BECKHOFF TwinSAFE - Manual (Page 74)

Circuit examples
72 Application Guide TwinSAFE
The following assumptions have to be made now:
The door switches S1/S2 must both be actuated. Since the switches have different values, but the
complete protective door switch consists of a combination of break and make contacts and both switches
must function, the poorer of the two values (S1) can be taken for the combination!
Relays K1 and K2 are both connected to the safety function. The non-functioning of a relay does not lead
to a dangerous situation, but it is discovered by the feedback. Furthermore, the B10d values for K1 and
K2 are identical.
The tumbler is mechanically connected to the switch S2 in such a way that a separation of the coupling is
impossible.
The restart is monitored, so that a signal change is only valid once the door is closed.
There is a coupling coefficient between the components that are connected via two channels. Examples
are temperature, EMC, voltage peaks or signals between these components. This is assumed to be the
worst-case estimation, where ß =10%. EN 62061 contains a table with which this ß-factor can be
precisely determined. Further, it is assumed that all usual measures have been taken to prevent both
channels failing unsafely at the same time due to an error (e.g. overcurrent through relay contacts, over
temperature in the control cabinet).
This produces for the calculation of the PFH value for block 1:
PFH
tot
=
β
* (PFH(S2|Lock|UNlock|EL2904|Tumbler)+ PFH(S1))/ 2 + PFH(EL1904) +
PFH(EL6900) + PFH(EL2904) +
β
* (PFH(K1)+ PFH(K2))/2+PFH(EL1904) +
PFH(Restart)
with
PFH(S2|Lock|UNlock|EL2904|Tumbler) = PFH(S2)+
β
* (PFH(Lock)+PFH(Unlock))/2 +
PFH (EL2904) + PFH (Tumbler)
to:
PFH(S2|Lock|UNlock|EL2904|Tumbler) = 8.4E-10 +
10%
* (1.68E-08+1.68E-08)/2 +
1.25E-09 + 8.40E-10 = 4.61E-09
PFH PFH PFH PFH
tot tot tot tot
= == =
10 %*
(4.61E-09+8.40E-09)/2 + 1.11E-09 + 1.03E-09 + 1.25E-09 + 10%*
(1.29E-09+1.29E-09)/2 + 1.11E-09 + 1.68E-09 = 6.96E 6.96E 6.96E 6.96E----09 09 09 09
The MTTF
d
value for block 1 (based on the same assumption) is calculated with:
1
ܯܶܶܨ
ௗ ௧௢௧
= ෍
1
ܯܶܶܨ
ௗ ௡
௡
௜ୀଵ
as:
1
ܯܶܶܨ
ௗ ௧௢௧
=
1
ܯܶܶܨ
ௗ
(ܵ2|ܮ݋ܿ݇|ܷ݈ܰ݋ܿ݇|ܧܮ2904|ܼݑ݄݈ܽݐݑ݊݃) +
1
ܯܶܶܨ
ௗ
(ܧܮ1904)
+
1
ܯܶܶܨ
ௗ
(ܧܮ6900) +
1
ܯܶܶܨ
ௗ
(ܧܮ2904) +
1
(ܯܶܶܨ
ௗ
(ܭ1))
+
1
(ܯܶܶܨ
ௗ
(ܧܮ1904)) +
1
(ܯܶܶܨ
ௗ
(ܴ݁ݏݐܽݎݐ))