Help: Circuit Design 4in NOR Gate

I own an Alfa Romeo 75 car and I have fitted an alarm. Unfortunately Alfa decided to make the doors switches opposite to usual. When the doors are open they supply 0v GND (ground/negative) and when shut they supply 12v GND. So this makes it difficult to hook up to my alarm as it wants a 12v GND or 12v POS (positive) supply when the doors are open.

So not being that savy with electronics I figured I could get by with 4 relays (with NO/NC terminals), one for each trigger wire. I could not use one as there would still be 12v supplied if the other doors where shut and only one was opened. I rigged these up but as you would expect they chewed through the cars battery power pretty quickly as the coils were nearly always active because as you realise doors spend most of their time shut!

I went to a local electronics shop (JayCar) in Geelong, Australia and the person behind the counter recommended the following circuit layout.

**broken link removed**
(Also attached below)

I have put his suggestion together, but I am not getting the reaction I had hoped for. The basic idea of the above circuit is that if door switch 1 or 2 or 3 or 4 are not switched/pulled/toggled/on/down/together then the NOR gate will tell the transistor to switch the relay on or off. Where the lamp is hooked into the above circuit the alarms positive door trigger wire would be hooked up in the actual application.

I hope that I have given you enough information to help me come up with a circuit that works.

Thanks in advance,
Simon

I am making an inspired guess that your car door switches and interior light are wired as shown below as "Standard.GIF".

In which case, I think the easiest thing to do would be to wire a relay across the interior light, and then wire the contacts to your alarm unit, as shown in "With Relay"

JimB

JimB said:

I am making an inspired guess that your car door switches and interior light are wired as shown below as "Standard.GIF".

In which case, I think the easiest thing to do would be to wire a relay across the interior light, and then wire the contacts to your alarm unit, as shown in "With Relay"

JimB

Click to expand...

Can't do that because there is a light delay on the wires to the roof light. That means that when you open the door and shut it the interior light will remain on, which in turn means that the alarm thinks the doors are still open. It then stops detecting door movment which is no good.

The next closest place to break into the circuit is the four wires running to the door.

Thanks,
Simon

OK.

Another good idea while it lasted!

JimB

1/ You are dealing with CMOS with high input impedances, so when any of the switches on the inputs are open, you should have resistors to pull up the gates to the +v potential. You never leave CMOS input gates floating in mid air!

2/ Also you will need a freewheeling diode over the relay coil as I don't see one in your diagram

3/ You are driving a PNP transistor from the CMOS output, First determine how high the level on the gate will be for the logic High. I bet at least a volt below 12v.

Your problem is here that it is not high enough to turn off the PNP transistor AND you have the 4.7K resistor going to GND that will keep the transistor on independent of the output of the CMOS gate (as there will be a constant flow of current through the emitter to GND). To overcome this problem, use a 1K resistor with a small diode in series, cathode side to CMOS output, anode to one side of 1K, other side of 1K to transistor base (This combination will now be in R1's position) and a 10k resistor from the base to the +12v to help switch off the transistor. This 10K may be increased in value if your relay needs more current to switch on.

You might be able to use a circuit like the one below. As the One stated, it has a pull-up resistor on the input of a CMOS inverter. As shown, the diodes are connected to the switches from the doors. When no doors are open, the resistor pulls up the inverter's input, so it's output is low. When any door is open, current will flow thru the corresponding diode to ground, grounding the inverter's input making it's output go high. The diodes are also there to isolate the inputs from each other, preventing any short circuits from 12v to GND when one door is opened and the others stay closed. As shown, the output from the inverter connects to your relay driver circuit.
Hope this helps.
JB

If your door switches indeed connect to GND when the door is open, and you want to sense when any door is open, then you need a NAND gate, not a NOR gate. Also, with the resistor values shown in your schematic, the PNP will always be on. Even if you had the correct resistor values, the NOR gate would keep the relay on unless all doors were open. See the schematic below for something that should work.

Don't get me wrong,I am always up for a good project but sometimes things can be done easier right out of the box.Most alarms these days have multiple inputs for both positive and negative trigger door systems.On top of this most are microprocessor controlled and can be programmed for the different inputs and time delays with just a few switch strokes,manufactures vary.All the new ones that I have seen also have a time delay built in where it will wait for the interior light to go off before securing the interior,generally the norm is 30-seconds.I am not sure what model or brand you have,but I recommend Audiovox-Prestige brand as they use quality Components and have many features that allow it to bypass many factory systems in cars,espically for the auto starts.Stay away from the viper product line as I always had false alarm problems when we carried them.Good luck either way,hope I could help......

Ron H said:

If your door switches indeed connect to GND when the door is open, and you want to sense when any door is open, then you need a NAND gate, not a NOR gate. Also, with the resistor values shown in your schematic, the PNP will always be on. Even if you had the correct resistor values, the NOR gate would keep the relay on unless all doors were open. See the schematic below for something that should work.

Click to expand...

My doors connect to GND when the doors are closed. The circuit is broken by the switch when the doors are opened.

So, when door open; V=0 (No connection to GND)
when door closed; V=12 (Connected to GND)

Thanks,
Simon

eminence said:

Ron H said:

If your door switches indeed connect to GND when the door is open, and you want to sense when any door is open, then you need a NAND gate, not a NOR gate. Also, with the resistor values shown in your schematic, the PNP will always be on. Even if you had the correct resistor values, the NOR gate would keep the relay on unless all doors were open. See the schematic below for something that should work.

Click to expand...

My doors connect to GND when the doors are closed. The circuit is broken by the switch when the doors are opened.

So, when door open; V=0 (No connection to GND)
when door closed; V=12 (Connected to GND)

Thanks,
Simon

Click to expand...

Well then, disregard my schematic with the NAND gate. Here is what you said in your original post:

When the doors are open they supply 0v GND (ground/negative) and when shut they supply 12v GND.

Click to expand...

I didn't understand your terminology (still don't). Your most recent post is unambiguous.

I have a solution to the problem from another forum: **broken link removed** Take a look if your interested

Thanks for all your suggestions,
Simon