need to create a ground for a moment just after a car ignition off

I am an absolute beginner so I apologize if my question is trivial or if I
use some wrong term or am verbose , but I would greatly appreciate help.

I have built in a delay circuit to my in-car cabin light so it will stay
switched on for 15 seconds after all my car doors are closed.

Now I would like to add to it an automatic activation of the cabin light
after the ignition key is set to the off position. My delay circuit needs
just one short ground impulse and it will be on for 15 seconds.

My cabin light is activated by the door switch that keeps wire at +12v while
a door is closed, and changes to a ground (0v) when the door is open.

I would need a simple way of creating a short ground connection on that wire
at the moment when my ignition is turned off (ignition on/off and the ground
wires are not a problem). Such a ground connection after the ignition off
should last less than e.g. 1 second and can be as short as practically
momentarily. When the ignition is on, this circuit should leave my wire
going from the door switch to the lamp undisturbed.

Could somebody, please, advise?


Regards,

Dave

 

If your delay circuit signal doesn't require lots of current to drive low,
you could use a simple capacitor between the ignition switch and the delay
circuit input. When the switch it turned off, it grounds, and the other side
of the cap is pulled low, triggering the delay.

If you can't trigger it with so little current, you can use an emitter
follower as shown here:

http://home.comcast.net/~rcmonsen/misc/ignition.gif

S1 is the ignition switch. When its opened, R2 pulls the cap to ground,
which pulls the base of Q1 low momentarily. This causes R3 to go to ground
(R3 is your delay circuit control).

It is a bit wasteful; however, it only uses as much current as your delay
circuit will sink, so if it has high impedance, then you'll only sink a
little current. Again, if its really high impedance, you can use a cap that
is pulled up by a resistor. It'll stay at 12V until the other side of the
cap is pulled low, at which time it'll glitch low. The product of resistor
and cap values determine how long it stays low.

Regards,
Bob Monsen

 

Thanks for the answer. I will try to understand it now Sorry, but I am
really new, so I am trying to learn through simple examples.

I like this solution with just a capacitor, I have to try it. What about the
vaules for the capacitor? Any non polarised? And connect one end to the +12v
that goes via the igntion and the other end to my wire that goe from the
door switch to the delay circuit?

Best place to take an ignition switch signal is from the radio supply wire -
when igntion is on, this wire is +12 v and radio works, when the igntion is
off, this wire is cut. This would go to the V1 position.

I am not exactly clear what is on the other side of the S1?
Where would my signal wire from the door switch be connected? Other end of
the R3 (ground).
Or R3 gows with R2 to the regular - of my car (I have a black - wire in
radio assembly).
I will build it today and will ty it out on my workbench.

Thanks a lot.