Braking

Alan Case

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Hi, A reasonably simple question, I have a high inertia load which I am accelerating to a speed with an Allen Bradley Powerflex 70 drive (2.2KW).
I need to decelerate and stop the load in a lot shorter time than the acceleration time. I have the external resistor unit and have the drive set for for bus regulation of resistor first and freq. regulation second. My question is: will a larger drive allow the load to stop quicker. Regards Alan Case
 
I think No

Alan
As i understand there is a parameter can be calculated known as t(dec)which means the optimum decelaration time .This time is totaly depend on the motor shaft inertia,the difference in the to speeds 1st you runnung in 2nd which you want to go to as well as the frictionn torque of the system and the braking torque.
If you want to decelrate your motor in shorter than this time you will need a braking resistor and braking switch which you already have.
As in the braking the motor act as generator cause the voltage of the intermediate circuit to raise up(Dc link).
all the drive have a protecion against DC link overvoltage the voltage at which the unit will trip at is root 3xsupply voltage so a larger drive with tha same input rating will not make the load to stop quicker because it's not depend on current rating of the drive.
I hope that this will help you
N.B. wait for Dick reply .He is the drive master here.
Good Luck
 
You gotta love Physics!

Alan,
Originally posted by Alan Case will a larger drive allow the load to stop quicker
I think that bigger is always better is only true in the movies or other settings that do not include industrial machinery. If you want to defeat a physical force, you have to adhere to the rules old what's his name came up with to explain our physical world. Fight fire with fire or in your "case", no pun intended, fight energy with energy.

Hesham said,
so a larger drive with tha same input rating will not make the load to stop quicker because it's not depend on current rating of the drive
and how true that is. So you have these built in or external resistors to absorb energy from the rotating inertia, then use them. If the drive can handle all the energy your load can throw at it, then by all means use the drive and set the decel rate of the drive to achieve that function. Another choice, although of lesser quality, would be to add some external form of braking. Either mechanical or electro/mechanical will do the job but you always add a few more worries when you add more stuff to your machine. I have never cared much for AC drives with what amounts to DB resistors on them cause it is a very different animal than the DC counterpart. If your application will allow, you might also try electric breaking by using a DC supply to power one winding of your motor after you disconnect the AC drive from it. That will make the load slow down much faster than freewheeling for sure. Your 3hp or 2.2 kw motor will not be very hard to adapt this to and the power supply will be fairly small. Set the DC supply to deliver up to the nameplate rated current to the winding while it is stopping. Or, you can even leave it on while the load is standing still if you the current rating is not over nameplate. Dick DV will doubtless have some other pearls of wisdom to offer so keep an eye out here for his post.
 
If your application will allow, you might also try electric breaking by using a DC supply to power one winding of your motor after you disconnect the AC drive from it. That will make the load slow down much faster than freewheeling for sure
Randy
Yes Dc injection can help in stopping the motor faster but with loads with high inertia this will also cause a trip "DC link overvoltage"
If there is a suitable breaking resistor to absorb energy.
Your 3hp or 2.2 kw motor will not be very hard to adapt this to and the power supply will be fairly small. Set the DC supply to deliver up to the nameplate rated current to the winding while it is stopping
I think that he don't need an external Dc supply ,Mostly all VFD nowadyas have a stopping mode named "Dc injection "mode and in this mode he can control the level of the applied Dc ,The frequancy of the applied voltage(it is not a must to be zero) as well as the time in which the dc will be applied to motor.
 
Points taken!

Originally posted by ME! If your application will allow, you might also try electric breaking by using a DC supply to power one winding of your motor after you disconnect the AC drive from it.

Careful inspection of my statement will indicate that the drive should not be connected at the point of external DC injection so the DC link overvoltage does not come into play. I am aware of the DC breaking supply that is in some drives, but not all have it. Alan may or may not have it on the Powerflex he is using. Trying to cover some bases with suggestions is all. Nice to know we can be colleagues on this level, even considering the distance between us.
 
Hi Alan,

The PowerFlex 70 does have DC injection braking. Take a look at the following parameter's of the PowerFlex 70 drive under the Dynamic Control File:

155 - Stop Mode A (Select the type of stop mode)
156 - Stop Mode B (Same as above)
157 - DC Brake Lvl Select
158 - DC Brake Level (Defines the level of DC to inject)
159 - DC Brake Time (Set the amount of time for DC inj.)
160 - Bus Reg Gain (Responsivness of bus)
161 - Bus Reg Mode A (Sets the method of DC bus regulation)
162 - Bus Reg Mode B (Same as above)
163 - DB Resistor Type (Set for internal or external brake)

In some of my applications (i.e. - automated hoisting), if we are using a drive for traverse control (horizontal travel), I generally, except under special circumstances, do not need to add external braking resistors.

If for example, I am using a drive to control hoisting (lifting), we absolutley must have a braking resistor to dissipate the regenative power that is sent back to the drive (a heavy load going down for example).

I have used DC injection with great results on all my applications to do what you are describing with drives matched exactly to the motor size.

Regards,
Chris
 
Thanks for the compliments, guys. Alan's application is a good example of the need to recognize that decelerating a load requires torque just the same as accelerating a load. People that manufacture industrial brakes even rate them in horsepower occasionally.

In view of that, when you are faced with an application that requires both motoring and braking, you must size the drive and motor to meet the largest of the two torque requirements. Most of the time, the motoring requirement is larger than the braking requirement but, every once in a while like I suspect here with Alan's situation, you have braking torque requirements larger than the motoring requirement.

At present, Alan has a 2.2kw motor on a drive of unknown size. Judging by the 2.2kw sizing, this is a metric or IEC motor. The first thing to do is obtain a torque speed curve for the motor. Find the peak torque point on the curve and determine if that is enough torque to decelerate the load in the required time period. If it isn't, the motor must be changed out for a larger one that can develop the needed decel torque. Nothing in the drive or braking system can make the motor produce more than its peak torque.

Once you have got the right motor for the job, determine the current requirement for the motor at that torque. That current level must be absorbed by the drive to produce that much braking torque in the motor. Size the drive to absorb that current or more (If the decel time is short, you may be able to use the short term overload current capacity of the drive rather than the continuous rating). Since braking current is higher in this application that motoring torque, the drive will automatically be large enough for accelerating.

Next, size the braking resistor to meet the time and KW requirements for braking and you've got a working system.

DC injection braking never produces as much braking torque in a given motor as snubber braking in a drive. This is even more true if the drive has a good sensorless vector or flux vector control system. That is true because the vector system can take the motor to its peak torque point on the curve and hold it there. A DC injection system simply shoots DC at the motor which is not synchronous to the rotor speed and may end up positioning the motor anywhere on the torque curve, maybe even at the minimum torque point. Clearly, that's not the way to get the most out of your motor.

Alan, hope I helped a bit here. If there are follow-on questions, ask away!
 
I, in the elevator busniess and we have to accournt for overhauling conditions, so when I read Allens post I started to salivate, but then Dick took all the wind out of my sails...Bummer!
 
You got to get up real early!

elevmike you now know why DickDV is the MAN when it comes to motors and drives. We are lucky to have his expertise here on the forum. Took all the wind did he? That was funny!
 
One client, an OEM (mechanical) has a system which accelerates a rotating load in about 5 seconds.

To Stop it he decelerates it in about 2 seconds and all is OK.

The motor is a 1.5 HP and the Inverter is a 2 HP sensorless vector...

To decelerate the load in 1 second he installed a 5 HP version of the same Inverter keeping the same motor BUT it did him no good.

He stills uses the 2 second decel. time.

Dick, can you tell us about what happens with the motor current. Is it just like a Stall.. i.e. the current rushes upward infinitly?

Tancks.
 
The power rating has mostly to do with the current capacity of the IGBT's and the size of the copper. It makes sense that there would be little effect on braking, unless it is a regen drive that dumps back up the mains.

I have a similar situation to Alan. I could take all the time I want to start, but as little as possible to stop. I tried DC injection, my decel time went from 21 sec to 12 w/o faulting the drive. I then bought a braking resistor and got the stopping time down below 2 sec. I used the ABB 550 drive with the built in brake chopper so this was easy to do. My ABB rep basically sized it for me.
 
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Pierre, I wish I knew how to post graphics on this BBS but I don't so here goes with a word description.

First, you need to examine a NEMA design B motor speed/torque curve. You will see that the torque is zero at motor sync speed (for a four pole motor at 60Hz, that would be 1800rpm). At the motor begins to produce torque, the shaft speed starts to drop slightly until, at the point where the motor output reaches its nameplate hp, the speed will have dropped to its nameplate speed (using the example above, at 1.5hp output, the speed will have dropped to, say, 1750rpm). Using the formula hp = torque x rpm/5252 we calculate that the torque output at that point would be 4.5ft-lbs. Note that the torque speed curve is close to linear between no-load and the rated load point.

At this rated load point, the manufacturer calculates that, in a 40 degree C ambient, the motor will, over a long time period, rise in temperature to whatever insulation temperature class the nameplate lists.

If you were to superimpose over the NEMA B speed/torque curve a speed/current curve, you would find that, at no load, there would be, not zero current but about 25% of nameplate current (this is magnetizing amps and, for the motor we are discussing, would be about .65amps at 460V). This current also rises about linearly from that point to the full load current of about 2.6amp. If you scale and position the two curves so the full load torque and the full load
amps occur at the same point then you will see that the torque and current co-incide pretty well except that as you get close to no-load, the difference becomes noticeable (the torque goes to zero and the current goes to 25% of full load).

As the motor is pushed into the overload range, the torque continues to rise almost linearly (it develops a slight arch toward the torque axis) up to about the 220% of full load point. The current will track almost exactly with the overload torque up to about 190% where it starts diverging on a less-arching path upward.

At the 220% point in the torque curve, the rotor begins to loose its ability to stay synchronous with the stator field and the torque falls off rapidly. Unfortunately, the current does not fall off similarly but continues on up until, at the point where the rotor is completely stopped (as at starting across the AC line with a contactor) the current has risen to 6 to 8 times the nameplate full load current. This is the starting inrush current that happens whenever a NEMA B motor is started across the line.

It is important to understand that, while the above description is for a motor that is motoring, the very same curve and torque/current behavior occurs when the motor is expected to act as a brake by absorbing torque.

In view of the above, when an inverter is feeding an induction motor, there can be no sensible reason for sizing the drive output current for more than 220% nameplate since the current falls beyond that point rather than increasing further. So, for Pierre's example with a 1.5amp motor (2.6amp full load) and a 5hp drive (at least 9amps short term output current) it can be clearly seen why the larger drive didn't help as much as expected---9/2.6 = 346% !

Pierre simply didn't have enough motor. Even tho he had more than enough drive, the weakest link was the motor and having a drive that could absorb a lot of current still didn't get the job done.

And, for Rick, drives usually have a continuous current rating and a short-term rating (generally spec'ed at 1 minute). Some manufacturers including ABB now include a 5sec rating also which is handy for fast braking applications and fast accels.

These ratings are far below the max limits for the drive semi-conductors. Instead, they are based on heat sink capacity. That is why, if you are running a drive at 50% of its rated output, the short term max current is higher than if you are running at 95% of rated output (the heatsink is cooler at 50% than at 95% and therefore will withstand a higher current burst for a few seconds before it reaches its max rated temp). It's not uncommon in ABB's drives for the semiconductors to be rated at four times the max short term current just for safety sake.

While you can understand that I think the new ABB ACS550 is an excellent drive, Rick's success is due to the drive/motor system being sized properly to handle both motoring and braking currents. If the motor had been too small as we were discussing above, not even the miracle-working ACS550 (snicker!) would have been able to save the job.
 
One other thing for Rick Densing. Whether the drive is four-quadrant (full regen) or simply uses snubber braking, the output current limitations are enforced by the drive software the same way. That business about heatsink temps applies the same, too.

The only difference between the two assuming they have the same output current ratings is that the braking energy goes back into the AC supply and is saved for reuse with regen where, with snubber braking, the energy is wasted as heat in a resistor.

In my experience, you have to do a lot of braking (such as tension control in a coil unwind stand) to justify four quadrant drives. Resistors are comparatively cheap as long as you've got somewhere to go with the occasional burst of heat.
 

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