In other words, if the load is great enough, will a 110V, 18A rated motor being run at 5V stall at the same current as if it were run at 110V? I'm asking because I'm currently working with a treadmill, which at its lowest speed (I read ~6V across the motor free-running) has enough torque to move a large amount of weight. I measured the motor's resistance to be 0.9 Ohm, so at its lowest speed free-running the current draw is around I=V/R = 6.7A? Or is that the stall current?

Would powering the motor on its own with a 5V supply be the same as the onboard controller providing 5V to the motor through PWM 120V?

I guess I'm just surprised a motor can provide so much torque with such little voltage. So what determines a motor's current?

  • \$\begingroup\$ "what determines a motor's current?" The motor's "motor constant". This defines how the motor reacts to applied voltage over it's coils. It's a factor which relates voltage with current and torque. You can get little brushless DC motors for quadcopters these days with ridiculously high "KV" values, which is sort of like their "motor constant". KV values in the many thousands, where 1000 KV means 1 volt will product 1000 rpm on the motor output. \$\endgroup\$
    – KyranF
    Mar 29, 2015 at 6:41
  • \$\begingroup\$ and once the motor has started (begin moving, overcome static friction and the coils have energized based on their inductance and inherent magnetics) it will pull current based on the voltage applied. \$\endgroup\$
    – KyranF
    Mar 29, 2015 at 6:42

1 Answer 1


Some motor basics.

A motor's stall current is determined by the resistance of the coil. (quoted at the nominal operating voltage). If the shaft is clamped so it cannot rotate then the 'motor' will act like any other resistor and follow Ohm's law - so yes, the stall current will increase with increasing applied voltage.

Once the rotor starts to rotate it will induce a back emf so the voltage 'seen' across the rotor will be reduced. The net current through the motor will be reduced.

Increasing the applied voltage will increase the speed (how much depends on the motor). With increasing speed comes increasing loss (friction, windage etc.) so there will be a corresponding increase in current.

Applying a load (torque) to the output shaft slows the rotation speed and reduces the back emf. This in turn will increase the motor current. Applying too much torque may reduce the shaft speed to 0 and this is once again back to a stall.


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