Electronic – Braking a DC Motor

dcmotor

I've been working on a lab for my electrical class, and I can't quite seem to find the answer.
I have a DC motor that plugs into a 120V supply,from the supply it is connected to the shunt field, then to a switch (sw3), which either connects the armature to the supply/shunt field, or disconnects it from the supply/shunt field and connects it to a light bulb.
Our first exercise was to simply unplug the motor after it is running at max speed, and time how long it takes for it to stop turning. This took about 5 seconds, as I assume the only braking torque is caused by windage and friction.
My second exercise was to let it get up to speed, then switch sw3 so it disconnects the armature from the shunt field and connects it the to light bulb. I understand this is "dynamic braking" and it took ~3 seconds for it to stop spinning.
The last exercise was to short out the light bulb, bring the motor to full speed, and then switch sw3 to the shorted light bulb. This stopped the motor almost immediately.

I have a few questions, I know the braking has to do with counter EMF, I'm just not sure exactly how/why it causes it to brake so much quicker without the resistance of the light bulb. I'm also not sure if both exercise 2 and 3 would both be considered dynamic braking?
Thanks a lot for the help!

Best Answer

In the first exercise, the motor winding was disconnected, therefore the effective resistance was infinite. This allows no braking current and therefore no elecromagnetic torque. The motor slowed down due to viscous damping torque.

In the second exercise, you have added a lightbulb, which is really a resistor. Hence, the motor bEMF forced current through the lightbulb, which made braking torque. So the motor slowed down quicker. The amount of braking torque is proportional to winding current.

Finally, in the third exercise, you shorted out the winding, which means that the only resistance left in the circuit was the winding resistance. Thus, the induced current was really large- giving rise to a very high braking torque. All the energy was absorbed in the motor winding itself.

I'm also not sure if both exercise 2 and 3 would both be considered dynamic braking?

Yes, they would. Dynamic braking is when the mechanical energy is transferred to the electrical subsystem.

The rotor energy can either be stored in energy storage elements such as capacitors or batteries (regenerative braking) or dissipated in resistors, such as is your case (dynamic braking). You have two series resistors - bulb and motor winding. Thanks @Charles Cowie for reminding me of the difference.

The electrical braking power is: \$ P_{el-braking} = R_{total} * I_{winding}^2 \$

Case No. 1 is not considered dynamic braking because the rotor energy is dissipated only through the means of mechanical losses.

\$ T_{braking} = K_T I_{winding} [Nm] \$

And the mechanical motor equation is:

\$ d\omega/dt = -T_{braking} - b_{viscous-damping} \omega \$

enter image description here

1) Infinite external resistance, zero braking current

2) Finite external resistance, some braking current

3) Zero external resistance, large braking current

Related Topic