speed regulation.

The speed regulation in per unit is given by the equation,

$Speed \hspace{0.1cm} \hspace{0.1cm} Regulation \hspace{0.1cm} of \hspace{0.1cm} DC \hspace{0.1cm} motor \hspace{0.1cm} = \frac{No \hspace{0.1cm} Load \hspace{0.1cm} Speed - Full \hspace{0.1cm} Load \hspace{0.1cm} Speed}{Full \hspace{0.1cm} Load \hspace{0.1cm} speed}$

Percentage speed Regulation is given by,

Solved Problem

Let us look at a problem.

A 220 V shunt motor runs at 1500 rpm at no load and takes 6A. Calculate the speed of the motor when it is loaded and takes 40 A. The armature resistance and field resistances are 0.5 Ω and 220 Ω respectively. Assume the flux to be constant.

Given Data: V = 220 V, N₁ = 1500 rpm, I₁ = 6 A, I₂ = 40 A, R_a = 0.5 Ω, R_sh = 220 Ω, φ₁ = φ₂

To find: N₂

Solution :

We know the formula for speed relation,

$\frac{N_2}{N_1} = \frac{E_{b2}}{ E_{b1}} --->(5)$