Electronic – Apart from speed, do VSD’s also alter current

Does the back EMF limit the voltage the motor can accept and hence it's speed?

Hmm. I think you're a little confused. Back-emf limits the motor speed because it dictates how much voltage you need to achieve a given speed.

Would an ideal motor have a very high Kt and very low Kemf?

No. (The symbol \$K_e\$ is usually used for the back-emf constant, by the way.)

If you use SI units (Nm/A for \$K_T\$, V/(rad/s) for \$K_e\$), then \$K_T = K_e\$ for DC motors and permanent-magnet synchronous motors (aka "brushless DC"), and depending on the type of motor and how you define \$K_T\$ and \$K_e\$, the ratio of the two should be a fixed proportionality constant.

Proof of why this is true for DC motors:

At constant operating point (constant speed, voltage, current, torque):

• \$V_T = K_e \omega_m + IR\$
• \$T_m = K_T I\$

(\$V_T\$ = terminal voltage, \$\omega_m\$ = motor angular velocity, \$I\$ = motor current, \$R\$ = motor resistance, \$T_m\$ = torque, including frictional losses)

Electrical power in = \$V_T I = K_e \omega_m I + I^2 R\$

mechanical power out = \$T_m \omega_m = K_T I \omega_m \$

Losses = \$I^2 R\$

Conservation of energy means electrical power in = mechanical power out + losses

This is true if and only if \$K_e = K_T\$.

What happens if the back emf voltage reaches the input voltage, does the motor stop running?

No -- what happens is that the ability of the motor to produce torque decreases with speed. Back-emf voltage "uses up" voltage from the electrical power source; the remaining voltage available to the motor is what's left for IR drop in the motor and the inductance drop \$L \frac{dI}{dt}\$, and since torque is proportional to current I, the available torque decreases. The system reaches equilibrium at some point where the electromagnetic torque matches the motor's mechanical load. If you increase the mechanical load, current will increase to match that torque as speed slows down, making more IR drop voltage available.

Is a bigger Kt always better?

No. Rule of thumb with DC motor selection (also true for brushless DC motors to a large extent) -- pick a motor with a \$K_T\$ and back-emf constant such that the supply voltage you have available is well-matched with the back-emf at your maximum speed. You usually want back-emf voltage to be 80-95% of the supply voltage, but the exact number depends on the load torque and the IR drop in the motor at that operating point.

If you pick a \$K_T = K_e\$ too high, you'll run out of voltage and won't be able to achieve the speed you need. If you pick a \$K_T = K_e\$ too low, the current needed to achieve the torque you need will be higher than necessary.

Just like it's simplest to learn about a lossless inductor first, so let's start with more or less lossless motor. We'll take account of losses when we have to, but they are not essential for basic understanding.

A motor is also a generator. Spin it, and it generates volts on the armature. It doesn't matter whether it's spinning because it's a motor, or spinning because you're driving it as a generator, speed = armature_volts/k.

Pass a current through it and it generates a torque. Torque = armature_current.k

You can think of a motor as a mechanical transformer. Power in = power out. Volts x amps in = speed x torque out. When equating power, that pesky k has cancelled out. If you change the value of k, the torque constant, then the motor gets faster and delivers less torque, or vice versa, but the power balance is the same. If you run the same machine as a generator, then speed x torque in = volts x amps out.

What happens if you apply a voltage source to a motor at rest?

Two things happen, at different speeds, the first so quickly you may not notice, the second rather more slowly.

A motor armature has inductance, Larm, and resistance Rarm. At the moment of switch on, we apply V to the armature. The current starts to increase. It initially increases at such a rate that Larm x dI/dt generates a back EMF equal to the terminal voltage. The current flowing through Rarm generates a voltage IRarm which opposes V, so there is less voltage across the inductance to drive an increase of current, so the rate of current increase slows down. Eventually, the current has increased to settle at V/Rarm, with a time constant of Larm/Rarm, typically in a matter of mS.

With a 'good' motor with a low Rarm, this current will typically be very large. It's known as the 'starting' current, for obvious reasons. Small motors are rated to be started like this safely. Big motors cannot be started like this, and need some sort of soft start controller.

So far, the motor still hasn't moved, the mechanical inertia means it's not rotating yet, or has barely started. But now there is a big armature current flowing, which generates a torque, and the motor accelerates.

Once it is turning, at any speed, it generates a back EMF proportional to its speed. This back EMF reduces the effective terminal voltage available to drive current through Rarm. The armature current therefore falls (with a time constant of Larm/Rarm), and so generates less torque.

Eventually the motor reaches a balance, where it's at a speed where the generated back EMF balances off most of the input voltage, and the small difference in voltage that remains drives an armature current through Rarm, which generates enough torque in the motor to match the load torque, plus loss torques like friction and air resistance.