DC motor: why is torque NOT DEPENDENT on the angle of rotation?

Question

I am reading a book on control systems where they take an example of an armature controlled DC motor. Because the goal is to describe constructing a block diagram from the system equations, the equations themselves have not been explained in more than a couple of lines. The circuit diagram provided alongside is that of a separately excited DC motor, almost identical to the one here:

Just for context I am providing some of the lines from the book:

In servo applications, the dc motors are generally used in the linear range of the magnetization curve. Therefore, the air gap flux is proportional to the field current. $$\phi = K_f i_f$$ where Kf is a constant. The torque Tm developed by the motor is proportional to the product of the armature current and the air gap flux: $$T_m = K_1K_fi_fi_a$$ In an armature controlled DC motor, the field current is kept constant $$T_m = K_Ti_a$$ ...

The torque equation is: $$J \frac{d^2 \theta}{dt^2} + f_0 \frac{d\theta}{dt} = T_M = K_T i_a$$

I decided to try to explore this with some of my own reasoning: I have assumed that the flux is being generated inside a solenoid (stator windings) so that $$B = \mu ni_f$$ which is a linear characteristic as mentioned in the book.

Now, the torque produced on the armature coil should be that of a magnet with a magnetic moment M placed in a field B:

$$\vec{T} = \vec{M} \times \vec{B}$$ $$\vec{T} = Ni_a \vec{A} \times \mu n i_f \vec{b} $$ $$\vec{T} = nNA i_a i_f (\vec{a} \times \vec{b}) $$ $$T = K i_a i_f \sin(\theta)$$

where theta is the angle between the normal to the coil (a-vector) and the direction of the field (b-vector), same as the angle through which the motor is turned at any point, right? So the 2nd order differential should actually become:

$$J \frac{d^2 \theta}{dt^2} + f_0 \frac{d\theta}{dt} = T_M = K_T i_a sin(\theta)$$ which would change the dynamics of the entire system?

Answer 1

By the time you're doing that sort of calculation with a DC motor, it's been commutated. It's either a brushed DC motor that is commutated mechanically with brushes and a commutator, or it's a brushless motor that is commutated electronically.

This commutation switches the current to the coils so that the torque on the armature is much more even. For an \$n\$-pole motor the torque has a position-based ripple with a spatial frequency of \$2n\$ for an odd number of poles, and \$n\$ for an even number of poles. In general, this ripple is greater the fewer poles the motor has (which is why, if you're in the habit of disassembling motors, you see that expensive "servo" motors have lots of poles). In brushless "AC" motors this ripple is diminished by a combination of the magnetic design of the motor and by controlling the current to each individual coil.

With this for our model, the armature torque would be more like $$T_M = K_T\,i_a\,f(\theta),$$ where \$f(\theta)\$ depends on the motor construction and generally has a period of \$\frac{\pi}{n}\$ for \$n \in \mathbb Z_{\mathrm{odd}}\$.

Depending on the motor and (if it's brushless) the driver, your motor may also suffer from cogging torque, due to the magnetic path having variable reluctance, which gives the motor preferred directions where it wants to "stick". Servo motors are designed to minimize this, but it's something to look for and pay attention to.

Answer 2

In DC motor angle between area vector of coil and magnetic field is always perpendicular to each other hence $$\theta=90$$ and hence torque is independent of angle.

This is due to curved shape of pole faces(electromagnet) , constant air gap and coils of rotor lie in a ferromagnetic core form a magnetic circuit and flux follows the path to minimize the reluctance seen by the electromagnet hence it take the shortest possible path through air (as reluctance of air is much higher than that of core) And shortest path would be perpendicular to the rotor surface