Nord Drivesystems SK 750E-751-323-A - Terminal Assignment; Control Terminals of Customer I;Os

3.5 Terminal assignment
BU 0750 GB Technical design subject to change 51
3.5 Control terminals of customer I/Os
Function
Data
Name/
assign-
ment
Customer intreface / special extension units
Terminal
Relay
N/O contact
I
max
= 2A
U
max
= 28V DC / 230V AC
BSC
STD
MLT
MLT 20mA
CAN-RJ
PBR
POS
ENC
REL 1.1
X3.1.01
X1.1.01
X2.1.01
X2.1.01
X7.1.01
X6.1.01
-
-
REL 1.2
X3.1.02
X1.1.02
X2.1.02
X2.1.02
X7.1.02
X6.1.02
-
-
REL 2.1
-
X1.1.03
X2.1.03
X2.1.03
-
-
-
-
REL 2.2
-
X1.1.04
X2.1.04
X2.1.04
-
-
-
-
REL 3.1
-
-
-
-
-
-
X10.1.05
-
REL 3.2
-
-
-
-
-
-
X10.1.06
-
REL 4.1
-
-
-
-
-
-
X10.1.07
-
REL 4.2
-
-
-
-
-
-
X10.1.08
-
Reference volt-
age source
+10V
I
max
= 10 mA
BSC
STD
MLT
MLT 20mA
CAN-RJ
PBR
POS
ENC
VREF 10V
X3.2.11
X1.2.11
X2.2.11
X2.2.11
-
-
-
-
GND reference
potential
Reference potential for the
inverter connected to PE via
resistor and contactor
BSC
STD
MLT
MLT 20mA
CAN-RJ
PBR
POS
ENC
AGND /0V
X3.2.12
X1.2.12
X2.2.12
X2.2.12
-
-
-
-
GND /0V
-
X1.4.40
X2.2.40
X2.2.40
X7.2.40
X6.2.40
X10.3.40
X11.1.40
X10.4.40
Analogue
inputs
AIN1 = differential voltage
input with 0V ... 10V
Ri 40 k
BSC
STD
MLT
MLT 20mA
CAN-RJ
PBR
POS
ENC
AIN1 -
X3.2.13
X1.2.13
-
-
-
-
-
-
AIN1 +
X3.2.14
X1.2.14
-
-
-
-
-
-
AIN1 + AIN 2 = -10V...+10V
Ri 20 k
AIN1 +
-
-
X2.2.14
X2.2.14
-
-
-
-
AIN2 +
-
-
X2.2.16
X2.2.16
-
-
-
-
Analogue
output
0V ... 10V
I
max
= 5 mA
Resolution = 8 bits
Accuracy = 0.1 V
BSC
STD
MLT
MLT 20mA
CAN-RJ
PBR
POS
ENC
AOUT1
-
X1.2.17
X2.2.17
X2.2.17
-
-
-
-
AOUT2
-
-
X2.2.18
X2.2.18
-
-
-
-
Digital input
Ri 4 k
High = 7,5V .... 33 V
Low = 0V ... 7,5V
Time of response =
5ms...15ms
BSC
STD
MLT
MLT 20mA
CAN-RJ
PBR
POS
ENC
DIG IN 1
X3.3.21
X1.3.21
X2.3.21
X2.3.21
X7.2.21
X6.2.21
-
-
DIG IN 2
X3.3.22
X1.3.22
X2.3.22
X2.3.22
X7.2.22
-
-
-
DIG IN 3
X3.3.23
X1.3.23
X2.3.23
X2.3.23
X7.2.23
-
-
-
DIG IN 4
-
X1.3.24
X2.3.24
X2.3.24
X7.2.24
-
-
-
DIG IN 5
-
-
X2.3.25
X2.3.25
X7.2.25
-
-
-
DIG IN 6
-
-
X2.3.26
X2.3.26
-
-
-
-
NOTE: >BUS< option use
only DIG IN 1 as temp. sen-
sor input, and with >MLT<
only DIG IN 6!
The following will apply:
Ri 2 k
High = 2.5V .... 33 V
Low = 0V ... 2.5V
DIG IN 7
-
-
-
-
-
-
X10.2.27
-
DIG IN 8
-
-
-
-
-
-
X10.2.28
-
DIG IN 9
-
-
-
-
-
-
X10.2.29
-
DIG IN 10
-
-
-
-
-
-
X10.2.30
-
DIG IN 11
-
-
-
-
-
-
X10.2.31
-
DIG IN 12
-
-
-
-
-
-
X10.2.32
-
DIG IN 13
-
-
-
-
-
-
-
X11.1.33
Voltage supply
+15 V
Sum of the currents of all
power supply units on a sin-
gle inverter:
I
max
= 300 mA
BSC
STD
MLT
MLT 20mA
CAN-RJ
PBR
POS
ENC
VO +15 V
X3.3.42
X1.3.42
X2.3.42
X2.3.42
X7.2.42
X6.2.42
X10.2.42
X11.1.42
Voltage supply
+5 V
BSC
STD
MLT
MLT 20mA
CAN
PBR
POS
ENC
VO +5 V
-
X1.4.41
X2.3.41
X2.3.41
-
X6.2.41
X10.4.41
X11.1.41