For a dc "brush" motor or a permanent-magnet synchronous (PMS), or "brushless dc," motor, under what conditions will maximum mechanical output power be achieved? The answer is not when the motor is driven at maximum current, though that will cause the most torque and power loss in the windings. To derive the answer, let's start with the basic torque-speed equation:

where T is torque, R is the winding resistance, V_s is the motor supply voltage, and w  = w _me is the mechanical speed (in radians/second). L is the torque or speed constant:

L  = vw /w  = T/i

where i is motor current and vw is the motor induced voltage, or "speed" voltage. The torque-speed equation is plotted below.

The maximum motor speed occurs when the induced voltage equals the supply voltage applied to the motor terminals, or when vw  = V_s. Then no voltage difference occurs across the winding resistance, R, no winding current results, producing no torque. (See the L equations above.) Consequently, at maximum speed, the motor has no torque load, and the mechanical no-load speed is,

w ₀ = V_s/L

Similarly, the maximum torque is produced at zero speed, or stall. The stall torque,

T₀ = L × (V_s/R)

where the stall current - the maximum motor current - is the current limited only by the winding resistance, and is V_s/R.

Mechanical power is torque times speed, or

P_me = T× w

Substituting for T,

This is the equation of an inverted parabola which can be plotted and the value of w found corresponding to its maximum. Alternatively, the above equation can be differentiated with respect to w and solved for w . In either case, the torque and speed values at which maximum power occurs are:

and

In other words, it is at the center of the range of both torque and speed that maximum mechanical power is delivered. If maximum output power is what you want, this is the desired operating point for your motor. The value of power at the maximum power point (assuming no losses) is:

Although the (w _max, T_max) point delivers maximum power, it is not the point of maximum efficiency. We can derive an equation for the efficiency using our simple (static) motor model, as shown below.

The mechanical power is the power delivered into the induced voltage source (right) by the motor supply (left). We can calculate this power from the above electrical model of the motor using Watt's law;

P_me = vw × i = (L × w )× i

The supplied electrical power is

P_el = V_s× i

Maximum mechanical power occurs when the loss in the winding resistance equals the mechanical (output) power, or (by the maximum power transfer theorem) whenever:

i^2× R = L × w × i

Divide through by i and the voltage drop across R equals the induced voltage. Consequently, the induced voltage is

vw  = L × w  = i× R = [(V_s – L × w )/R]× R

Solving,

vw  = V_s/2

and the mechanical speed,

w  = w ₀/2

is the speed at maximum power.

The efficiency at maximum power is

A 50 % efficiency is better than 0 % at stall or at the no-load (zero-torque) speed, but is usually much less than that at slightly less than the no-load speed.

In passing, the expression for maximum mechanical power can alternatively be derived from the electrical motor model equation for P_me by substituting V_s/2 for L × w :

which agrees with the previous derivation. With 50 % efficiency, the electrical power loss in the windings (in R) is the same as the output power. The derivation of efficiency did not include magnetic or mechanical losses, which in practice reduces the efficiency to below 50 %.