Keyless Entry Shutdown Controller

**Sinorm** · 04-07-2004, 06:30 PM

My car has keyless entry, and I thought it would be cool if the computer would start as soon as I unlocked the car, and turn off after I locked the car again. This way, the computer should be completely booted up by the time I have the car started, and it won't turn off even if the car is off, as long as I am still in the car. (So yes, you will need a power supply that can survive a crank).

The circuit I have designed works in conjunction with the shutdown controller here: http://home.comcast.net/~danielsoh/shutdown.htm

My circuit will activate the shutdown controller (turning on the car) as soon as the doors unlock. After the car is turned on, you can unlock and lock the doors at will, the computer will not turn off while the car is on. When you turn the car off, the computer will stay on until the doors are locked. After locking the doors, the shutdown controller will wait a set amount of time before turning your computer off (so you can go in the store real fast without having to restart your computer).

THIS CIRCUIT IS UNTESTED!!!! It should work, but I have not actually tried it out yet.

Depending on how your car is setup, getting to the lock and unlock wires could be somewhat difficult, but due to my cars aftermarket alarm, it shouldn't be hard.

You should be able to find a smaller Hex Schmitt Inverter so that you don't have to waste 5 inverters (or, find a larger one and combine this circuit with the shutdown controller's inverter).

If anyone sees anything I have missed, or has any comments, please share. Thanks.

EDIT: see below for revised version.

**Ricky327** · 04-07-2004, 08:04 PM

Ok heres what I can see wrong...but maybe you can explain why you did it that way.

The shutdown controller circuit already have a pontential divider (3K, 1K) at its VACC input, this accept 12V and reduce it to a level that the logic circuit can take.

On your diagram you are again reducing the VACC voltage down by the (100K, 47k with another 47K). By the time the shutdown controller receive the original VACC its gonna be too low to trigger it.

You are using 2x47K in parallel...no need unless you really want this resistor value.

Im not too sure why you are directly feeding back the output of the gate back to the other one. Your circuit input and output seems to be clashing.

**Sinorm** · 04-07-2004, 08:30 PM

You are correct about the voltage into the shutdown controller. I had forgotten about that aspect. The 47k resistor is there for the comparison between the unlock wire and the chip. The 100k resistor shouldn't have been there originally though (but is now needed because of a change I mention below). There is a problem though, because the chip only outputs 5v instead of 12v.

The resistors on the shutdown controller will need to be changed from 3k and 1k to 1k and 1M respectivly. That will allow plenty of voltage to be dropped to trigger the chip. Due to the change in the resistors on the shutdown controller, I will keep the 100k resistor on Vacc (although it was incorrect in the 1st revision) to bring the voltage down to 5v.

I forgot the mention the other flaw you pointed out in my original post. While there appears to be a wire shorting out the 2 outputs on the chip, notice there is not a square there. In the real implementation, you would need to use a wire or something along those lines to pass over the other connection without touching it. I just found a better way to draw that part of the schematic, and have changed it on the revised version.