NORD Drivesystems BU0530 - Manual (Page 30)

Functional safety with NORD SK 500E frequency inverters
30 BU 0530 GB-3712
3.4.1 Connecting the safe pulse block (X8 – VIS 24V, VIS 0V)
SK 51xE SK 53xE and SK 54xE series frequency inverters are equipped with the option "Functional safety".
Connection of the corresponding control cables is made via terminal block X8. Up to and including size 4, this
terminal block and all other control connections are located under the front cover (See Section 3.4.2
"Connection
terminals "). For sizes 5 to 7, the terminal block X8 is located on the underside of the frequency
inverter (motor output side).
A two-wire shielded cable must be used for the safe pulse block. The shield must be connected at on both
ends. For mechanical switching devices the line drop in the cable must not exceed
ΔU
Cable
≤ 3 V and for
electronic protective switching devices it must not exceed ΔU
Cable
≤ 1 V. For the calculation, a peak current of
I
IN,Peak
= 400 mA should be assumed.
Operation at OSSD
The safe pulse block is specially designed for use with an OSSD.
The capacity between the wires (including the shielded capacities) must not exceed a value of 20nF for each
inverter connected.
Additional restrictions may apply here with regard to the protective switching device.
EMC
When the „safe pulse block“ has been wired correctly for EMC, the recommended EMC values (See BU 0500
GB) can be complied with up to a cable length of 20m between the safety switching device and the inverter.
Multi-device operation
When operating several inverters at one safety switching device, the switching capacity of the switching device
and the load rating of the 24V power-feeding supply must be observed.
The shield must be applied to all inverters.
The permissible line drops on the cable must be observed!
Example:

5x frequency inverter at an electronic safety switching device!

The 5x FIs are arranged next to each other in a switching cabinet

20m must be bridged between the FI and the switching device!

Shielded cable 2x1.5 mm
2
q
l
R
CU
*
 
with
km
mm
CU
2
* 19
  
Double the length of cable must be used because line drops occur in both wires.
 
5 , 0 R
V A I number R U
Peak IN FU cable
1 4 , 0 * 5 * 5 , 0 * *
,
    
=> OK!