Drive Braking Resistors

CT782

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Jan 2004
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Is there a formula or rule of thumb for sizing external braking resistors for AC drives equipped with the option to use this feature?
 
Welcome to the forum!

As you probably already know, the size of a resistor is based on how much energy you have to dissipate. The wattage of the device must be at least as great as the total energy you have to dump so the only thing left to consider is how long does the bank have to endure this energy and what should the ohmic value of it be. I would give my motor requirements to the drive manufacturer and let them size the resistors for the drive. They will doubtless have a better method for doing this than you or I. But to get close, just fall back on the old power equations, P=I X E or the other two variations depending on what your data is. To find R you need to know power and current or power and voltage so there you have it. Some things are just better left to those folks that do it for a living. There is one guy on this forum, DickDV, who has a vast knowledge base on this subject so perhaps he has a rule of "thumb" that we can all learn from.
Check out this thread:
http://www.plctalk.net/qanda/showthread.php?s=&threadid=7133&highlight=breaking+resistors
 
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Another bad case of Deja-Vue!

CT782,

I reply to your post only to re-enforce randylud's, referal to the that thread. Very carefully read all of DickDV's responses.

However a more direct response would be as follows:

1. Calculate duty cycle (DC)= breaking time/total cycle time

2. Calculate deceleration torque.

Dcell tq = RPM change x (Inertia (ft lbs.) / 308 * time (sec)) - Friction (ft lbs).

3 Calculacate wattage:

W = Dcell tq x (Speed to start braking + Speed after braking) x Duty cycle x .0712

Add 20 to 25% just to be sure!

Hoisting applications require a slightly different calculation. So please double check this with your manual. Conslult your drive manufacture. You can usually purchase the resistor bank from the drive manufacture.

Good luck with it!

Mike
 
I guess what's really cloudy to me is how AC drives use them. I gather from what I've read here and there that when you use the resistor with an inverter, it's only there as a path for the drive to dump the DC bus overvoltage during hard decel. and DC braking. Am I correct? If so, this really wouldn't be alot would it? I have only set up one drive so far using this feature. It was a Lenze 8210 with an external chopper and Braking resistor, but this stuff was bought before I started working there so I don't know how it was selected.
I may have some projects coming up in the near future where I'll need to consider this method. It's a lift conveyer, they have tried using drives before and say it always trips on overvoltage during decel on the down-stroke. It's lifted and lowered by a big cam. I thought I'd start investigating early, Thanks alot, guys the info was helpful and I'm looking forward to reading DickDV's response as well.
 
CT782,

If your dealing with a lift type device, and its not somehow counterweighted, the difference in the mechanical inertia may be the problem. If your motor and drive are only sized to pick the load, and not to lower it, it may be that the drive and motor does not have enough power to stop it in the down direction. You may have unwittingly undersized the drive and the motor for the application, if you only considered the HP required to pick the load, but not to stop it in the down direction..

Figure out the inertia of the load in the down direction, compare that with the lifting requirements, and choose the motor and drive according to the highest load. Also if you are using an open loop inverter, and this doesnā€™t provide the control required, you may want to switch to a vector drive with feedback. There are many good drives out there, however in my opinion Magnetek is the master of vector drives in lift type applications. We have had great success the Baldor, and ABB drives also. Call any one of these manufactures, and since yours is a lift application be sure to talk to the elevator drive guys and nobody else.

Whadoya think Dick?

Mike
 
I would agree with the others who have said to go to the drive supplier and let them supply a resistor. However, if you want to size the resistor yourself, I copied the following formulas from the Control Techniques Unidrive manual.

[attachment]


Most drive manufacturers list a minimum resistance for the brake resistor. This value will give approximately 150% braking torque. You should never use a resistor with a lower resistance than the specified value.

Another important point frequently overlooked is insuring the resistor can be disconnected from the power in the event of a fault. The fault could be a shorted brake chopper transistor or a higher than normal line voltage which turns on the chopper. I have seen both conditions occur. Generally a temperature sensor is mounted near the resistor or a current sensor (overload relay) is connected to the resistor. These sensors should drop out a contactor on the line side of the drive if a fault occurs. The resistors are not rated for continuous duty and will burn up if the power is not shut off.

Mike
In a hoisting application without a counterweight, the lowering torque is the same as the raising torque. If the motor and drive can raise the load, it can lower the load if the drive can produce the same torque while braking.

resistor.gif
 
The lift is already in operation, it's using an across-line start contactor and electromechanical brake, but the speed it moves up and down are way too fast using the direct start method, hince the drive idea, to slow it down while lifting and lowering. The motor was changed about 2 years ago as well as the contactors. They put a 5HP motor on it and resized the contactors, wire, etc.. Since that's what they had in stock. It was designed and built with a 3HP. It works fine with the Drive, but you get the overvoltage fault when it goes down, I was going to try using the external resistor, but wasn't sure about sizing it. Don't have a drive picked out yet.
 
Vic,

Thank you for your correction.(y)

No, I'm really not that dumb, and you dont have to be afraid to ride on my elevators.. utoh

I was thinking about the possability of an asymetrical operation of which I would have no experience with but have seen on some RVCs. I totally faild to mention what I was refering to in my last post. I am SWAGing as to why he is overshooting in the down only as I have little to no knowelge of the application being refered to.

HP= (LBS x FPM x (100-ocw))/ (33000 x EFF), so if he has no counterweight it would be figured for more HP. šŸ™ƒ

Our Regen wattage calc for hoisting equipment is:

Watts = (Duty Cycle x LBS x FPM (100-OCW) x EFF)/ 44

Duty cycle = Lowering time/total cycle time

As far as I know, there are different regen watt requirements for hoisting applications, than in general machinery applications such as machine tools and horizontal conveyers etc. If his drive and motor are sized properly, it may be that the requirements for the regen resistors have been calculated for general machinery application and not hoisting applications thus providing a lower wattage & capacity to absorb the regen. Due to the difference in inertia between the up stop and down stop, this problem would most likely manifest it's self in the down direction in a more dramatic fashon.

In CT782's first post he faild to mention that he was dealing with a hoisting application.

To be honest, I dont have to do this often because my drive suppliers fully understand my application, (1 only) and supply all the components, (drive, motor, and regen bank etc..).

Ok Vic, did I talk my way out of it, or did I dig a deeper hole? :oops:

Mike
 
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CT782,

In your last post you said:

"It works fine with the Drive , but you get the overvoltage fault when it goes down, I was going to try using the external resistor, but wasn't sure about sizing it. Don't have a drive picked out yet ".:confused:

I am totally confused. Your post said twice that you are currently using contactors, and you are using a drive?? Please clairify that for me. Do you or Dont you have a drive on the machine. Is it overshooting withe the contactors, or with a drive??? :confused:
 
I think CT782 is mixing some past and present. From what I can tell they currently use contactors and mechanical brakes to handle the application. At some point in the past they tried a drive but ran into bus overvoltage problems (imagine that). CT782 is looking to try the drive angle again but wants to avoid the past problem of overvoltage.

Mike and Vic-
It looks like you are looking at two different facets of the same problem, and you are both correct.
It seems Vic is assuming the same accel and decel rates in all four of the possible accel/decel cases (accel up, decel up, accel down, decel down). In this case Vic is right in saying that if the drive/motor has the torque to raise the load it has the torque to stop it in the down direction. Accelerating while raising will require the most torque.
Mike is looking at this purely from an energy disipation standpoint. In that case he is correct in saying that the greatest energy disipation case occurs during the decel in the down direction. This is where you are taking out the energy due to motion as well as the change in potention energy due to the height change.
Also, I think Mike is using the term 'inertia' a little loosely to mean the total energy that needs to be accounted for to stop as opposed to only the energy component related to mass and accel/decel.
One point I would bring up is Mike's equation:
Our Regen wattage calc for hoisting equipment is:

Watts = (Duty Cycle x LBS x FPM (100-OCW) x EFF)/ 44
According to this equation (assuming I'm reading it right), if you are perfectly counterbalanced you have no regen energy. So if you have a 2,000lb load that you counterbalance with a 2,000lb weight you can stop it without disipating any energy. That's where I get lost.
As for the 'rule of thumb', that is a pretty tough one. I have to agree with everyone else up to this point and recommend you talk to your drive manufacturer. This really depends alot on duty cycle. At the very least you should put together some numbers on total energy to stop. Account for both the total potential energy due to the change in height as well as the kinetic energy you need to absorb to stop this thing. Also come up with some solid numbers for your cycle time information (total lowering time and decel time). The good news is you can buy a whole lot of resistor wattage for very little money if the chopper transistor is already in the drive. That bad news is that alot of resistor wattage will take up alot of room.

Keith
 
Keith,

Let me re-enter that:

Watts=((Duty cycle x Lbs x FPM x EFF)/44)

Thanks for catching that..I got my HP & Regen calculation mixed up togeather in all the excitement.:oops:

Mike.
 
You're correct, Keith, sorry for making is so confusing. I did some other work for this customer and they asked me to take a look at this thing. It IS using contactors at present. In the past they have installed a drive and tried that appraoch. The drive was a Fuji (F9-EVS) if I remember the numbers correctly, it's still mounted in the cabinet without power. It was sized correctly(for the old 3HP motor). They did not try using a DB resistor, they thought I was joking when I asked, they have only seen those on DC drives. They got overvoltage during decel fault every lowering cycle.

OK I'll try to explain this thing, I wish I had a picture, but here goes nothing. You have a motor-gearbox assembly (I have the ratio, but not here with me), on the end of the output shaft of the gearbox you have a piece of round steel approx. 20cm in diameter x approx. 3cm thick. This piece of steel has a pin approx. 3-4cm(diameter) x approx. 5cm(length) sticking out of it, mounted at the edge of the larger ring. If it was a clock the smaller 'drive pin' would be mounted about where the 12 goes and the whole clock turns.
Well the pin goes into a slot in the frame of the conveyer. When the gearbox shaft turns, this pin raises the conveyer, It stops at top position, (pin at 12:00)the whole conveyer then moves to another line,(it lifts and lowers with it's load in place) and lowers the load (pin in 6:00) position, and offloads it's load. It's during the lowering that they get the fault. At 60hz that they're using now the lifting/lowering cycle takes approximately 1-second. So they don't have much leeway with ramp times, although they CLAIM to have tried making it ramp up and down for 5 seconds and still got the fault.

The resistor idea popped into my head, but I'm not sure how to research it enough to be sure it will work in this application. It's simple enough to install, and they have plenty of room. I thought about suggesting that they try it and assisting with setting up the parameters, but if it doesn't work I've wasted their tech's and machine time. I wanted to see what some of you who are alot more drive application savvy thought about it. It would be a learning experience for me so not wasted time, but they might not see it that way.

I think I have alot yet to learn about drives! This site is a big help to us less-experienced controls folks. I know I've learned alot here. Thanks guys!
 
They tried the drive back when they had the 3HP motor on there, it didn't work, so they went back to the contactor system and just didn't ever take the drive out of the cabinet. That's why I haven't considered using that drive. The 3HP is what was put on it from the OEM. When that motor failed, they had a 5HP in stock, so they put that on there so they could get the machine back into production.
They have not tried using a drive on the 5HP that I'm aware of.
The problem since this machine was installed is the load moving up and down so fast is causing the stacks to become uneven. The OEM told them this would stop when the brake disk 'wore in', which ironically takes about the same length of time the warranty takes to expire.
 

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