Goal = Interior lamp on until start’n car

Hi,

I tried to do a search, but couldn't find clear information on this.  I'm trying to have the interior dome lamp stay on once I open the door until I turn the key to the start the car.  Then with the motor on and key in the "ON" position, the lamp should also turn on when a door is opened,  but turn off immediately after the door is shut.

With the motor off and the key in the "OFF" position, the lamp should stay on for a short period of time and then fade off by itself.

Most of the newer cars' lamps function like this and it would be nice to get my car to do the same.  Currently, it turns on and off with the opening and closing of the door.

Thanks in advance for your help.  This is a great site.

Simpson

I'm a newbie to all this, so work with me.  Would this work, or am I WAY off?

Man, I just realized that diagram is WAY off. 

Does anyone have a proven diagram?  Thanks again...

Here is one.

hotwaterwizard wrote:

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Here is one.

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Thanks!  But since I'm EE illiterate, could you please explain the elements of the diagrams in terms of a SPDT relay pinout - where they go and what they are?  Thanks again...

You will need a Soldering Iron.

Thanks John!

A couple of questions: 

-what will energize the relay to give continuity between pins 87 and 30?

-and I'm guessing if the ignition switch isn't in the "OFF" position, the capacitor will not get charged up?

The Transistor conducts and sends a voltage to the Relay.

 As the capacitor discharges thru the Diode on pin 85 it slowly looses voltage until the Transistor finaly turns off and the Relay in turn shuts off too.

hotwaterwizard wrote:

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The Transistor conducts and sends a voltage to the Relay.

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I see.  So basically when current flows throw the wires leading to the light switch (pins 87 & 30),  the Transistor activates the Relay?

Questions:
- What role does the lead going to the ignition switch and its Diode play?  If it's to prevent the light from staying on when the key is not in the "OFF" position, how?
- How does the discharged current from the Capacitor "know" to go through pin 85 and not pin 86?
- Do pin 86, pin 85's Diode, and the Capacitor all get wired to the same ground?

Thanks again for the EE education!

Here is the original Text From Bills website.

The two circuits below illustrate opening a relay contact a short time after the ignition or ligh switch is turned off. The capacitor is charged and the relay is closed when the voltage at the diode anode rises to +12 volts. The circuit on the left is a common collector or emitter follower and has the advantage of one less part since a resistor is not needed in series with the transistor base. However the voltage across the relay coil will be two diode drops less than the supply voltage, or about 11 volts for a 12.5 volt input. The common emitter configuration on the right offers the advantage of the full supply voltage across the load for most of the delay time, which makes the relay pull-in and drop-out voltages less of a concern but requires an extra resistor in series with transistor base. The common emitter (circuit on the right) is the better circuit since the series base resistor can be selected to obtain the desired delay time whereas the capacitor must be selected for the common collector (or an additional resistor used in parallel with the capacitor). The time delay for the common emitter will be approximately 3 time constants or 3*R*C. The capacitor/resistor values can be worked out from the relay coil current and transistor gain. For example a 120 ohm relay coil will draw 100 mA at 12 volts and assumming a transistor gain of 30, the base current will be 100/30 = 3 mA. The voltage across the resistor will be the supply voltage minus two diode drops or 12-1.4 = 10.6. The resistor value will be the voltage/current = 10.6/0.003 = 3533 or about 3.6K. The capacitor value for a 15 second delay will be 15/3R = 1327 uF. We can use a standard 1000 uF capacitor and increase the resistor proportionally to get 15 seconds.