Braking a DC Motor

Thread Starter

scorca

Joined Oct 31, 2008
36
Hi all,

I am doing a project that make use of a dc motor. There is a 24 VDC motor with nominal current of 1.17 A.

I have a problem with emergency stop. The requirement is that when E-stop button is pushed or safety hood is open, the motor has to stop on a dime.

My question is: how to stop the motor electrically in short period of time without damaging it?


scorca
 

bertus

Joined Apr 5, 2008
22,270
Hello,

You can brake the motor with a power resistor (to limit the current) and a switch contact
(that breakes the power and switches the resistor across the motor).

Greetings,
Bertus
 

hgmjr

Joined Jan 28, 2005
9,027
If you are using an H-bridge to drive your motor, you can brake the motor by driving both ends of the motor to the same potential. That is to say you can drive both ends with 0V or conversely you can drive both ends with +24V.

In effect you are shorting the two motor terminals together. This will bring the motor to a very rapid halt. The technique is referred to as self-braking.

hgmjr
 
Last edited:

mik3

Joined Feb 4, 2008
4,843
A resistor will get very hot if it doesn't have the proper power rating. It is better to use a MOS and PWM to control the braking current of the motor.

If you use an H-bridge rectifier you can control the on-off time of the two lower MOS (the upper MOS will be off) as to control the braking current. Note that the MOS has to have a built in protection diode otherwise you need to put external ones.
 

Thread Starter

scorca

Joined Oct 31, 2008
36
I'm using a built in motor controller and don't know what kind of mos they put in there so any changes on the H bridge wouldn't be possible.

My only way (so far I know) is to use a power resistor but I'm still looking for the way to calculate the power rating and how long it takes untill it stops.
 

kubeek

Joined Sep 20, 2005
5,794
You could try to persuade the motor controller into the braking regime.
Because this is an emergency stop and not the main brake, you don´t have to worry too much about the resistor, just try it without.
But you need to know what energy is stored in the system at full RPM to be able to tell if the motor and the driver will hold the shorted current. Knowing the internal resistance of the motor would help too.
 

kubeek

Joined Sep 20, 2005
5,794
So it takes about 9 seconds for the motor to reach the full speed when starting? Is that right, or is it longer?
I know the voltage and current on the motor when stopping and starting will be exponential, so if I assume linear behavior for calculation, the actual results for stopping should be longer by the above time ratio.

If you shorted the motor, the inital power dissipated is cca 260W, so it should stop in about a second, with the current peaking at 11A. Is the driver capable of taking that current, and what is it´s internal resistance when the fets are on?
 

kubeek

Joined Sep 20, 2005
5,794
E=W.t so making 250J with 28W takes 250/28=8.9s
But this is only true when the motor does 24W all the time from zero to full speed.

I=V/R so 24V/2.19, when the motor is shorted at full speed, makes 11A.
 

Thread Starter

scorca

Joined Oct 31, 2008
36
ok, now I have 250 J brake energy, my regenerative power is 260 W and my internal resistance is 2.19 ohm. I want the motor stops after 2 sec, The dynamic brake resistor would be:

Pbrake= 250 J/2 sec = 125 W

Rbrake= 24^2/125 = 4.6 ohm

Rextra = Rbrake - Rinternal = 4.6 - 2.19 = 2.4 ohm

is that right?
 

kubeek

Joined Sep 20, 2005
5,794
It seems right, but all my assumptions were based on the motor spinning up in 9 seconds, which you must confirm first.
Do you have access to the motor, so that you could try it?
 

Thread Starter

scorca

Joined Oct 31, 2008
36
Nope, I dont have. We are still in designing phase. What I got from my coleague is that the acceleration of the motor is 360 rad/s^2. The stall torque is 0.234 Nm and the mass inertia is 0.00065 kgm^2.

Based on that, I think it wont take 9sec to get the full speed but it would be shorter.
 
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