The release voltage (7.5 V) is more than 2.7 V. The wire break detection current leak will not activate this relay.
Relay coil resistance calculation example 2
For a 12 V supply, a relay with a 0.6 V release voltage and an 848 Ω coil is proposed.
The right side of the equation is then (12 V - 4.5 V) / (3900 Ω + 848 Ω) × 848 Ω = 1.3 V.
The release voltage (0.6 V) is less than 1.3 V. The wire break detection current leak will activate this relay. Use a bigger
relay, or use an external resistor to prevent relay activation.
Using an external resistor to prevent relay activation
If you do not need to detect a wire break in the stop coil, you can install an external resistor to stop the wire break detection current
leak from activating the relay.
Figure 6.23
Wiring example for external resistor to stop the wire break current leak from activating the relay
Use the following formula to calculate the maximum resistor size (in ohms):
R
resistor
< R
coil
× V
release
× (2 × R
coil
+ 7800) / (2 × R
coil
× V
supply
- 9 × R
coil
- 7800 × V
release
- 2 × R
coil
× V
release
)
CAUTION
This formula does not include a safety factor.
INFO
If you get a negative result on the right side, then you do not need a resistor.
External resistor size calculation example
For a 24 V supply, a relay with a 1.2 V release voltage and a 3390 Ω coil is proposed. The wire break detection current will
activate this relay, and so an external resistor is required.
The external resistor must have less resistance than:
3390 × 1.2 × (2 × 3390 + 7800) / (2 × 3390 × 24 - 9 × 3390 - 7800 × 1.2 - 2 × 3390 × 1.2) = 517 Ω
INSTALLATION INSTRUCTIONS 4189340909I UK Page 80 of 144
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