Omron Sysmac CJ Series, Operation Manual

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65
Using the Heater Burnout Alarm Section 3-11
3-11-3 Determining the Heater Burnout Current
• Use the following equation to calculate the average of the normal current
and the current with a heater burnout:
• If more than one heater is connected through the CT, set the heater burn-
out current to the current induced when the heater with the smallest cur-
rent consumption is burned out. If the heaters all consume the same
current, set the heater burnout current to the current induced when one of
the heaters is burned out.
• The difference between the normal current and the heater burnout current
must meet the following minimum levels:
Heaters drawing less than 10.0 A:
Normal current
−
current with heater burnout
≥
1.0 A
(Operation will be unstable with a difference less than 1.0 A.)
Heaters drawing 10.0 A or more:
Normal current
−
current with heater burnout
≥
2.5 A
(Operation will be unstable with a difference less than 2.5 A.)
• The setting range for the heater burnout current is 0.1 to 49.9 A. Heater
burnout will not be detected if the heater burnout current is set to 0.0 or
50.0 A. The heater burnout alarm will be OFF if the heater burnout current
is set to 0.0; it will be ON if the heater burnout current is set to 50.0.
• The total heater current during normal operation must not exceed 50.0 A.
3-11-4 Example Applications
SV =
Current in normal ope current after heater burnout
2
Heater
1 kW
×
3
CT
200 VAC
Control output
Example 1:
In this example, a single 1-kW 200-VAC
heater is controlled through OUT1.
Normal current =
1,000
200
= 5 A (< 10 A)
Current when control output is OFF = 0 A
5 + 0
2
= 2.5 A SV =
Normal current - control-OFF current = 5 - 0 A = 5 A
(Operation will be stable with difference
≥
1.0 A.)
Heaters
CT
200 VAC
Control output
Example 2:
In this example, three 1-kW 200-VAC
heaters are controlled through OUT2.
Normal current =
1,000
×
3
200
= 15 A (
≥
10 A)
Current with 1 burnout =
15 + 10
2
= 12.5 A SV =
Normal current - control-OFF current = 15 - 10 A = 5 A
(Operation will be stable with difference
≥
2.5 A.)
1,000
×
2
200
= 10 A
1 kW