The basic dilemma is where to mount the sensor. If you use a rotary sensor, you get a 4 micron
backlash error. On the other hand, if you use a linear encoder, the backlash in the feedback loop
will cause oscillations due to instability.
An alternative approach is the dual-loop, where we use two sensors, rotary and linear. The rotary
sensor assures stability (because the position loop is closed before the backlash) whereas the linear
sensor provides accurate load position information. The operation principle is to drive the motor
to a given rotary position near the final point. Once there, the load position is read to find the
position error and the controller commands the motor to move to a new rotary position which
eliminates the position error.
Since the required accuracy is 0.5 micron, the resolution of the linear sensor should preferably be
twice finer. A linear sensor with a resolution of 0.25 micron allows a position error of +/-2 counts.
The dual-loop approach requires the resolution of the rotary sensor to be equal or better than that
of the linear system. Assuming that the pitch of the lead screw is 2.5mm (approximately 10 turns
per inch), a rotary encoder of 2500 lines per turn or 10,000 count per revolution results in a rotary
resolution of 0.25 micron. This results in equal resolution on both linear and rotary sensors.
To illustrate the control method, assume that the rotary encoder is used as a feedback for the X-
axis, and that the linear sensor is read and stored in the variable LINPOS. Further assume that at
the start, both the position of X and the value of LINPOS are equal to zero. Now assume that the
objective is to move the linear load to the position of 1000.
The first step is to command the X motor to move to the rotary position of 1000. Once it arrives
we check the position of the load. If, for example, the load position is 980 counts, it implies that a
correction of 20 counts must be made. However, when the X-axis is commanded to be at the
position of 1000, suppose that the actual position is only 995, implying that X has a position error
of 5 counts, which will be eliminated once the motor settles. This implies that the correction
needs to be only 15 counts, since 5 counts out of the 20 would be corrected by the X-axis.
Accordingly, the motion correction should be:
Correction = Load Position Error - Rotary Position Error
The correction can be performed a few times until the error drops below +/-2 counts. Often, this is
performed in one correction cycle.
Example motion program:
Instruction Function
#A Label
DP0
Define starting positions as zero
LINPOS=0
PR 1000
Required distance
BGX Start
motion
#B
AMX
Wait for completion
WT 50
Wait 50 msec
LIN POS = _DEX
Read linear position
ER=1000-LINPOS-_TEX
Find the correction
JP #C,@ABS[ER]<2
Exit if error is small
PR ER
Command correction
BGX
JP #B
Repeat the process
#C
EN
DMC-1600
Chapter 7 Application Programming
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