Hey all!
So I've been working on a circuit that would essentially continue giving power to a load for a set period of time after a switch is thrown. A guy named "Manfred von Steinborn" made a post about a project similar to what I'm trying to accomplish here. However, the way he does this is using a relay to both reset the circuit and power a load, which is all fine and good, except I can't get my hands on a relay. I have many transistors and MOSFETs that have similar specs (as far as max voltage and amperage) to the relay. Is there any way to replace the relay with a transistor?

Additional information about the project:
I will be using the exact same schematic as "von Steinborn", but hopefully with something other than a relay. The purpose of this circuit is to keep my 12v dc LED lamp on for a few seconds (maybe 4 or 5) after I flip the switch so that I can get out of the room without tripping over anything . I am wide open to any other methods someone may have as to accomplishing this end result even if that means using something other than a 555 timer. I'm still a bit of a beginner in electronics, so I apologize in advance if I go a little bit slow in understanding what you may be saying .
Thanks a bunch!
TimmyTimer55

 

How much current does your LED lamp take?

Here is how I would do it:

 

It only draws about 0.5 amps

 

You could drive the gate of the FET in the illustration below with pin 3 of the NE555. RIN could be 100 ohms (to prevent high frequency oscillation during the switching transistions), RGS could be eliminated.

Source of the illustration is http://www.electronics-tutorials.ws/transistor/tran_7.html

If the time delay is not critical and the current through the lamp is not very high compared to the current rating of the MOSFET, you can skip the NE555 and just choose a capacitor to place across Vin which is connected to button or switch to the battery. When Vin goes away, RGS would discharge the capacitor after a while. I've done this a few times and it works. If there is a load from the switch to ground, you will need to use a diode in series with the RIN.

The reason this is not a good idea with a very high current load is that the MOSFET would gradually turn off and might become warm in the process. In its basic form,it would look something like this:

Source of the illustration: http://easy-electronics4u.blogspot.com/2012/02/simple-dc-timer-using-mosfet-onoff.html

 

Dick, minor nit about your circuit. When the switch is closed, there is nothing to limit the instantaneous charging current into C1. That is an invitation to weld the contacts on S1 (or at least pit them). If you notice, I put a resistor in series with the switch to limit the current spike.

 

Alright! Huge thanks to both of you for your quick and awesome answers! This is exactly what I'm looking for! I guess I was overthinking this a little bit too much...

But again, I really appreciate your answers and I'm glad that I actually understand them too!