I am very new to DC motors and to stackexchange. Please correct me if anything I said does not make sense.

For DC motors, the equation looks like this:

$P = \tau\dot{\theta}$

where $P$ is power, $\tau$ is torque and $\dot{\theta}$ is angular velocity. The power can be expressed as $P=VI$ (voltage times current) and $\tau=K_\tau I$ ($K_\tau$ is the torque constant).


$VI = K_\tau I \dot{\theta}$

It seems from the above equation that the angular velocity is directly proportional to the voltage applied to the motor and the current flowing through the motor is irrelevant.

For a PWM controlled motor, the effective voltage is $V_{in} \times \textrm{duty}\%$. For different loads applied to the motor, the angular velocity should be the same if the PWM duty is fixed. However, in reality, when the load increases, the RPM goes down for a given PWM duty.

So, the question is:

  • Is $K_\tau$ constant across different loads for a motor (in this case, a DC shunt motor)?
  • What is the 'effective voltage' that should be used in the equation?

If $K_\tau$ is constant with different loads for a given motor, then the 'effective voltage' must not be a simple $V_{in} \times \textrm{duty}%$. When the load increases with a fixed PWM duty, there must be a decrease in the 'effective voltage' since the RPM goes down.

If the 'effective voltage' is the voltage difference between Motor+ and Motor-, I did measure it and the voltage difference drops when the load increases for a given PWM duty. Is it that the back EMF causing the reduction in the voltage difference?

Thank you in advance.

  • 1
    $\begingroup$ Hi - I re-did the math formatting in your post. Feel free to tweak it if it's not quite to your liking. This is a starting point to learning the LaTeX markup used on SE: physics.stackexchange.com/faq#notation $\endgroup$
    – Kyle Oman
    Feb 9 '13 at 5:42

I can't see anything wrong with your reasoning, so I'd have to conclude that you're correct and the angular velocity is proportional to voltage.

However when you increase the load on the motor and slow it, the back EMF decreases and the motor draws more current from your power supply. The angular velocity will only stay constant if your power supply can provide the extra current without its voltage falling i.e. if the power supply has zero internal resistance. In practice all power supplies have some internal resistance so the voltage will fall as the motor draws more current.

If you can ignore the resistance of the cotancts and windings in the motor then the voltage you need to consider is the voltage across the motor terminals, and I note you say this does fall. That's what you'd expect if your PSU doesn't have zero internal resistance. If you adjust your power supply to keep the voltage across the terminals constant you shpuld get a constant angular velocity. It would be interesting to hear if this works with your motor.

  • $\begingroup$ Thank you for the answer. So, I guess the drop in the voltage does not have anything to do with the back EMF. I forgot that fact that the power source in our case is not a voltage regulator and the voltage drop due to the internal resistance seems reasonable. $\endgroup$
    – Brian Wang
    Feb 10 '13 at 4:12
  • $\begingroup$ dear sir the way you explained is really correct....but i feel you can better explain with the blok rotor test..... $\endgroup$
    – user25602
    Jun 10 '13 at 10:48

The previous answer is correct up until the motor draws more current. This 'More Current' generates MORE TORQUE since the torque is proportional to the motor speed. Finally your increased load balances with the motor increased torque.
The torque constant stays same if you ignore loss, such as copper & iron loss, or magnetic saturation inside the motor.


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