Hi,
I've got two siemens 1FL3042 motors I would like to use for my x/y-axes. However, I'm not sure how to control/drive these steppers.
any suggestions or diy circuit would be great
Best regards
Thomas
Hi,
I've got two siemens 1FL3042 motors I would like to use for my x/y-axes. However, I'm not sure how to control/drive these steppers.
any suggestions or diy circuit would be great
Best regards
Thomas
Your going to have significant difficulties finding a driver to work with these... they are 3-phase high voltage motors (325v @ 2A per phase). The proprietary Siemens FM_Stepdrive system has a step control unit operating from 24v DC and a motor power unit operating from 115/230v AC. Some details regarding typical application/wiring to a PLC can be found here and the user manual for the overall Siemens FM_Stepdrive system here
Hope this helps (probably doesnt if you see what I mean )
Hi irving2008, thanks for your reply
yea...I know it won't be easy to find a driver that works..I also noticed the FM-stepdrive unit and so on..
I am used to read electronic circuits/schematics but I am not educated in electronics. However, I'll try to build my own driver. 325v/2A supply is not a problem, I just need a driver schematic for a high voltage AC stepper. I've found the term 'half H-bridge' and guess that's the starting point.
Thanks
Thomas
Have a look at the Sanken 7600D datasheet. This is a 3-phase stepper control IC with everything on-board except the MOSFET output stages. Given the voltages in question a discrete output stage is probably a must and some care in design and layout will be needed... transient voltages will be potentially problematic. As the 7600D includes the high-side drivers some thinking is required to handle the voltage differential between the 7600D high-side outputs and the actual high-side drivers - I think a simple resistive arrangement might work OK if the Vbb rail is the same as the Vcc logic rail but given the output voltages I'd consider implementing opto-isolation for the output stage which would also overcome the voltage differentials.
The actual MOSFETs would need to be driven by hi-voltage drivers such as irf2114 as these are one of the few devices that can handle that offset voltage.
Attached is an outline concept diagram... datasheets should give you more ideas and specifics but as I said before the layout of the output stage will be critical - there will be >500v transients floating around here...
I doubt you'll find an off-the-shelf design for this requirement and be prepared to fry a few devices getting this to work...
Also I dont need to reiterate the dangers of DIY and >100v, do I?
Hi again, thanks very much for your help irving2008
I'm about to give up on this...I think I'll go with something more usual.
I misunderstood the 325/2A per phase. Such a power supply would kill my budget. And since 4Nm/3Nm is available in lower voltage range, I think I'll go with that instead and just sell the siemens on ebay. The drivers are also available.
However, I'm new to steppers and have tried to read the FAQ docs found on the net.
My machine is constucted of aluminium - solid 30mm x 80mm flat bar. I would like to be able to route in aluminium bar and maybe light steel plate. Can you give me a quick answer on what to look for when choosing the motors?
btw, the table is 800mm x 500mm and the x-axis has two leadscrews
Best regards
Thomas
Thomas,
You'll need to do some calculations to tell you what torque etc. you'll need. There are several threads on this subject, however if you answer the following I could give you a clue...
type/pitch/# starts of leadscrew on X, Y, Z
type of linear bearings on each axis and travel
weight of cutting head on Z axis
weight of Z assembly on Y axis
weight of Y+Z on X axis (I'm assuming moving gantry here and not moving table)
expected rapid traverse speed on each axis
max cutting speed on each axis
max dia of cutter planned
max depth of cut planned
router spindle speed
stepper to leadscrew connection (direct, belt/geared?)
I think thats all... I'd prefer metric values (its so much easier to work out!)
thanks
type/pitch/# starts of leadscrew on X, Y, Z = trapezoidal ISO TR14x4 guess 0.004m/turn one start
type of linear bearings on each axis and travel = 20mm Sinterbronze for x,y and 16mm for z
weight of cutting head on Z axis = around 2-3kg incl. router
weight of Z assembly on Y axis = 5kg max (16mm x 320mm steel rail x 2 + C-construction z).
weight of Y+Z on X axis = Yes, moving gantry - around 20kg (20mm x 600mm steel rail x 2 on y-axis)
expected rapid traverse speed on each axis:
max cutting speed on each axis: if wood x,y = 0.1 - 0.2m/s - z = not important
max dia of cutter planned: need help on this one , depends on material
max depth of cut planned: this one too
router spindle speed: depends on material - up to 30.000rpm
stepper to leadscrew connection (direct, belt/geared?): Direct, via L050 jaw coupling
I want an 'overall machine' with enough power - to cut wood, aluminium and light steel plate.
I was reading a document - 'what motor and gear should i use'. but don't understand all what it's saying. I guess 3 x 3Nm (x,y-axes) and 1Nm holding torque for the z-axis will do the job. But it would be nice to be sure, of course
Thanks again
Thomas
Btw, I found these - £20 each.
http://www.slidesandballscrews.com/p...H88-3008DF.pdf
Or you could probably use berger lahr drivers.
http://www.schneider-electric-motion...01&language=de
can be found on ebay sometimes.
Hi irving2008, hello gotis
I'm looking forward to see your calculations , thanks
gotis, thank you for the info!
best regards
Thomas
Btw irving2008, I'm not sure about the travelling/cutting speed issue. I've read that 1000RPM is the usual need.
Thomas,
Here's my thoughts...
Torque is required to overcome the bearing friction, which for oilite bearings is roughly 0.3x the load and for linear/ball bearings is 0.1x. This is M * g * L * u/(2pi * e) where M= mass of load, g = gravity, L is lead/pitch, u = friction coeff and e = efficiency of screw (.8 for ball, .2 for acme)
Taking worst case, which is the X axis, your working load is 12kg which gives 12 * 10 * .004 * .1/(6.28 * .2) = 0.04Nm
Torque is also required is to resist the cutting forces. For TR14x4 leadscrew this is approx .00256W Nm (see here for method) where W is applied load in N
Cutting forces can be estimated as the force due to the maximum spindle torque which is related to spindle speed and revs...
assuming a 600W spindle at 12000rpm (=200rps) the torque produced is 600/(200 * 2* pi) = 0.5Nm, so with a typical 10mm dia cutter the maximum available cutting force is in the order of (Torque/cutter dia) = (.5/.005)N = 100N
So we have to resist a 100N cutting force, which requires 100 * .00256 = 0.256Nm of torque.
So the total torque when cutting is 0.04 + 0.256 = 0.3Nm
This torque is needed at the cutting speed. A typical cutting speed would be 2m/min or .03m/s (roughly 72 inches/sec)
To maintain 0.03m/s on a 4mm leadscrew requires 0.03/0.004 revs/sec = 7.5rps or 7.5 * 200 steps/sec = 1500 steps/sec (with a resolution of .004/200 = 0.02mm/step)
To accelerate we need to overcome the inertia of the screw and load in order to accelerate it to the required speed without losing steps. For the leadscrew the rotational inertia is pi/32*density * length * diameter^4 = 3.142/32*7850*.8*.014^4 = 2.37x10^-5 Kgm^2 and for the load the rotational inertia is W * (L/2pi)^2 Kgm^2 (W= mass, L = pitch of the screw) = 12 * (.004/6.28)^2 = 4.87x10-6 Kgm^2 and the inertia of the motor itself is typically 4x10^-5Kgm^2 so total inertia (given a direct drive or 1:1 gearing) is 6.86x10^-5 Kgm^2. Typical acceleration for a small to medium machine is 2000 rad/sec^2 (equivalent to a linear acceleration of 2000L/(2pi) m/sec^2 or 2000*.004/6.28 = 1.27m/sec^2. The required torque is 2000 * 6.86x10^-5Nm = 0.14Nm.
This accelerating torque is added to the motion torque as it represents the ability of the machine to change direction when cutting, so total torque = 0.3 + 0.14 = 0.44Nm
At 1500steps/sec the motor will be operating at about 40%of holding torque, so the required nominal holding torque is 0.44/0.4 = 1.1Nm.
So, a motor capable of a minimum of 1.1Nm holding torque is required. Allowing for some inefficiencies and a margin of error 2 - 3Nm would do nicely. This is as you surmised, but its nice to know that its right...
Hi irving2008!
That is some heavy calculations you came up with! Thanks very much! I just stored your reply in my documents.
I think I will go with the 'CNC driver kit 1' from mcpltd.
I've got an Astrosyn L189/2 Phase stepper rated 1.6A, 5.4V /phase which I want to use for the z-axis. However, the 'CNC driver kit 1' only supports three motors - and the breakout supports three axes. If I buy an extra driver for the L189 is it then possible to connect two drivers to the same slot (the x-axis) at the breakout board?
Thanks
Thomas