In of our partner's access control devices they're using MF-MSMF200-2 Bourns Resettable Fuses - PPTC 2A 8V 0.02ohm for protection. However their boards have gone burned and reasons we received is 'Voltage Spike or Short circuit from external body'. How much current & voltage above fuse will bear in open circuit & close circuit states?
As partner is not ready to cover damaged items in warranty to I'm trying to put back on them that. However their devices is supposed to work on 12Volt, 0.7A.
It's not clear whether the fuse has failed, other components on the same board as the fuse, or other downstream equipment.
A PTC fuse has to stay cool enough when carrying current, I < Ihold. Which your device meets.
When tripped, a PTC has to have low enough dissipation (= high enough resistance & low enough voltage) to not destroy itself. Its resistance will only rise so much, either before it gets to a maximum temperature, or before the resistance change mechanism tops out. If it's fed excessive voltage when 'open', then there's the chance it could overheat and destroy itself. Whether it would fail short or open is another matter.
When tripping, the internal dissipation will be very high. This could cause local overheating, rather than the general overheating above, depending on the homogeneity of the construction. Parallel paths will self balance, but series paths could run away. Limiting the applied voltage will limit the power available for local heating.
The fuse is rated for 2A carry, and is specified to open in less than 3 seconds at an 8A overload. This may or may not be close enough to protect the rest of the circuit, or any downstream equipment.
The fuse is rated for 8v withstanding when open. Using an 8v rated component in a 12v system is evidence of engineering incompetence, it is not adequate to protect the fuse. Whether or not they will be embarrassed enough about it to reimburse you is another matter.
