Brushed and Brushless DC motors have two traits that make them easy to control for several applications. The first is the relationship between voltage and speed, and the second is the relationship between current and torque. Both relationships are separate from each other but below you can see how they are used together to provide a system to control the speed of the motor over a wide torque range.

Datasheets for DC motors show how the motor performs at its rated voltage. If the voltage is increased the speed will increase proportionally. Also if the voltage is decreased the speed will decrease. If we have a motor that runs at 2500 RPM at 24 VDC and the voltage was decreased to 12 VDC the motor would run at 1250 RPM. The speed/torque curve would actually be parallel to the datasheet published by the manufacturer. The relationship between the speed and voltage is often termed K_e. K_e is expressed in the units of V/kRPM or V/rad/sec.

In DC motors, output torque is also directly related to the current supplied to the motor. This is designated by the term K_t. K_t is expressed in the form of torque unit per A. This could be in.lbs., N-m, or any other unit used to measure torque. The datasheet shows the stall torque of the motor at (using the example above) 24 VDC. If the motor above was only supplied 12 VDC then the stall torque would be reduced by 50% (24 VDC / 12 VDC = .5). This is because at stall the motor acts like a resistor. If the resistance is fixed and the  voltage is reduced by 50% then only 50% of the current can be consumed per Ohm’s Law.

Is it possible to get the slower speed but still manage to achieve the torque required for an application? Simple motor controls have been made to accomplish this. We now know that speed is proportional to voltage and torque is proportional to current. Let’s assume the motor above which is rated at 2500 RPM can output 10 in.lbs. of torque at continuous duty. However your application needs to be run at 1250 RPM with varying torque and a peak of 10 in.lbs. is required. By just reducing the input voltage you would not be able to achieve the required peak torque. However, by implementing a current compensation loop you can.

(V_r – V_n1) / (I_r – I_n1)

• If V_r = Rated Voltage
• V_n1 = No Load Voltage (Voltage required to get desired speed)
• I_r = Rated Current (or current at peak load)
• I_n1 = No Load Current

Then the result would be volts/amps. This ratio is important because it shows how you can achieve constant speed over a varying torque range. As the current increases in the system the voltage needs to be increased at the rate shown above to maintain constant speed. As an additional note, as the voltage increase, once the variable voltage reaches the rated voltage the motor performance will follow the speed/torque curve on the data sheet.