# Braking a DC Motor

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!

• Another way to think of it is that when the armature is connected to a load, the motor is acting as a generator. All of the power delivered to the load has to come from the rotation of the motor. There's a fixed amount of energy available from the rotation of the rotor; if you pull energy out faster, then rotation slows faster. Nov 2, 2015 at 19:42

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:

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

1) Infinite external resistance, zero braking current

2) Finite external resistance, some braking current

3) Zero external resistance, large braking current

• I believe that dynamic braking generally means systems in which the braking energy is dissipated in resistors including the motor itself. Braking in which the energy is stored or returned to the supply system is called regenerative braking.
– user80875
Nov 2, 2015 at 19:38
• @CharlesCowie Good point, let me update the answer. Nov 2, 2015 at 19:42
• Thanks! Just one more thing I'm a little confused on.. I have a formula for C-EMF or bEMF as you call it. Ec = Va - Ia x Ra.. So in procedure 2 and 3.. Is the braking torque higher in 3 because Ra would be considered higher due to the light bulb, making bEMF higher? Or is it higher because the reversed current would be higher due to less resistance to the bEMF? Would bEMF be the same for both procedures, or higher in procedure 3? Nov 2, 2015 at 20:48
• Ra is the armature resistance only. In the beginning, when the motor is spinning, the bEMF is dictated by the speed : V = omega * Kv. The moment the windings are shorted, either through Ra or (Ra + bulb), the bEMF does not change immediately so you have this voltage across the resistances, causing the braking torque. Removal of the lightbulb makes the total resistance smaller and since bEMF is the same initially, it drives much higher braking torque. So you are right, the bEMF is the same and the current is higher when the lightbulb is shorted. Nov 2, 2015 at 21:05