Issues with a stepper motor drive

Hello folks,

I am facing some issues regarding a stepper motor drive. The stepper motor I have is a 20KG 12V DC L/R Full step driven one. I need it to drive a Lead screw which eventually builds up pressures of the likes of 400Kg cm in a cylinder. I have attached a gear box of ratio 12:1 to the stepper and tried boosting the torque. With the direct coupling the pressure used to reach around 100-150 Kg cm but even with the gearbox which I presume would boost the torque by a factor of 12 the pressure fails to rise above 200. And the stepper slips if I try to run it at high RPMs (15RPM ) even without any load. What can I do to find a solution to these 2 problems? What could be going wrong in this whole exercise? Thanking you in anticipation. Regards, Chetan.

Reply to
chetanthegreat
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On Feb 17, 7:21=EF=BF=BDpm, chetanthegreat wrot= e:

be

you dont say what you are driving the stepper with. its not so easy to get performance from a stepper.

Reply to
cbarn24050

Using a stepper is wrong in the first place. You'll need a proper servo controller and a servo motor. The next thing you need to do is determine the maximum amount of torque required to achieve your goals. However, I guess the required torque depends highly on the amount of friction in the lead screw so it may be difficult to calculate the amount of torque precisely. You may want to get rid of the lead screw and opt for a system with a more constant friction.

--
Programmeren in Almere?
E-mail naar nico@nctdevpuntnl (punt=.)
Reply to
Nico Coesel

The problem with the gear box could be as simple as friction losses thriugh the gear train.

The problem with the stepper motor seem to be that you haven read the application notes carefully enough.

A stepper motor is just a synchronous AC motor. The torque you get out of it is proportional to the AC current that you drive through motor coils,and the faster the motor rotates, the harder it is to drive current through the motor coils.

You have to allow for the back emf of the motor - this can be found from the manufacturers data sheet, though some of them assume you know that the back emf 0f a motor - in volts per radian per second - is the same as the torque constant (when expressed in newton metres per ampere) and you also have to remember that the AC impedance of the coil is the vector sum (root means square sum) of the coil resitance and the coil inducance. The inductive impedance increases in direct proportion to the stepping rate.

This means that you have to apply a lot more than twelve volts to a twelve volt motor to get its rated torque when it is spinning fast,and the drive circuit has to be designed in such a way that the motor won't burn out if you happen to stall it.

See

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Nico Coesel is not entirely right to tell you not to use a stepper motor for your job - the right stepper motor with the right drive could do it - but a brushless DC motor (which is just a stepper motor with a built-in stepper drive) is a lot easier to use.

-- Bill Sloman, Nijmegen

Reply to
bill.sloman

Everyone has given you great advice. Steppers are difficult to drive fast and tend to lose torque when you do. One simple way to improve high speed performance is to use a much higher voltage power supply and drop the current with a resistor. Lots of lost power but faster performance. Generally for what you are doing, you don't want a stepper at all! Get an nice strong DC motor and use some servo scheme to monitor the shaft position. Usually it's some kind of digital encoder. You can buy these as total systems. The beauty of this is the DC motor can have huge torque and the encoder does not "slip". Downside is complexity and expense. A stepper can be simpler and cheaper but you really have to over-rate the torque spec to make it reliable. I had to increase the motor size on a device I built that only moved a drill X-Y with leadscrews and didn't build up any pressure at all! It also used high speed, high voltage drivers so you could get some speed out of the thing.

Reply to
Benj

-- snip --

Be careful with your terminology, though. Many brushless motors don't come with a built-in drive, even ones marketed as "brushless DC".

I wouldn't characterize a brushless DC motor as 'just a stepper with built-in drive' -- there are physical differences in the way they're built that makes each one better for certain tasks. When you get down to the bottom, though, a brushless motor and a stepper are both synchronous AC machines.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html
Reply to
Tim Wescott

You've been given a lot of good advice. Something that has only been alluded to, however, is that you need to figure out what torque is necessary on your lead screw to develop the pressure you need, then figure out what gear box to use to turn your available torque into the torque you need.

Gear boxes (and lead screws) have gear ratios (or lead/turn ratios), and ideal gear boxes will, indeed, multiply torque by the gear ratio while an ideal lead screw will transform a torque into a force in a predictable way (force = 2 * pi * torque / pitch, if I'm not mistaken, but check).

But gear boxes and lead screws also exhibit friction. Lead screws (and worm gears) have _lots_ of friction. You need to check not only the gear ratio or lead, but the efficiency of the gear box and lead screw -- basically, an 80% efficient gear box will deliver your input torque, times your gear ratio, times 0.8.

So check the efficiency of the lead screw, or better yet get an indicating torque wrench and _measure_ the torque it takes to develop the pressure you want. Then use a gearbox with a known efficiency and enough of a ratio to develop the torque you need with the amount of torque that you can trust to get out of whatever motor you use, whether you stick to your stepper or go to another sort of motor (a brushed DC motor is much lower tech, by the way, and may be easier to implement if you're doing the drive electronics).

Good luck. If you succeed, you'll come out of this smarter than you were going in.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html
Reply to
Tim Wescott

Thanks for the help people! Thing is I am kind of stuck with the stepper I bought for this application. And for the first few rotations I don't need precise movement or much torque but the last revolution is highly demanding as far as torque and precision is concerned. Today, I tried hanging weight on the cross bars handle attached to the lead screw and found out that at around 20-25 Kg weight the lever moves in order to develop the ultimate 400 Kg/cm pressure in the gauge. As the distance from the centre was 16 cm, the calculated torque needed would come around 20 / 25 x 16 Kg cm. Thus I am hoping that a 60 Kg Stepper with a 12:1 gearbox will be suffice to drive the lead screw in. Please comment if you could point out a flaw in my logic. Regards. Chetan.

Reply to
chetanthegreat

Perhaps you could use a counterweight which is equal to half the maximum pressure you need on the rod (assuming the cylinder is standing). This way the stepper engine needs to deal with half the load.

--
Programmeren in Almere?
E-mail naar nico@nctdevpuntnl (punt=.)
Reply to
Nico Coesel

On Feb 18, 6:09=EF=BF=BDpm, chetanthegreat wrot= e:

ing

First you need to fing the torque needed at the motor shaft at the speed you need. That means you'll need to measure it or try guessing. Stepper performance depends on the drive, you seem to keep ignoring that, so you need a motor/drive pakage, ie. you have to buy one or guess. Then look at the torque/speed curve published for said package to see if it exceeds your minimum requirement. Repeat untill you have the best match.

Reply to
cbarn24050

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