I have a magnetic “alarm” switch that is normally closed. It opens when the garage door is open. When the door is open I want it to close as circuit to light an indicator bulb inside the house.
I went down to the surplus store and bought a relay, a lamp, and a (judging by the weight) transformer type 12 volt DC wall wart. I wired it up and took some measurements. The relay draws 90 ma (after it’s been on for a while it edges down a couple???), the lamp draws 60.5 ma, the power supply is rated at 150ma.
Good idea or no? Is there a better way to do this than a relay?
Magnetic switches are usually available in both NC and NO configurations. Wouldn’t it make better sense to just add a separate NO switch for the secondary circuit rather than try to invert the logic of the NC switch you have? Assuming your lamp is driven directly by the 12V supply you could do this with simple wiring and no relay.
Apologies if I’ve misunderstood what you’re trying to do, but this is how I’d do it.
The switch is a relatively expensive special one designed for garage doors, it has a lot more “gap” than a typical alarm type switch and was only available normally closed. Maybe other brands offer normally open, but it’s installed and works so I’d like to stay with what I have. Northing else is on the switch, the only purpose is to light an indicator light in the house.
The lamp is incandescent, but maked “12 volt DC” for some reason. (I thought incandescent lamps didn’t care if they got AC or DC). I wanted something brighter than the typical indicator LED and visable at a wider angle.
Instead of a relay, I would use a power transistor configured as an inverter. (When the switch is closed it grounds the base or gate, thus keeping the transistor off. When the switch is open, a pull-up resistors turns on the transistor.) Should be fine for powering a 12 VDC bulb. I would use a transistor in a TO220 package. A FET or TIP120 bipolar transistor would work. I can draw up a schematic if you like.