E2S D1xB2X Series, Инструкция по эксплуатации

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European Safety Systems Ltd. Impress House, Mansell Road, Acton, London W3 7QH sales@e2s.com Tel: +44 (0)208 743 8880
www.e2s.com Fax: +44 (0)208 740 4200
Document No. D191-00-601-IS Issue 1 20-03-2020 Sheet 10 of 8

16-2 SIL 2 system wiring for fault detection in standby mode only – 2 wire installation
The customer is required to wire into power supply terminal only. The unit will be monitored in standby mode only, via an customer
installed system EOL resistor (2.2k
Ω suggested customer EOL and default 2.2k Ω fault resistor will draw a total current of 35.9mA
@ 24Vdc as shown in table 2).
In the event of a fault, The SIL 2 unit will automatically place the power supply terminal fault resistor across the power terminals
which already has customer EOL resistor (2.2k
Ω ) in place. This will result in a total fault detection current of 41.8mA @ 24V but
can only be detected when unit is in Standby Mode.
If the customer chooses to use this configuration within their system, it must be noted that the factory default settings for the unit
does not have an EOL resistor installed. The customer can request E2S to install an EOL resistor and this will be depicted in the
product code. See section19 for further information on EOL and fault resistor value choice.
Important
: - This configuration will not warn of a fault whilst in Active mode as the PLC will be supplying the unit with power.
The PLC will only be able to see the fault when in standby mode, by measuring the fault detection current.
Important
: - This configuration requires the customer to set J2 header pin to be set to position B (see figure 9), as the units
default position is A.
Figure 7 - Schematic of SIL 2 system wiring for fault detection in standby mode only – 2 wire installation
To evaluate the total current drawn from the SIL 2 unit, use the equation below.
In standby mode, where there is no fault, RLY 1-1 is open. This means the voltage only passes through the customer EOL resistor
and the current drawn from the SIL 2 board is 25mA. Therefore, the equation for a No Fault scenario is then:
In standby mode, where there is a fault, the circuit is closed. This means the voltage passes through both the customer EOL
resistor and current sense resistor and the current drawn from the SIL 2 board is 20mA. The customer must first calculate the
resistance of the two resistors in parallel before applying the currents to the equation. The equation for a Fault scenario is then:
I
(Total Current
drawn)
=
I
(Current drawn from
Fault Resistor)
+
I
(Current drawn from
Customer EOL resistor)
+
I
(Current drawn
from SIL board)
I
(Standby Mode, Total Current drawn - No Fault)
=
I
(0mA)
+
I
(See table 2)
+
I
(25mA)
I
(Standby Mode, Total Current drawn - Fault)
=
I
(Total Resistance when EOL & FR in parallel)
+
I
(20mA)
Standby
Mode
Power Supply Fault Resistor Customer EOL Resistor (Fault Mode Only)
Current
drawn
from SIL
Board
Total
current
drawn
Resistor
Value
Current
drawn ( I )
Resistor
Value
Current
drawn (
I
)
Total
resistance
Current
drawn (
I
)
No Fault
2.2 k Ω
0 mA
2.2 k Ω
10.9 mA - - 25 mA 35.9 mA
Fault - - 1.1 k Ω 21.8 mA 20 mA 41.8 mA
No Fault
1.0 k Ω
0 mA
1.0 k Ω
24.0 mA - - 25 mA 49.0 mA
Fault - - 500 Ω 48.0 mA 20 mA 68.0 mA
No Fault
2.2 k Ω
0 mA
3.3 k Ω
7.3 mA - - 25 mA 32.3 mA
Fault - - 1.3 k Ω 18.2 mA 20 mA 38.2 mA
No Fault 1.8 k Ω 0 mA 3.9 k Ω 6.2 mA 25 mA 31.2 mA