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Posted - Apr 28 2004 : 11:09:34 AM
I recently disabled (permanently) my Automatic Headlight System (with daylight sensors) by inserting a 1k ohm resistor which effectively jumps the circuit from sensor to output. The "automatic headlights" were a nuisance , not to mention I added sensitive HID xenon headlights that can't take the quick-flicking on-off when I go to & from underground parking. By disabling the system everything is manual via rotary dial, & I lost the auto-off timed/timer headlight function. Since I am an absent minded buffoon who always forgets to turn off my headlights upon exiting the vehicle, I badly need a practical schematic so that I can build my own circuit (dumbed down, please, I just learned what a transistor does last night!) that provides a timed delay feature on the headlights so that once the ignition is turned off and the headlight manual switch is accidentally left in the on position, it would power down by itself per timer AND which resets itself so that I can have headlight power upon restarting just like a typical OEM modern day headlight circuit. I'm guessing it needs to have input from ignition switch fuse and possibly car door sensor/or dome lamp so it doesn't turn off immediately with ignition-off, giving me time to leave the vehicle with headlights on for a brief period and illuminating my path for a little while upon exiting the vehicle. i'm assuming also the output will be tied in with the powerwire going to headlights.
Oh how I yearn for this circuit!!! On the web, I easily found the ever-present and well-covered circuit on how to hook up a "left-your-headlights-on alarm", but I need to take it one step further as described above. I'm pretty much a novice but very comfortable with following well-labeled simple schematic using laymens terms and practical parts so that I can go to Radio Shack and actually find the parts without having to resort to hardcore wholesale electronics place. Please help me. Thanks
Edited by - reanime on Apr 28 2004 11:23:03 AM
Edited by - reanime on Apr 28 2004 11:59:43 AM |
| 5 L A T E S T R E P L I E S (Newest First) |
| ecm |
Posted - May 05 2004 : 8:25:56 PM
I have a couple of 555 timer circuit for car head lights .
http://members.rogers.com/adesorm/555headlighttimer.jpg
http://members.rogers.com/adesorm/555timers.jpg
to do or not to do
Edited by - ecm on May 05 2004 8:28:23 PM |
| YS |
Posted - May 01 2004 : 11:56:08 PM
Feeling lonely? sorry about that. As of myself, I do not go often to "auto" forum, so just missed your topic..
as for your problem: you do not need 555 there. The circuit you provided is pretty simple and should work OK. It is hard to give any easier explanation, but if you are interested in theory, there are many books there.
Practically, just build it. I would use circuit on the right and probably use 10K potentiometer (variable resistor) to adjust timing. Do not go too low on resistance, or you will kill the transistor. The good way is to connect 1K resistor and 10K pot in series.
As for relay, you can use any small-signal (generic) relay with 12V coil, providing it needs the current less than 50 mA. arrows on the right side show relay contacts. Relay should have switching contacts or normally open - so when relay is powered, contacts connect together . Your headlights are connected to the battery not through the switch directly, I believe, but through some power relay; anyway, your small 12V relay contacts should be connected parallel to "light on" switch. Second diode may be the same 1N4002. Both may be 1N4001
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Posted - Apr 30 2004 : 11:45:30 PM
Allow me to respond to my own message. i suppose everyone on this board is sick of the classic IC 555 timer didn't find this challenging enough to bother answering, or perhaps the message was too wordy, so I had to rely on other or should I say more helpful resources and precious time to educate myself on this most basic of circuits. Ah-hem, thanks.
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Posted - Apr 28 2004 : 10:23:41 PM
oh well blow it off. I'm ordering the circuit from the UK. Guess its out of everyone's league here......kidding
Edited by - reanime on Apr 30 2004 11:37:29 PM |
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Posted - Apr 28 2004 : 12:03:49 PM
I found the circuit and made a link to my webshot foto site, see circuit here:
http://community.webshots.com/photo/138389954/138393202RBbGWW
The man who scribed this circuit said this, and now I ask for anyones help to translate this into laymens terms, basically I just need a simple explanation of the schematic in practical terms to achieve a delay of 15-30 seconds and perhaps recommended parts in that scenario, cause he left it pretty general (i.e., what size diode, what is a 12v DC relay coil, and what is the unlabeled arrow diagram to the right of photo?)
many thanks to any takers.:
"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."
Edited by - reanime on Apr 28 2004 12:06:04 PM
Edited by - reanime on Apr 28 2004 12:32:26 PM |
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