# Resistive braking of series wound DC motor

I am looking at braking a series wound DC motor by putting a resistor across the motor feed terminals.

All of the sources that I have looked at state that the polarity of the field winding needs to be switched as you start braking. I understand how this works and how it applies braking.

What I do not understand is what happens if you do not change the field coil polarity. My instinct is that this will apply a reverse current to the field coil reversing the polarity of the output which will reverse the field.... ad infinitum. The result will probably be the field collapsing to 0 as an under-damped second order response totally failing to brake the motor.

If there is anyone out there with any experience of what actually happens I would be happy to hear it.

The torque of an ideal series wound DC machine is $$\T = k i^2\$$. So the torque is always in the same direction, regardless of the current direction.

Regardless of how you're generating current, if you want to reverse the direction of the torque you need to change the constant $$\k\$$. You can only do that by changing the polarity of the field coil with respect to the armature.

So, if you want to brake -- by any means -- you need to reverse the direction of the field coil.

** Edit **

Going to the next level of detail, by conservation of energy*, in an ideal motor, $$\v_a i = T \omega\$$, where $$\v_a\$$ is the armature voltage and $$\\omega\$$ is the shaft speed. So $$\v_a = k i \omega\$$.

Model the motor with some inductance, so the inductive contribution is $$\v_L = L \frac{di}{dt}\$$.

$$\E = I\, R\$$, and if you're careful with your sign conventions you'll see that the resistor opposes the coil voltage, so $$\i = -\frac{v_a + v_L}{R}\$$.

Separating that out, and doing more work in one step than would be acceptable in a homework assignment: $$i = -\frac{k \omega}{R} i - \frac{L}{R} \frac{di}{dt}$$

Rearrange terms and again doing more steps than I ought to, I get $$\frac{di}{dt} = - \frac{R + k \omega}{L} i$$.

So -- if you just put a resistor on the thing, the current will decay exponentially based on the speed of the motor; i.e., the ideal motor running at constant speed "looks" like a resistor. This makes sense for energy conservation, because $$\P_{out} = T \omega\$$, and $$\P_{in} = i^2 R\$$, and the energy stored in the inductor is a function of $$\i^2\$$. Since the power out is a function of the shaft speed, it would make sense that the energy in the inductance would get sucked out faster the faster the motor is turning.

I didn't try modeling the dynamics of the motor itself ($$\\dot \omega = T/J\$$, with the angular acceleration, $$\\dot \omega\$$, depending ultimately on current) because I assume that they are slow enough that the motor current decays long before the motor speed has a chance to. So for the purpose of this problem, you can take the speed as constant.

* Thank you Emily Noerther! This makes it so easy!

• Tim thank you for taking the time to answer. I am aware of the theory and how this applies. I am looking for the next level of detail as to what is happening within the motor if you attempt to brake the motor with the field coil connected "forward". – RoyC Sep 29 at 7:19
• I edited the answer with more detail. – TimWescott Sep 29 at 15:21