PFH Calculator for High Demand Safety Functions
For safety functions demanded more than once a year or running continuously, IEC 61508 uses PFH, the average frequency of dangerous failure per hour. This calculator gives PFH and the SIL band for simple 1oo1 and 1oo2 architectures.
- PFH ≈ λDU = 5.00 × 10^−7
- PFH = 5.00 × 10^−7 /h → SIL 2
IEC 61508 high demand / continuous mode
| SIL | PFH (per hour) |
|---|---|
| 4 | ≥ 10⁻⁹ to < 10⁻⁸ |
| 3 | ≥ 10⁻⁸ to < 10⁻⁷ |
| 2 | ≥ 10⁻⁷ to < 10⁻⁶ |
| 1 | ≥ 10⁻⁶ to < 10⁻⁵ |
PFH Equations (Simplified)
1oo2: PFH ≈ 2[(1 − β)λDU]² × tCE + β × λDU, tCE = T/2 + MRT
λDU is the dangerous undetected failure rate per channel, β the common cause factor, T the proof test interval and MRT the mean repair time. Detected failures are assumed to bring the process to a safe state. These are simplified forms of the IEC 61508-6 equations.
Worked example
A 1oo1 function with λDU = 5 × 10⁻⁷ /h has PFH = 5 × 10⁻⁷ /h, which is SIL 2 in high demand mode. As 1oo2 with β = 5 % and a yearly test, PFH ≈ 2.7 × 10⁻⁸ /h (SIL 3).
Frequently Asked Questions
When is PFH used instead of PFDavg?
When the demand rate is higher than once per year or the function operates continuously, as defined in IEC 61508 and IEC 61511.
What is the unit of PFH?
Dangerous failures per hour, the average frequency of dangerous failure of the safety function.
Why is 1oo1 PFH equal to λDU?
In high demand each undetected dangerous failure is soon followed by a demand, so the failure rate becomes the hazard rate.
Learn more on the blog
What is SIL (safety integrity level)
A practical, field focused guide on instrumentationblog.in that explains the theory behind this calculator and how engineers apply it on real plants.
Read Full Article