BECKHOFF TwinSAFE - Manual (Page 67)

Circuit examples
Application Guide TwinSAFE 65
The following assumptions must now be made:
The door switches S1/S2 are always actuated in opposite directions. 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!
The proximity switches S3/S4 are monitored for plausibility (temporal/logical) and are type A systems
according to EN61508 (simple components whose behavior under error conditions is fully known). The
safe position is driven to once per shift.
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.
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(S1|S2| EL1904)+ PFH(S3|S4| EL1904)) /2 + PFH(EL6900) +
PFH(EL2904) +
β
* (PFH(K1)+ PFH(K2))/2
with
PFH(S1|S2| EL1904)=
β
* (PFH(S1)+PFH(S2))/2 + PFH(EL1904)
PFH(S3|S4| EL1904)=
β
* (PFH(S3)+ PFH(S4))/2 + PFH(EL1904)
to:
PFH(S1|S2| EL1904) = 10% *(1.68E-09+ 8.4E-10)/2 +1.11E-09 = 1.24E-09
PFH(S3|S4| EL1904) = 10% *(8.4E-10+ 8.4E-10)/2 +1.11E-09 = 1.19E-09
PFH
PFH PFH PFH
tot tot tot tot
= == =
10 % *
(1.24E-09+1.19E-09) /2 + 1.03E-09 + 1.25E-09 + 10 % * (1.29E-
09+1.29E-09) /2 = 2.53E 2.53E 2.53E 2.53E----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
ௗ
(ܵ1) +
1
ܯܶܶܨ
ௗ
(ܧܮ1904) +
1
ܯܶܶܨ
ௗ
(ܧܮ6900) +
1
ܯܶܶܨ
ௗ
(ܧܮ2904)
+
1
(ܯܶܶܨ
ௗ
(ܭ1))