# Dynamic Braking of 3-Phase DC Motor

1. Feb 24, 2008

### Phrak

To brake a 3ph DC, the best I've come up with, apparently, is to place a pair of power FETS, back-to-back across each pair of leads.

I would like to manage a hard short without additional load beyond the motor winding resistance, then duty cycle to get controlled braking.

As far as having the (positive) load current in a NPN MOSFET directed from source to drain, I have only heard of this done once, and that was at low currents.

Is this at all feasable to do this up near the rated current, or am I completely wrong about this?

2. Feb 24, 2008

### rbj

uh, what's a 3-phase DC?

what does phase have to do with DC?

3. Feb 24, 2008

### NoTime

4. Feb 24, 2008

### Phrak

NoTime, I need to dump about 800 joules in 3 seconds. At the maximum rated torque the motor disipiates about 120 watts. That's 6 or 7 seconds worth of power. It could have a heating problem at high braking duty agrivated by high torque duty.

5. Feb 25, 2008

### rbj

to repeat, what is "3-phase DC"???

i don't work in electromechanical energy conversion (a.k.a. "power") but i used to teach EE at the university level. i used to know what 3-phase power is, what 3-phase motors were about, what DC motors were about, what variations like stepper motors or other electronically driven motors were, what "regenerative braking" is (at least in the context of an electric car), but i never even heard of the term "3-phase DC". it sounds like an oxymoron to me.

Last edited: Feb 25, 2008
6. Feb 25, 2008

### zeitghost

Is it not an electronically commutated dc motor?

7. Feb 25, 2008

### stewartcs

It's not "3-phase DC", it's a "3-phase DC Motor". It just means that the DC motor has three phase windings, not that it is powered by 3-phase DC, which as you already noted is nonsensical.

The phase windings are switched in a specific order to make the motor rotate.

CS

8. Feb 25, 2008

### stewartcs

Why?

Define low currents.

CS

9. Feb 25, 2008

### NoTime

Think of what happens if you short the output of a generator.
Say a motor is 90% efficient. So for your 120 watt motor, 100 watts goes out the shaft and 12 watts gets dissipated as heat in the motor.

800 joules in 3 seconds is nearly 300 watts for a motor designed to get rid of 12w.

10. Feb 25, 2008

### stewartcs

Sounds like you need more time to dump the load, or you need some mechanical braking to assist.

CS

11. Feb 25, 2008

### RonL

The FETs are neat, and I'm guessing your using something like hard drive motors. Can you say anything about how big the system can be ?
As stewartcs suggested something mechanical might be needed to absorb that much wattage. Are you looking for ideas ?

12. Feb 26, 2008

### Phrak

Sorry, NoTime. I wasn't clear about that. The power dissipation of 120 watts is the motor's power loss at maximum continuous rating (55 amps through 30 milliohm). The maximum shaft output is 800 watts or so.

My application is for a vehicle that takes about 3 seconds to stop. If this occurs once every 7 seconds, the motor itself will be dissipating 113 watts, average in heat.

This assumes that the heat capacity of the wire is sufficient to absorb excess heating at the low pulse rate of 1/7th Hz. Also it involves the thermal resistance of the windings to ambient, and all that, but I'm not sophisticated enough to understand how all that is calculated.

13. Feb 26, 2008

### Phrak

Hey, stewartcs.

Why do I want a dead short?

I'll be using the motor to both power and brake a vehicle. I can't obtain an acceptable deacceleration rate otherwise. .. Maybe I could supplement with dissipating up to 20% with an external load, but not much.

It's been a long time, but I believe the current in question was less than -10% of I_{D}max. at a V_{GS} of 15 volts (full on).

Last edited: Feb 26, 2008
14. Feb 26, 2008

### Phrak

I'm sure you guys must be right. I'm comming around to combined mechanical and dynamic braking.

I don't have a great deal of diameter to do mechanical breaking.
The total available volume for braking is a cylinder 2" in diameter by 5 inches long with the wheel shaft passing through the center. In addition it could have as much as 1 inch protrusions off two sides of the cylinder over it's entire length.

Any suggestions would be greatly appreciated.

and I am still clueless about FETs...

-deCraig

Last edited: Feb 26, 2008
15. Feb 26, 2008

### stewartcs

Wrong why. I was asking why not load it beyond the resistance of the windings. In effect, I was asking why not use a resistor bank to dynamically stop it...Not why do you want manage a dead short.

CS

16. Feb 26, 2008

### stewartcs

I seem to remember it being around 3-5 amps. Not 100% sure though.

CS

17. Feb 26, 2008

### stewartcs

There is always Plugging, but that can possibly over-stress the motor.

CS

18. Feb 26, 2008

### NoTime

Here are some FET circuits that may or may not be useful.
http://www.discovercircuits.com/F/fet1.htm

Most dynamic braking systems use resistor banks to dissipate the energy or in the case of regenerative braking charging the battery.

19. Feb 26, 2008

### Phrak

CS, the rate of deceleration is inversely proportional to the total series resistance. The series resistance is the motor resistance + load resistance + FET channel resistance.

To be precise, Power_of_deceleration = (E_motor)^2 R_total.

E_motor is the voltage generated by the motor obtained by operating the motor as a generator.

So you add an external load, and your rate of decelertion decrease. I'd love to dump the heat elsewhere but it makes for a very soft brake unfortunately.

20. Feb 26, 2008

### stewartcs

I wasn't asking a question, it was rhetorical.

If you use a resistor bank as a dynamic brake the resitor bank is generally sized to provide an armature current that will approximate 150 to 300 percent rated current.

The motor behaves as a generator and feeds current to the resistor, dissipating heat at a rate equal to $$I^2R$$.

You'll need a mechanical brake once the motor has slowed dynamically to completely stop and hold it still.

If you need to stop it VERY quickly, then Plugging will do the trick (with the Caveat I mentioned previously). Just remember to use a Plugging resistor in series with the armature circuit to limit the current.

CS