Back emf and Torque Equations of a DC motor

The effect of the force in a magnetic field may be used to cause the rotation of a motor armature. A single loop of a conductor shown in the diagram above has its ends connected to a simple commutator; which comprises two copper segments insulated from each other. The commutator and loop are fixed to a central shaft which enables the whole assembly to be freely rotated. Two fixed carbon brushes bear on the surface of the commutator, enabling a supply to be connected to the loop.

In practice, the dc motor comprises an armature of many loops revolving between electromagnetic poles. The commutator has many segments to which the ends of the armature coils are connected, and the armature core is laminated to reduce eddy currents.

Back emf

When the armature revolves, its coils cut across the field flux, but we know that if a conductor cuts across lines of force, an e.m.f is induced in that conductor. This is the principle of the generator. So applying Fleming’s right-hand rule, we see that the induced e.m.f is opposing the supply voltage. This induced e.m.f is called the, ‘’back emf’’. If the back emf were of the same magnitude as the supply voltage, no current would flow and the motor would not work. As the current must flow in the armature to produce rotation, and as the armature has resistance, then there must be voltage drop in the armature circuit. This voltage drop is the product of the armature current (I_a)and the armature circuit resistance (R_a).

Armature voltage drop = I_a x R_a

It is this voltage drop that is the difference between the supply voltage and the back emf.

Therefore, E = V – (I_aR_a)

We know that, the induced emf is dependent on the flux density (B); the speed of cutting flux (V) and the length of the conductor (L)

Both L and a, for a given conductor will be constant and V is replaced by n (revs/second) as this represents angular or rotational speed:

Therefore E α nΦ

So, if the speed is changed from n₁ to n₂ and the flux from Φ₁ to Φ₂, then the emf will change from E₁ to E₂ :

Torque

We know that, work = Force x Distance

Therefore, turning work or torque = Force (F) x Radius (r)

We also know that:

Force = B x L x I

Therefore, Torque T = B x L x I_a x r

Once again, for a given machine, a, L, and r will all be constant.

Hence, T α Φ x I_a

Also, mechanical output power in watts is given by

P = 2ℼnT

Where P is the output power in watts, n is the speed in revs/second and T is the torque in newton metres.

If we multiply E = (V – I_aR_a) by I_a we get EI_a = VI_a – I_a²R_a

VI_a is the power supplied to the armature and I_a²R_a is the power loss in the armature; therefore EI_a must be the armature power output. Hence

EI_a = P

Therefore, EI_a = 2ℼnT

From the above, it is clear that torque is directly proportional to armature current and inversely proportional to speed i.e. if the mechanical load is lessened, the torque required is less, the armature current decreases and the motor speeds up.