# Door open warning light

Here's a question for all those electrical guru's out there.
I'm a volunteer fire fighter in a small department. We are working on
getting an equipment truck operational. This truck will carry extra pumps, suction hose, drop tank, hazmat supplies, heavy extrication tools, cribbing, etc. We installed 12 volt fluorescent lights in each cabinet. Each cabinet has a door switch that controls the light in the cabinet. All the cabinets are on the same branch circuit.
What we would like to do is install a warning light in the cab that illuminates when a cabinet is not closed. Any suggestions on how this can be done?
TIA, Mark Gramlich Sky Valley / Scaly Mountain Volunteer Fire & Rescue
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On Sat, 16 Oct 2004 20:06:06 GMT, mark gramlich

Assuming you will be using the door switches to determine if the cabinets are open or closed AND assuming that the current draw on the branch circuit is zero when all doors are closed, you will need a circuit that senses current flow in the branch circuit.
How you design this depends on what the value is. Connect a good quality ammeter (like a Fluke) to the common point and open each door one at a time and record the current reading. What you are looking for is the minimum current reading for any on any door.
Then you need something like a Hall Effect DC Current Sensor. If you don't want to build a circuit, here is a finished product that looks like it might work:
http://www.crmagnetics.com/newprod/ProductView.asp?ProdName=CR5395
(It does need at least 24 VDC for the power supply, but there are DC to DC voltages doublers that will do this for you.)
Don't want to sense current? You could run wires to each of the door switches and OR Function the output to light the "Door Open" light.
Beachcomber
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On Sat, 16 Oct 2004 21:14:01 GMT, not snipped-for-privacy@xxx.yyy (Beachcomber) wrote:

Yes - the current flow is 0 when all the doors are closed. Our equipment truck has 16 doors and 8 lights on it. I would much prefer to use something like the current sensor.
Question - Would some sort of solid-state relay work for this application?
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mark gramlich wrote:

Yes, but...
You can connect the diode (sensor) part of an opto-isolator in series with the lamp supply - the voltage drop across it will not affect the lights but will trigger the other half of the isolator when any light is on.
You will need almost certainly need to limit the current through the diode part by shunting it with some high power diodes connected in series and putting a current limiting resistor in series with the diode part. The combination will provide an output whenever one or more lamps is on. You can use the output direct to light a warning lamp or use it to operate a relay.
If you can tell me the truck voltage and the approximate current drawn by a lamp, I can give you a circuit complete with values. The components are widely available and very cheap and the circuit will cope with a wide range of variability in battery voltage and lamp currents.
I will need to either email you the circuit or post a web page link as I can't post binaries here. If you prefer the former, email me an address to use - just leave off the invalid bit of mine.
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Sue

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On Sat, 16 Oct 2004 20:06:06 GMT, mark gramlich put forth the notion that...

Run a red +12 volt lead to the warning light, then run a black wire from the other side of the warning light to a common auto alarm switch mounted where the cabinet door will activate it. When the door is closed, the switch is pushed in. When the door is open, the switch grounds out, completing the circuit.
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Checkmate

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mark gramlich wrote:

Yes. Are the door switches in the hot lead to the cabinet illumination or on the ground side?
Either way, you can connect a single 12V dashboard warning lamp to all of the door switches through blocking diodes. The diodes will prevent a sneak circuit where one door switch feed all the cabinet lights through the warning circuit. The diode polarity will depend on whether the door switches control the hot or ground leads.
This approach is better than current sensing in that a failed cabinet lamp will not prevent the warning lamp from lighting.

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Paul Hovnanian mailto: snipped-for-privacy@Hovnanian.com
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Put an "always on" SSR in the circuit with perhaps a cut off switch in the cab. When it is drawing current you will have a .7v drop across the SSR. This should be easy to detect.
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Paul Hovnanian P.E. wrote:

Yep, this is arguably the best solution. If the blocking diodes are connected close to the cabinet switches, a single wire can be run as a "bus" chained from one diode, to the next, to the next and so on, finally running up to the cab and the warning lamp. This may be simpler and easier (and cheaper) to install than running wires from each switch to a diode assembly in the cab. It also has the advantage, as stated by the proposer, that it will still work even if one or more lamps are u/s.
The opto-isolator suggestion has the advantage that only the power line feeding all the cabinet lamps has to be accessed - and this line may already be accessible in the cab, so making wiring simple.
The Hall-effect sensor has the additional advantage that no existing wiring has to be broken or connected to - which may be better for insurance purposes.
However, neither the opto-isolator or Hall-effect switch will detect a door open if the interior lamp is defective. That the fluorescent lamps will have internal invertors that will draw current even when the tube is broken, may make this less of a problem.
Fitting auto alarm switches, as suggested, probably involves the most work but they can be fitted so that they are activated by the door bolt and not just the door - if so fitted, they are the only method that will show not only that the doors are closed, but that the catches are engaged. This solution also needs no connection or break in existing wiring - which again may be needed for insurance purposes.
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HTH

Sue

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Seems to me you would want to be able to control the "branch" circuit via a switch in the cab in any case. (E.g.: temporary circumstances may make it desirable to be able to reduce battery loading and keep several doors open at the same time.) IF that the case, you can put a SMALL led with current limiting resistor across the switch. When the switch if OFF, the LED should be OFF. IF the LED comes ON, when the switch is OFF, one of the doors is open.
The switch will save your battery from discharge, the LED will warn you that your "stuff" may not be secure.
After a few uses, seeing a lamp come "ON" when you turn the switch off will serve as a nice warning that not all is well. If that switch is near switches that control some of the "perimeter" lighting that's usually only used when you are parked, it will routinely be switched off when you head back to the station.
Of course, if your fluorescent fixtures don't draw enough juice to light the LED ....
You might just put a 12 volt panel lamp across the switch contacts. The 12 volt bulb might pass enough juice to positively get the inverter for the fluorescent lamps to attempt to draw current.