# Brushless motor controlled via current source math model

I'd like to model a brushless motor's (BLDC) dynamic response using Matlab. I'm controlling the motor by an Elesctonic Speed Controller (ESC). I'd like to derive a transfer function of such a circuit with shaft's angular speed being its output (as here) and with current being its input. I'm powering the whole plant using a constant 11.1V battery source. In the picture above, one could write equations for the electrical part of the model as:

$L \frac{di}{dt} + Ri = V - K \dot{\theta}$ (1)

with K being the back-emf constant and the mechanical part can be modeled as (Newton's second law):

$J \ddot{\theta} + b \dot{\theta} = Ki$. (2)

It's easy to connect those two equations and derive a transfer function $\frac{\dot{\theta}}{V}$. I'm also expecting it must be possible to derive a transfer function of the form $\frac{\dot{\theta}}{i}$. I imagine I could solve my problem by substituting $V$ for $i$ in equation (2) - I just don't know how.

• Are you using constant current or constant voltage? In the title soy say CC and then in the body of the question CV. Jul 11, 2012 at 10:10
• The title says "current source". I'm using a 11.1V battery and control motor's speed by differentiating current.
– mmm
Jul 11, 2012 at 10:29
• You can either model the angular velocity as a function of the applied voltage (CV) or as a function of applied current (CC). You can't have both at the same time since the voltage and current are interdependent. Jul 11, 2012 at 10:33
• Yes, I expect it to be possible and the solution to be pretty easy. I'm just stuck on it and don't see how. I rephrased the question and added some details, to be more specific.
– mmm
Jul 11, 2012 at 10:58

If you are driving the motor with a known voltage, the page you linked to gives you the answer so I assume you want to drive the motor with a controlled current.

Since the motor's torque is proportional to current, you can forget the resistance, inductance and back-emf constants.

$T = K.i$

and

$T=J\ddot{\theta}+b\dot{\theta}$

this can be expressed as

$s(Js+b)\theta=Ki$

$\theta = \dfrac{Ki}{s(Js+b)}$
$\dfrac{\dot{\theta}}{i} = \dfrac{K}{Js+b}$