Electrical – Maximum value of back emf in dc motor

You can think of the motor as behaving like a resistor connected to a generator (the latter corresponding to the back EMF). It produces a torque that is proportional to the current through the resistor.

The higher the difference between the back EMF and the input voltage, the higher the current. The higher the current, the higher the torque.

If the motor is stalled, the entire input voltage appears across the resistor and a lot of torque is produced. Since it only takes a bit of torque to overcome losses such as bearing friction and windage and magnetic losses when there is no load, the motor spins up until the back EMF is almost equal to the input voltage and the torque has dropped until the motor is no longer accelerating significantly. The current resulting from the remaining difference (multiplied by the input voltage) gives you the power consumption with no load. Efficiency is zero, of course, with no load since it is producing zero output power at the shaft.

The losses in the copper windings are the current squared times the resistance (which is the winding resistance) aka \$I^2R\$ losses.

If the windings had zero resistance the back EMF would always be equal to the input voltage, (even under load) since any difference would allow infinite current to flow, generating infinite torque.

50% back-emf occurs at maximum power output, at which point the motor is slightly less than 50% efficient.

Why? Because at 50% rpm half the voltage going into the motor is being dropped across the resistance of the armature windings, so half the power is wasted. This becomes obvious when you look at the equivalent circuit of a DC motor. Here is an example:-

^{simulate this circuit – Schematic created using CircuitLab}

Back-emf is caused by the armature spinning and generating a voltage which opposes the supply voltage. The rest of the supply voltage is dropped across Rm.

In this case back-emf is 3.5V, so Rm must have 7V-3.5V = 3.5V across it. 3.5V / 0.21Ω = 16.7A. 3.5V * 16.7A = 58.3W of 'copper' loss. 0.75A is used up in 'iron' losses, leaving 15.95A to produce output torque and power. 3.5V * 15.95A = 55.83W.

Efficiency = output power / input power. 55.83W / (7V * 16.7A) = 47.9% efficiency.

Maximum efficiency is achieved when copper loss equals iron loss. This usually occurs at around 80-90% of no-load rpm. If you know Io and Rm then calculating this point is simple.

Iron loss = Vm-(Io*Rm) * Io. In this case that works out to 5.13W. At peak efficiency copper loss is the same, so Im² * Rm = 5.13W. Solving for Im we get 4.94A. Now calculate the voltage drop across Rm at this current, 4.94A * 0.21Ω = 1.04V. That leaves 5.96V for back-emf. Now calculate the no-load back-emf. Voltage drop across Rm is 0.75A * 0.21&ohm = 0.158V. 7V - 0.158V = 6.84V. 5.96V/6.84V = 87% of no-load rpm.

Here's an actual dyno test of the motor simulated above. Peak efficiency occurred at 5A (very close to the 4.94A we calculated) at 7360rpm. This is 86% of the no-load speed, 8550rpm.

You wouldn't want to run this motor at maximum power for long because it only weighs 50g. With over 50W of heat to get rid of it would quickly burn up. DC motors are generally operated somewhere between peak efficiency and maximum power output, as this provides the best compromise between power, efficiency, and size.