The DC motor in question: adafruit product 4641
The corresponding datasheet: N20 DC Motor

My motorshaft is connected to a pulley, and so I want to do some calculations to see what load I can apply given a voltage source to my motor controller (rated 4.5-48V). Pretty much every resource I referred to has indicated that I need some key data from the manufacturer to calculate things like motor torque, speed, and max applicable load. The equations I've seen refer to Kt (torque constant), Continuous Stall Torque, among other important values. This is the kind of datasheet that I expected for a DC motor, but the datasheet for the N20 motor doesn't seem to have much of the same information.

Am I misunderstanding how to use the datasheet? Or is there an alternative way to calculate things?

Equations I want to apply to the N20 DC Motor:
torque = Kt * current


2 Answers 2


The data sheet provides the current and rotational speed at 1000g.cm. 150mA, 20RPM. This gives a torque constant of 6.66 kg-cm/Amp.

It also indicates the no load current and speed. 60mA and 33RPM.

The stall current at rated voltage (6V) is 0.35A. This indicates an armature resistance of ~17 Ohms.

În conjunction with the no-load current and speed this indicates that the resistive drop is 0.06 x 17 = 1.02V.

The back-emf at no-load is therefore 6-1.02 = 4.98V at 33RPM giving a back-emf constant of 1000*4.98/33 = 150V/K RPM.

It is high because of the gear reduction built into the motor.

Since it is a permanent magnet motor the best assumption is that the torque is proportional to current and the speed is proportional to the effective voltage. (terminal voltage minus the voltage loss due to the armature current flowing through the armature resistance)

If you do a linear estimation from those three points you can predict the speed and current at other loads. Note that this will not account for other non-linear losses such as windage, viscous friction and eddy-current losses.


The back emf constant and the torque constant are created by the same magnetic field. Looking closely at the units shows that when SI dimensional units are used \$k_b\$ amd \$k_\tau\$ have the ssame value. The units are expressed differently but the magnitude is the same.$$1\frac{N-m}{A}=1\frac{V}{rad/s}$$

As Kevin White has shown:$$R=\frac{6V}{i_{stall}}$$

Kevin White has shown that with the no-load speed (\$\omega_{NL}\$)and the no-load current (\$i_{NL}\$) given at the rated applied voltage (\$V_{r}\$):$$k_b=\frac{V_{r}-i_{NL}R}{\omega_{NL}}=k_\tau$$

Now you can use the formula $$\tau=k_\tau i$$

For the most part, where data sheets display these value in SI units, they are the same or very close. I have also measured them in different ways and obtained very close results. Measuring \$k_b\$ is quite easy. Then using that value for \$k_\tau \$ is convenient.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.