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This DC motor has a power range of 9.0W-300W, while its voltage range is from 6 to 20 volts. Does this mean that its internal resistance is somehow dependent on the load? Or does the user determine the actual power with using a resistor in the circuit?

RZ-735VA-9517 DC motor


Performance Table

MODEL   ---- VOLTAGE ----  -- NO LOAD ---  ---- AT MAXIMUM EFFICIENCY -----  ----- STALL ------
        OPERATING NOMINAL  SPEED  CURRENT  SPEED  CURRENT   TORQUE   OUTPUT    TORQUE   CURRENT
RZ-735VA  RANGE      V     r/min     A     r/min     A    mN·m  g·cm    W    mN·m  g·cm    A
9517     6 - 20     18     20400    2.8    17990   20.9    149  1523   281   1265  12895  156
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    \$\begingroup\$ The link is dead. Maybe include relevant data? \$\endgroup\$
    – Jeroen3
    Commented Aug 31, 2018 at 13:16
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    \$\begingroup\$ @Jeroen3: The link is not dead. Perhaps it is blocked at your location for some reason. \$\endgroup\$
    – Dave Tweed
    Commented Aug 31, 2018 at 13:29
  • \$\begingroup\$ @DaveTweed they are indeed using some cdn that isn't working properly. \$\endgroup\$
    – Jeroen3
    Commented Aug 31, 2018 at 13:38
  • \$\begingroup\$ That link's specs are fishy. The motor only weighs 280 grams but is allegedly most efficient drawing 21 amps?! And has a stall/starting current of 156 amps. All over wires that look to be #20 gauge, tops. \$\endgroup\$ Commented Aug 31, 2018 at 21:13

3 Answers 3

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Yes, of course the effective resistance of the motor depends on the load. This is true of all motors.

When a motor is lightly loaded, it generates a high level of back EMF, which opposes the flow of current. You can think of this as a high effective resistance, although the actual resistance of the coils doesn't change. What is actually changing is the net voltage (= source voltage - back EMF) across that resistance.

When a motor is heavily loaded, the back EMF is reduced, allowing more current to flow. This decreases the effective resistance by increasing the net voltage across the internal resistance of the motor.1

The wide range of power ratings for this motor indicate that it is both efficient (low power consumption at low loads) and robust (can handle the current associated with high loads).

The methods for controlling a motor depend on how you're using it. If you're primarily interested in controlling its speed, you regulate the voltage, allowing the current to vary (within limits) with the load. If you're primarily interested in controlling its torque, you regulate the current, and the voltage varies with load. A resistor is NOT a particularly useful way to accomplish either of these.


1With AC (e.g., induction) motors, the situation is a bit more complicated. The magnitude of the back EMF doesn't change as much as its phase relationship to the source voltage. This still has the effect of increasing the net internal voltage.

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  • \$\begingroup\$ We use a resistor with components like LEDs to limit the current maximum and so to protect them. Shouldn't a resistor be used with a relatively low power DC motor for this purpose? As was explained to me here, the current reaches it's maximum when the motor is stalled. \$\endgroup\$
    – apadana
    Commented Aug 31, 2018 at 14:16
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    \$\begingroup\$ An LED is a much more stable load than a motor. Its characteristics do not depend on external loads, so it's easy to match it to a resistor in order to get the desired result. Sure, a resistor will limit the current in a motor to a maximum value, but that severely limits the torque available at all speeds, which is generally undesirable. \$\endgroup\$
    – Dave Tweed
    Commented Aug 31, 2018 at 14:22
  • \$\begingroup\$ Is the back emf actually reduced? Or is it just propotionally smaller ie its the same but the power fed in is larger so the apparent resitance figure is smaller. \$\endgroup\$
    – joojaa
    Commented Aug 31, 2018 at 18:00
  • \$\begingroup\$ @joojaa: Yes, the back EMF really is reduced. It is a function of rotor speed in a DC motor (and also "slip" in an AC motor). The source voltage remains the same, and the power consumption goes up as a result of the increased current. \$\endgroup\$
    – Dave Tweed
    Commented Aug 31, 2018 at 18:39
  • \$\begingroup\$ Yes i know that its a function of rotor speed. But being loaded does not mean that the speed changed as such. \$\endgroup\$
    – joojaa
    Commented Aug 31, 2018 at 18:44
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Think of the motor as of a device that is meant to convert electrical energy into mechanical energy. If your mechanical load needs a lot of energy, the motor needs a lot of electrical energy to generate that. That's why it doesn't have a fixed power rating.

Does this mean that its internal resistance is somehow dependent on the load?

Yes. You can't really use a resistor model for a motor, because its resistance varies due to a lot of factors : mechanical load, input voltage, etc.

Or does the user determine the actual power with using a resistor in the circuit?

No. Using a resistor in any power electronics application is generally a bad idea. Resistors turn electrical energy into heat. That's a massive waste of energy unless you are making a heater. However, you can use a motor speed controller that basically varies the voltage given to the motor which in turn changes the power delivered to the mechanical load.

Now some extra info about the datasheet that you provided, just to make things clear:

  • In the performance table, you have a current of 2.8 A under no load. That's what the motor draws at the nominal voltage of 18 V if there is nothing attached to it.
  • There is a certain operating point (dependent on the load) for which the motor gives the maximum efficiency, which is also shown in the table : 149 mNm of torque for which the motor draws 20.9 A.
  • And finally, if the motor is stalled, meaning that it cannot spin at all, it draws 156 A.
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A typical DC brush motor will behave like an ideal motor-generator in series with a resistor and an inductor. Ideal motor-generators may be characterized by either of two parameters (either of whose value would be implied by the other): the amount of voltage required to produce a certain rotational speed, or the amount of current required to produce a certain amount of torque. At any moment in time, the voltage on an ideal motor, scaled by the appropriate parameter, will match its rotational speed, and the current, scaled by the appropriate parameter, will match the applied torque. These relationships hold in both directions.

Any time the voltage of a practical (as opposed to ideal) motor doesn't match what the rotational speed would imply, there will be some voltage drop across the modeled series resistance and/or inductance. Voltage drop across the series resistance will convert electrical energy into heat, but that and mechanical friction on the bearings are generally the only two aspects of the motor that would cause the amount of electrical energy flowing in (or out) to differ from the amount of mechanical energy being put out (or fed in).

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