Failsafe with battery. - Page 1
Forum section: electronics / projects

author: Mikey
author_rank: Regular Contributor
timestamp: September 03, 2013, 01:04:13 pm
content: I am planning to make a system to keep a door locked.The door will be locked by a electromagnet, but I want it to be be able to 100% fail safely.My idea is that the system will be powered by mains when running normal, but if something happens it will draw from a battery instead. So if something goes wrong, relay, mosfet, software... anything, it would fail safe and unlock the door after about 2 hours.But how do I know how big a battery I need to be able to make such a thing?Battery should preferably be something that can handle getting quite low on voltage, as the system might have completely failed and not be able to shut off as intended.

author: mariush
author_rank: Super Contributor
timestamp: September 03, 2013, 01:11:05 pm
content: But how do I know how big a battery I need to be able to make such a thing?You measure how much energy your circuit uses (mA, uA, whatever) and then you know how much a battery will last... ex if your circuit draws 10 mA, a 1.5v AA 2000mAh battery will give you 1.5-1.65v for about 200 hours.

author: amyk
author_rank: Super Contributor
timestamp: September 03, 2013, 01:25:50 pm
content: If it's the type of electromagnetic lock that holds closed a building door, they take around 6W per coil.

author: crisr
author_rank: Contributor
timestamp: September 03, 2013, 10:18:01 pm
content: What is your definition of "safely" in this case? Keeping the door locked or open? And what type of electromagnetic lock is it? What does it do when energized - keep the door locked, unlocks the door? Is the action momentary or continuous?

author: 8086
author_rank: Super Contributor
timestamp: September 03, 2013, 10:30:04 pm
content: Need to know more about the locking method you intend to use.The only way to make it truly failsafe is to have a system that only locks when continually active and unlocks in the event that power is removed.

author: edavid
author_rank: Super Contributor
timestamp: September 03, 2013, 11:19:22 pm
content: Here's one that draws 2.4W:http://www.ebay.com/itm/Door-Fail-Safe-access-control-Electric-Strike-v5-NC-/350256744192

author: SeanB
author_rank: Super Contributor
timestamp: September 04, 2013, 04:23:22 am
content: A typical maglock draws just about 1A from a 12V supply, so you wlll need a 2AH lead acid battery to give you the 2 hour unlock you want. Drive the lock with a 12V relay as it will be a lot more reliable than a transistor switch.

author: SArepairman
author_rank: Frequent Contributor
timestamp: September 04, 2013, 06:33:17 am
content: leave a sledge hammer, energy bars and bottled water inside of the room. And poop bags.

author: fcb
author_rank: Super Contributor
timestamp: September 04, 2013, 07:35:46 am
content: Quote from: crisr on September 03, 2013, 10:18:01 pmFail-safe: fails to a safe situation (i.e. firedoor fails closed and unlocked).Fail-secure: fails to a secure state (i.e. bank vault fails locked).

author: crisr
author_rank: Contributor
timestamp: September 04, 2013, 08:40:50 pm
content: Oh, yes... in Portuguese there is only one word ("seguro") for both "safe" and "secure";I don't know about Danish though...I guess the only real fail-safe in the OP case is using a lock that only keeps the door locked while energized, as the ones other people have suggested. In that case you have to calculate the necessary battery capacity (Ah) based on the system (circuit + lock) current draw (A) times how many hours you need the door to stay locked, also accounting for a margin for overly-optmistic battery specs (some of them are given for a dead-battery voltage of <6V for a 12V battery; a nominal 12V battery ideally should not get discharged under 10.5V or it might get damaged) - you would also need the minimum cutout voltage for the lock, whether from published specs or through experimentation.If it didn't need to be "100%" safe, a more complicated, less safe option, but which could yield a much longer battery autonomy, would be using a lock that unlocks the door when energized and having a circuit that unlocks the door when the battery voltage drops below a certain threshold (that is still above the lock's minimum guaranteed actuation voltage). Other methods of redundancy and self-monitoring could be added to improve the safety margin.

author: G7PSK
author_rank: Super Contributor
timestamp: September 06, 2013, 07:57:27 am
content: If that's a fire door that opens in the event of a fire , a person could get rather burnt in 2 hours while they are waiting for the battery to die.

author: komet
author_rank: Unknown
timestamp: September 06, 2013, 08:23:04 am
content: What exactly are you trying to do? The problem with doors is that they have multiple failure modes: a) letting thieves in; b) not letting authorized people in; c) not letting anyone out in case of fire; d) letting prisoners out. So you need to define your problem before looking for a solution.Having said that have you considered electromagnetic locks that can be opened from one side with a door handle but not on the other side? That might cover it.

author: Mikey
author_rank: Regular Contributor
timestamp: September 06, 2013, 07:24:58 pm
content: Quote from: komet on September 06, 2013, 08:23:04 amI actually never asked about the solution for how the door locking should be done, only how to size the battery to what I needed...But to all the people wondering, the door is held locked by a strong electromagnet, the entire system is powered by mains on normal use, but if something goes wrong I can shut the power off, and it will fail safe when the battery runs out of power (because there are no way the electromagnet can hold the magnetic field without power). At the same time this also provides extra safety if thieves want to get in, because they can't just kill the power and then get in, they need to also wait until the battery is flat...The actual question seems to have been answered here, if that is the way to do it.Quote from: mariush on September 03, 2013, 01:11:05 pmNext question is what kind of battery should it be? Preferably one that is not too picky about getting fully discharged. Size is not important, as long as it is within reasonable dimensions.

author: dfmischler
author_rank: Frequent Contributor
timestamp: September 07, 2013, 09:28:29 pm
content: Quote from: Mikey on September 06, 2013, 07:24:58 pmWhat temperature range will the battery be exposed to? Is it OK if the battery has to be replaced after a deep discharge event? How long does the battery have to last at full capacity if no deep discharge event occurs? How soon after power is restored must the battery be ready to function for the full 2 hours again?

author: edavid
author_rank: Super Contributor
timestamp: September 08, 2013, 04:59:31 am
content: Quote from: Mikey on September 06, 2013, 07:24:58 pmSealed lead acid is most common for security applications, but doesn't respond well to deep discharge. Why do you need that though? Are you expecting lots of long power failures? Even if you are, can't you size the battery for 4 hours, and discharge it to 50%?

author: Mikey
author_rank: Regular Contributor
timestamp: September 08, 2013, 07:44:04 am
content: Quote from: dfmischler on September 07, 2013, 09:28:29 pmIt will be from like +30c to -5c.It would be prefered if the battery could survive to be fully discharged without totally failing, a bit loss of capacity is expected.The battery is only in case of everything else failing, so it might be connected to power for months or years without having been in use, maybe even its entire lifetime. It is only meant to be used if everything else goes wrong.Quote from: edavid on September 08, 2013, 04:59:31 amAs I wrote earlier in this thread, it is for when things go wrong. Hopefully it will never be used, because that means something did something unexpected. It got nothing to do about power failures, it got something about to failing to a safe state, but delaying that so it does not happen right away, because that means you could just cut the power and then open the locked door. Discharging to a certain level means there would have to be something to keep track of that too, which also could fail and then there would be a 4 hour delay instead of 2 hours, plus the battery would get too low again.