I need clarity on quadrature encoder (by Ziggy)
OK I need clarity on quadrature encoder
Here is the task
A lead screw of 4mm pitch needs to transport a bracket over a distance of 500mm in a time of 7 seconds.
The distance is therefore equivalent to 125 revolutions of the screw.
The distance is traversed in 7 seconds therefore the screw runs at 125 revolutions per 7 seconds or 1071 RPM.
The screw will be driven by a DC motor which has a quadrature encoder on motor shaft of 13 pulses per phase or four times this rate 52 pulses out of the quadrature encoder.
The motor drives the screw through a 5: 1 reduction gearbox which means motor will run at 1071 * 5 =5355 RPM
So in operation the quadrature decoder will have to feed 5355 * 52 pulses / min.
This translates to 4641 pulses per second.
What should the PLC loop time be?
A bit better than 1/4641 seconds? =0.2mS
OR four times quicker? = 0.05mS
Looking forward to Your comments
(no subject) (by MGP)
4641Hz and you need at least 2 cycles to detect a high and a low signal, that means 9282cycles/sec minimum.
For not missing one puls you have to scan it faster, minimum X3.
As you can see it's not possible with LDmicro.
Why not 1 puls/rev?
(no subject) (by Ziggy)
These are small brush dc motors with a gearbox.
The encoders are 13 pulses per revolution quadrature outputs.
I could select to use only a single channel and operate at reduced resolution.
This would mean going away from 0.077 mm resolution to 0.31mm resolution.
0.31 is not going to be too bad.
1 Pulse per revolution would be too coarse.
(no subject) (by Ziggy)
The project has been deferred indefinitely although just today the interest has been revived... i will be seeing the fellow who needed this solution in a few days ( january 15th after Christmas ),
Will know more what to think about this issue.