Table 25: Module 765-1103/100-000 - Currents In Supply Line 1; Table 26: Module 765-1103/100-000 - Currents In Supply Cable 2; Table 27: Rules For Supply Cables 1 And 2 - WAGO 765-1103/100-000 Manual

Planning
Pin
4
The device is designed in such a way that it can be used in overload operation according
to the following table "Module 765-1103/100-000 – Currents in Supply Cable 1" without
time limit. The following applies to all pins:
If the maximum load capacity (upper limit of overload operation) of the pin is exceeded,
there is a risk of destruction of or damage to the pluggable connector or PCB of the de-
vice.
CAUTION
Note significance of pin 3
Note that the sum of the currents of pins 2 and 4 flows to pin 3.
Product-Specific Information
For supply line 1 of the 765-1103/100-000 Digital Output Module, the currents declared in
the following table must be taken into account:
Current
I
X31_1L
I
X32_1L
I
logic
I
module_1L
For this module, supply cable 1 is only used for transmission between the PWR IN (X31)
and PWR OUT (X32) connections.
For supply cable 2 of the 765-1103/100-000 Digital Output Module, the currents declared
in the following table must be taken into account:
Current
I
X31_2L
I
X32_2L
I ,
X01_2L
I ,
X02_2L
I
module_2L
During operation of the 765-1103/100-000 Digital Output Module, the rules below for the
currents in supply cables 1 and 2 must be followed at all times:
Current
Total current for the supply cable
(rule 1)
Permissible current feedthrough
(rule 2)
36 Product manual | Version: 2.2.0
16DO FLD PN DC 24V 2.0A
I
... , X08_2L
Normal Operation
2 A
Table 25: Module 765-1103/100-000 – Currents in Supply Line 1
Description
Current to connection PWR IN (X31): current 1L+ / reverse current 1L−
Current to connection PWR OUT (X32): current 1L+ / reverse current 1L−
Logic supply
Module current I = 0
module_1L
Table 26: Module 765-1103/100-000 – Currents in Supply Cable 2
Description
Current to connection PWR IN (X31): current 2L+ / reverse current 2L−
Current to connection PWR OUT (X32): current 2L+ / reverse current 2L−
Total current for supply cable 2 on digital output X0i (i.e., port X01,
X02, ... , X08); corresponds to current I
current is the sum of the currents on pins 2 and 4 of digital output X0i:
I = I = I + I
X0i_2L X0i_pin3_2L X0i_pin2_2L X0i_pin4_2L
Module current I I
module_2L = X01_pin3_2L
Supply Cable 1
I ≤ 16 A
X31_1L
I = I + I I
X31_1L logic X32_1L + module_1L
I ≤ 16 A − I − I
X32_1L logic module_1L
765-1103/100-000
Overload Operation
2.4 A
on pin 3 (ground). This
X0i_pin3_2L
+ I + ... + I
X02_pin3_2L X08_pin3_2L
Table 27: Rules for Supply Cables 1 and 2
Supply Line 2
I ≤ 16 A
X31_2L
I = I + I
X31_2L X32_2L module_2L
I ≤ 16 A − I
X32_2L module_2L