Help me get started with electronics
April 2, 2009 1:31 AM   Subscribe

DIY RFID door lock.

I've been wanting to play around with RFID for a long while, and so I've bought an electronic cabinet lock to teach myself a bit of soldering and basic electronics.

Facts about the cabinet lock:
-Works on 12v ac/dc
-Has five contacts: 1+2 12v; 3 n/open; 4 n/closed; 5 common

I'd like to buy an RFID kit to control the lock. But I have a couple of questions.

1) Would I need two power supplies, one for the RFID kit and one for the cabinet lock, or would one power supply power both?
a) What search term do I need for a power supply? All the power supplies I've seen don't have cables with soldered tips, they have plugs for sockets like mobile phones
b) Could I run it from batteries, how would I work out the life-span of the battery when I don't know the amperage draw of the cabinet lock?

2) Does n/open and n/closed refer to normally open and normally closed? Should the device be wired so that that 1+2 (12v) constantly gets supplied with 12v, but either 3 or 4 (depending on wished for functionality) gets supplied with current briefly and the lock does its stuff. I have no idea what common might be. Apart from that the lock apparently has a built-in microswitch to output to LED etc and that it could conceivably be that.

Could anyone recommend an RFID kit to power this? I've seen an $18 kit on eBay but it required modifying the PCB to actually do anything other than just recognise key fobs. But that's ideally the price bracket I'm looking in! It's what the cabinet lock cost!

posted by dance to Technology (5 answers total) 1 user marked this as a favorite
1. One power supply should suffice. If the voltages are different (and I think a lot of RFID kits require 5V), then you could lower the voltage for whichever item needs the lower voltage. This is typically done (for DC) using a switching regulator. If that sounds scary then maybe just use two supplies for now.
a) Use a knife to cut off the plug. Strip the ends of the wires. Tin the wires with a little solder.
b) Batteries are probably not the way to go here, unless you have some kind of circuit that will beep or flash an LED when the battery needs charging/replacing.

2. I'm fairly sure the device should be wired so that the 1+2 (12V) gets one wire from your power supply via a switch, and the common contact gets the other wire. As it's not fussy about AC or DC, polarity won't matter. A voltage is only supplied when you want the lock to open (i.e. when you activate the switch). The n/open, n/closed and common connections are used to monitor the lock status, so that's where your LED output is. Presumably you just treat common + n/open as the two pins of a switch in a circuit containing an LED (which will be lit when the lock is open). Similarly common + n/closed act as the two pins of a switch that makes a circuit when the lock is closed.
posted by le morte de bea arthur at 3:00 AM on April 2, 2009 [1 favorite]

and the common contact gets the other wire

Correction: The power supply goes to the two (1+2) 12V connectors. As I said later, the common connector is for monitoring the lock status. Sorry.
posted by le morte de bea arthur at 3:51 AM on April 2, 2009

Response by poster: I just wanted to say thanks, le morte, and that'll I'll read more thoroughly soon!
posted by dance at 4:43 AM on April 2, 2009

You may need to put a diode backwards across the solinoid that drives the lock. When you power up a solinoid you create a magnetic field. When you cut the power you have a coil and a magnetic field - aka a generator. The resulting spike can deep fat fry your RFID reader.

It's gonna drive up the price of your project, but you might want to pick up a copy of Practical Electronics for Inventors. You don't have to not fry too many chips for the book to have paid for itself.
posted by Kid Charlemagne at 5:37 AM on April 2, 2009 [1 favorite]

Additional thought - if it were me, I'd use a relay to switch the lock open and closed. They way you isolate the lock and RFID circuits, and hopefully eliminate any risk of frying the more delicate RFID circuit.
posted by le morte de bea arthur at 4:24 AM on April 6, 2009

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