# Modelling a Braking DC Motor Speed Controller

I'm trying to simulate a brushed DC motor. The model comes from the following circuit diagram KVL yields: $$v = Ri + L \frac{di}{dt} + K_b \omega_m$$

My rudimentary understanding of speed controllers is basically as a variable voltage source. Therefore I solve for the new current each time step like so:

$$i = \frac{v - L \frac{di}{dt} - K_b \omega_m}{R}$$

where v, i, and omega are modeled/provided by the simulation (other physics concerns are taken into account elsewhere) and everything else is a constant.

The speed controllers I'm using have a "brake" and "coast" mode. In brake mode, a neutral signal will, as the manual puts it, "effectively short the motor terminals." I know from experience that brake mode results in the system slowing much quicker. How is the short in brake mode different from applying 0 volts? What equations model the behavior of shorting the terminal?

My main confusion comes from trying to model the short-circuit as such:

$$\Sigma V = 0$$ $$0 = Ri + L \frac{di}{dt} + K_b \omega_m$$

but this is the same as applying 0 volts in the original model.

• Why do you think applying zero volts is wrong? The model you have is fine aside from ignoring the mechanical aspects of the problem, as this effects the back-emf. The other thing that would be needed is some initial conditions. – jramsay42 Nov 26 '18 at 7:58
• I know empirically that the behavior when supplying a neutral signal in brake mode is drastically different from coast mode. – Lytigas Nov 26 '18 at 16:38
• What does your manual say coast mode does? – jramsay42 Nov 27 '18 at 0:11
• @jramsay42 "When a neutral signal is applied in Coast mode, Back-EMF will not be generated, so the motor’s rotation will not be affected by the Speed Controller." I'm not quite sure how to interpret that, because all examples I've seen online do factor in Back-EMF like in my current model. – Lytigas Nov 28 '18 at 3:48