# DC motor model issue

I use the following DC drive model: with the following parameters of the motor: Matlab variables for these parameters:

L = 0.343*10^(-3);                  % H
R = 11.4;                           % Ohm
Kt = 11.2*10^(-3);                  % Nm/A
Kemf = 1/(2*pi*849/60);             % 849 [rpm/V] -> [V/(Rad/s)]
J = 0.993*(10^-3)*(0.01)^2;         % [g * cm^2] -> [kg * m^2]
B = 0.00;                           % Friction coefficient
G = 35;                             % Gearing coefficient

Jload = 0.0;                       % N*m^2


Transfer function which converts torque into omega is derived from the following considerations(here I omit damping B): What makes me feel that model does not work properly is its' output. With Jload = 0 I have the following current and speed responses:  And the settled speed actually is close enough to that of declared in No load speed field(note that field is in RPM while plot is in rad/s). However if I change Jload the behavior seems to be wrong. When I increase the load I expect that motor will shift along speed/torque curve leading to an increased current and reduced speed. Nevertheless making Jload = 0.05 gives me the following output:  Here is how output is generated: As you can see the only thing that changes is settling time. What am I missing here?

Thanks!

• Missing some units ? Is RPM shown? Computed power/rated power? Or I ratios? Or actual RPM/torque curve Feb 3, 2019 at 14:56
• @SunnyskyguyEE75 yep, sorry for that. Two plots:I(t) [A] and Omega(t) [rad/s]. Feb 3, 2019 at 14:59
• @SunnyskyguyEE75 I have also revealed that motor link is broken so I have added screenshot with motor parameters from maxon web page. Feb 3, 2019 at 15:04
• Define ALL inputs and ALL outputs with units for graph. I sense should not go to 0 for 12Vin, while L/R ratio is one time constant and torque current with gear reduced speed G=35 and load inertia another T Feb 3, 2019 at 15:07
• G reduces RPM and increases output torque so define new W range. Feb 3, 2019 at 15:14

$$\J_{load}\$$ is the moment of inertia of the load; mathematically it's the equivalent of mass in a system where you're dealing with linear motion. Here's the details: $$T = J \alpha \\ F = m a$$ where $$\T\$$ is torque $$\J\$$ is moment of inertia, $$\\alpha\$$ is rotational acceleration ($$\\ddot \theta\$$), $$\F\$$ is force, $$\m\$$ is mass and $$\a\$$ is linear acceleration.
Increasing $$\J_{load}\$$ will slow the response of the system to changes in voltage, because it takes more torque to spin the motor up. But $$\J_{load}\$$ does not change the amount of torque used by the system once it is at speed. That is why when it eventually settles, it does so to it's previous speed and current.