HELLO; I have V350-35-T2 PLC controler. At imput 3&4 (input I2 & I3) is shaft encoder (signal A and signal B from shaft encoder). Shaft encoder value is defined in program and this value decreas and increase, depends on witch way, shaft encoder turn. Reset shaft encoder value is in software at 360° (1440 - x4). With this value i can count turns + i have control over angle. frequency (max.) for this PLC is 20kHz (at 10m of cable, i have 5m of it). My first question is, if frequncy is halfed (10kHz) becuz I use channel A and ? I can then use only 10kHz? Problem occures at 100RPM (its the lowest RPM of motor) width subtract angle. When first 360° is over at my PLC i see 359, 358,357..depending on speed of motor. Moore time shaft is turning the lowest is number. The error is integreated over the time period. So at 30 turns i have left only with around 330° (depends what speed I use - arround 120 RPM).
Shaft encoder have 360 PPR and he is at 100RPM...This is 600Hz.
I dont get it, why 600Hz is to fast for my PLC. I configure input 2 & 3 as shaft encoder (x4) + reload. Reload accure at 360° rt at 1440°(cuz number is multiplayed x 4) (at full circle).
This shaft is mounted on mashine. This mashine create one product in 360° and stoped at exactly 360° (0°) -> full circle.
its seems that shaft encoderis is not able so quickly read angle? But i dont geet it - shaft encoder is high speed encoder for max speed 6000 RPM. Maybe there is something wrong with input of PLC? I configure it as shaft encoder etc...
Shoud i wrote PLC program in 2.5 mS interrupt? My program have just 3 lines of code (linearization, counting pulzes, reset at 360°). That's it. Should i use immediate function blocks?
Can i use them in "normal" program, or they must be in interrupt, otherwise they will not work properly?
thx for the answers
Best regards Dejan Rožič