List of principal symbols:

Time |
t |

Location |
x |

Velocity (Speed) |
v |

Acceleration |
a |

Force |
F |

Mass |
m |

$F=F_{pwm}-F_{bemf}=ma$

$F=C-Kv=ma \qquad with \qquad a=\frac{d v}{d t}$

C represents the Force originated by the PWM voltage, and Kv the BEMF (Back Electromotive Force).

We are going to suppose that **F**_{pwm} is constant and **F**_{bemf} is the only counteracting force and proportional to the speed.

At the beginning of the startup sequence, time=0, we have v(0)=0 and x(0)=0.

The maximum velocity of this motor will be **V**_{max}=C/K.

$\frac{d v}{d t}=(C/m)-(K/m)v\qquad .$ **(1.0)**

$\frac{d^2 x}{d t^2}=(C/m)-(K/m)\frac{d x}{d t}$

$\frac{d x}{d t}=(C/m)t-(K/m)x+Const1,\qquad Const1=0$

$x=\frac{1}{2}(C/m)t^2-(K/m)xt+Const2,\qquad Const2=0$

$x=\frac{1}{2}(\frac{C}{m})t^2/(1+\frac{K}{m}t)\qquad .$ **(1.1)**

From (1.0):

$v=(C/m)t-(K/m)vt+Const3,\qquad Const3=0$

$v=(\frac{C}{m})t/(1+\frac{K}{m}t)\qquad .$ **(1.2)**

If we ignore **F**_{bemf} (if K=0) we get:

$x=\frac{1}{2}(\frac{C}{m})t^2\qquad and\qquad v=(\frac{C}{m})t\qquad .$ **(1.3)**

And we can simplify this way because **V**_{max-startup}<V_{max}/10 (typical). We need this model only for little startup speeds (t near zero).

At Motor CTRL Basics we said: "We control mainly 2 variables: the time for the next commutation and the voltage.". At startup we will begin with PWM around 30% or 40%, and the time for the next commutation is derived from (1.3):$t_n=k1. \sqrt{x_n}$

to be continued…

## Links

Mountain Engineering- Sensorless Motor Control