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Considering that the electromagnets you are linking are 12V devices, it will already be at a significant disadvantage.
Even if it was a 9V device, P=5 , V=9 thus I = 5/9 = ~556mA ,a significant draw for a 9V battery. Perhaps a minute or less. Not an advisable position.
You'd be better advised to consider the opposite, door is locked by mechanical latch, electromagnet simply unlocks the door long enough for you to open it. In fact, you might not even need an electromagnet, just a strong permanent magnet.... (not to say you coudln't opt for an electromagent)
I wasn't aware it was for a gun cabinet. I don't think a magnetic catch is a good idea for such a purpose, whether using a permanent or electromagnet.
So would this make more sense? If so, what would be the best way to power it discreetly (I was thinking 9v or a string of AA's, hence the original question) but with this locking mechanism if the battery dies how do I open it? If the standard electromagnet dies it would just release.
It's for a locking gun cabinet so I do not want to write out to the house where ask one would have to do is cut the power to bypass.
Well... Skip the electromagnet altogether, as I suggested in the last sentence of my original response. Voila, no worries about battery failure. You could lock the cabinet today anmd come back in 50 years and the permanent magnet would still unlock the lock.
That should work fine. I have seen solenoids of relays that can latch one way of another and only use current while switching between 2 states (locked and unlocked)
As for power, I can only think of a power supply since you surely would not want the battery to die if it is built into the inside of the cabinet. If the batteries can be charged externally, a small SLA battery or 10 NiMh batteries in series (or 10 AA's) will probably be fine.
You can get electromagnets that latch one way, then you reverse the wire and they latch the other.
So if I'm to understand this correctly, I should use a solenoid that is in the locked position, then use electromagnet (or some sort of switch) to temporarily unlock it.
When you speak of "permanent magnet" is that just a normal magnet? Because I do want this to be locked, not just held closed.
Thanks in advance for everyone's help.
For DC, it is reasonable to model a solenoid as a constant resistance, just a long piece of wire. The data sheet does not actually say what the DC resistance of each solenoid is, but it is reasonable to guess P=V^2/R, and thus R= V^2/P, with V=12V.For the 5 watt electromagnet, in your data sheet, this calculation gives R = 12*12/5 = 144/5 = 28.8, or about 30 ohms.The next step is to look at the data sheet for a typical 9-volt battery, like this one:
Then look at the graph labeled, "Constant Resistance Performance"At 30 ohms, the time to drain the battery to 6.0V, is about 1 hour, and the time to drain it to 4.8V is about 2 hours.Note that your solenoid's holding force is not the same at 9V,or 6V, as it is at 12V. You know, the quote of 100 newtons, that's for the solenoid with a full 12V across it. I'm not sure how much weaker the holding force is at lower voltages. Proportional might be good guess.Have you considered using a permanent magnet instead. That way you could keep your cabinent locked forever.;-P
minutes, perhaps seconds.
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Posted:Dec 31, 2014
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