Need to design a 12 volt timed circuit

Thread Starter

KCAutosound

Joined Oct 6, 2009
4
Hey guys. Stumbled across this forum after some long searching for what I need.

Here's the deal. I have vehicle security systems that will allow negative inputs to be able to arm and disarm them. This comes in handy with the Ford vehicles since they have a keypad on the door. You can hit 7 and 9 to lock the doors and hit your five digit code to unlock the doors. I use the motor lock and unlock wires to arm and disarm the alarm system. The problem is that these are positive outputs and my alarm needs negative inputs. On MOST vehicles I can use a simple pair of relays to flip the positive from the locks to be a negative the alarm can see. A problem arises when I do certain vehicles like say the 04 to 08 F150s. The positive pulse from the lock/unlock motor wires is so fast that the relays don't have much time to respond and put out either a full ground output or a long enough output. I need it to be about a second or longer.

To correct this I have been using pulse timer relays from either Code Alarm or Directed Electronics. The problem is that I need to use two of them and these guys are about $15.00 apiece. I am using them more and more and my distributors inventory can't keep up with my demand. I would like to make a circuit that I can use that will keep cost down as well as allow me to have them in inventory when I do these vehicles. I do alot of them!!!

Basically I'm needing a circuit that will take the 12 volts positive input from the lock wire and then flip it to 12 volts negative for about say 2 seconds to be safe.

From my reading around I have seen people say you could use a capacitor to make the relay stay on longer but I have never done anything like that.
 

spacewrench

Joined Oct 5, 2009
58
Basically I'm needing a circuit that will take the 12 volts positive input from the lock wire and then flip it to 12 volts negative for about say 2 seconds to be safe.
Do you mean that you have a wire that is usually at ground, but jumps up to 12V to briefly to indicate "Lock" or "Unlock?" And what you need instead is a wire that is usually at 12V, but falls down to 0V (for about 2 seconds) when there's a 12V pulse on the other wire?

What will the second wire be connected to? Will it have to supply current while it's at the 12V level? (For example, is there a light bulb or something that's on most of the time, and blinks off when the lock pulse happens?)

Will the second wire have to sink current during the 2-second pulse? How much? (For example, will you connect the wire to one end of a relay coil, with the other end connected to 12V?)

This is probably a pretty simple circuit, you should be able to build them for less than $15.
 

Bernard

Joined Aug 7, 2008
5,784
Something like this? You can also replace R2 with a relay. If using a relay, shunt coil with a diode , cathode to +V. !N4002 or sim. Just about any N-ch MOSFET should do.
 

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Thread Starter

KCAutosound

Joined Oct 6, 2009
4
Bernard I kinda make sense with your drawing there but I'm really digging back to my Devry days from about 1993...lol Went for two semesters for Electronic Technition then kinda went another direction. I can make out the diodes and the capacitor symbols but I get lost on the other ones in your schematic.

Basically the locking system on a car is a reverse polarity system. The lock motors have only two wires. They flip polarity. So like what spacewrench was asking... yes the wire initially sits at ground but when the lock signal is sent to it it will pulse 12 volts positive. After the pulse it goes back to ground. I am trying to take this 12 volt pulse signal and turn it into a 2 second ground pulse. Sound simple enough?
 

spacewrench

Joined Oct 5, 2009
58
yes the wire initially sits at ground but when the lock signal is sent to it it will pulse 12 volts positive. After the pulse it goes back to ground. I am trying to take this 12 volt pulse signal and turn it into a 2 second ground pulse. Sound simple enough?
The circuit Bernard drew should work for you (plus it's a lot simpler than anything I could design!) The components, from left to right, are:

Diode
Capacitor (vertical, 1uF, "microfarad")
Resistor (R1, vertical, 2 megohm)
MOSFET (N-channel, looks like a sideways T stuck to a capital E)
Resistor (R2, vertical, above the MOSFET, 1Kohm "kilohm")

The input is the "arrow" side of the diode; the output is the junction between R2 and the drain of the MOSFET.
 

Thread Starter

KCAutosound

Joined Oct 6, 2009
4
Is there a certain wattage for these resistors that I need?

Little lost on this drawing. On this mosfet I see there are three prongs similar to a transistor. The datasheet shows D G S. Is my drawing I've attached look right from the schematic?
 

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spacewrench

Joined Oct 5, 2009
58
Is there a certain wattage for these resistors that I need?
R1 (2Meg) can be any size. 1/4 watt is very common, small & cheap. R2 (1K) could be 1/4 watt in a pinch, but 1/2 watt or 1 watt would be better.

Little lost on this drawing. On this mosfet I see there are three prongs similar to a transistor. The datasheet shows D G S. Is my drawing I've attached look right from the schematic?
Close, but no cigar. The 3 legs of the transistor are Drain, Gate and Source. (Looking at the flat front side, legs down, round side away from you.) You have (almost!) swapped D and S, and put some of the stuff that should be on G, on D instead. Here's a picture of the circuit with your MOSFET:
 

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Thread Starter

KCAutosound

Joined Oct 6, 2009
4
I like that drawing there. Good job. I'm gonna pick up the parts tomorrow from my local Electronics Supply. I checked their online catalog and it shows them all in stock. I throw it together and will post my findings.
 
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