Pulse output and CJ2m

vermaut

Member
Join Date
Oct 2010
Location
belgie
Posts
21
Hi

Newbie with servo and pulse output

I use pulse output cj2m and junma pulse .
I have read the manual but do not understand it
they say:
Output frequency 1 pps to 100 kpps (1 pps units), two pulse outputs × 2 Pulse I/O Modules.
Frequency acceleration and deceleration
rates
Set in increments of 1 pps for acceleration/deceleration rates from 1 to 65,535
pps (every 4 ms).
The acceleration and deceleration rates can be set independently only with
the PLS2 instruction.
Internal pulse control cycle 1 ms or 4 ms (Set in the PLC Setup.)
I have read that the junma driver 1000 pulses per revolutie.


How can i calculate the acceleration and decelleration in pulses?
And the travel distance?


Can some one help me? I want to learn it

Thanks

Regards
 
Welcome to the forum.
I deduced from your post that you are using a CJ2M CPU with a CJ2M-MD21 Pulse I/O Module.

"Output frequency 1 pps to 100 kpps (1 pps units), two pulse outputs × 2 Pulse I/O Modules."

This means the module can control 2 axes.

"Frequency acceleration and deceleration
rates set in increments of 1 pps for acceleration/deceleration rates from 1 to 65,535
pps (every 4 ms)."

Minimum pps is 1 and maximum is 65,535. Response time to a pulse rate change is 4 ms.

"Internal pulse control cycle 1 ms or 4 ms (Set in the PLC Setup.)"

Don't mess with this. Ignore it.

"How can i calculate the acceleration and deceleration in pulses?
And the travel distance?
"

See attached pdf.

"And the travel distance?"

Travel is based entirely on the mechanical components connected to the servo motor and how much they move for each rotation of the motor..
 
Last edited:
Thank you for the reply and pdf file.
I use junma pulse

I try to calculate.

move 10mm
1 revolution= 6 mm
10/6 = 1.6667 revolutions
2048*4 pulses = 1 revolution
1.6667*2048*4 = 13653 pulses

so i think i need 13653 pulses to move 10mm
and i try to calculate with pdf file

Regards
 

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