Is the torque dependent on current and voltage?
Support » FAQ » Stepper motors
The torque is directly proportional to the current (P = I²/R) or generated magnetic flux provided that the magnetization is within the linear section of the magnetization characteristic curve, i.e. not yet in saturation. This means that if the current is doubled, the power is quadrupled. (Example of ST4118L1206-1W = 0.6 A² * 3.1 = 1.11 W at nominal current = 1.2 A² * 3.1 = 4.44 W ). As the voltage has to “drive" the current through the winding (P = U/R) in order to generate a current flow and the electrical power is also P = U * I or P = U * (U/R) = U²/R, the power or the torque (P = Md * n = P/n(constant) = Md) also increases in proportion to the voltage. This means that if the voltage is doubled, the power is quadrupled. (Example of ST4118L1206-1W = 1.855 V²/3.1 = 1.11 W at a nominal voltage = 3.71 V²/3.1 = 4.44 W).
At a constant voltage this applies to control as long as the current can increase with the voltage and is not reduced by the impedance (square root of R² + X²). (Here, however, a series resistor must always precede the current limiter at voltages higher than the motor voltage.) With constant current control, the current can be kept constant up to its torque inflection point and we have only a proportional increase of the power in this area whereas the torque remains constant. Only far above the torque inflection point does the torque greatly decrease, specifically at 24 V, (nominal current decreases due to the increasing counter EMF, the greater are impedance proportional asymptotically to the speed) and only here does the proportional increase of the power or torque from the voltage then start. (Example ST5918M3008–P: at approx. 500 rpm the torque is approx. 0.9 Nm at both 24 V and at 48 V whereas at 1000 rpm and 24 V the power = 0.65 Nm * 1000 rpm * 3.14/30 = 68 W, and the power at 2000 rpm and 48 V = 0.65 Nm * 2000 rpm * 3.14/30 = 136 W. This means that the power can be accordingly increased with proportionally higher voltage. The problem is only that the eddy current losses, and hence also the power dissipation, increases quadratically with the speed and, due to the temperature increase, sets limits on the voltage increase.
