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Adding A Delay To The Dome Light


NRX

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Guys. I was thinking of modifying the dome light of the car to add a fade off effect using a capacitor(s) in parallel to the bulb ( not LED). Now my physics knowledge is a bit rusty :( but if my memory serves right, a capacitor added in parallel to the light bulb should light it up and slowly fade off when the power is cut off, ie: closing/locking the doors. Does anyone here have done that ? What could be the required capacitance ? Imagine the bulb is a 10W one.

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Will not work with just 1 cap, You'll probably going to need few resistors as well. For 12 v 10 w bulb 47000uF capacitor will be enough for 2-3 seconds. Larger capacitor will increase the fading time. Good luck & keep us posted!

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There are kits available for this if you don't want to go through the trouble of creating the circuit yourself.

For example:

http://www.jaycar.com.au/productView.asp?ID=KC5392

With installation instructions:

http://www.autospeed.com/cms/A_3068/article.html

If you want to create the circuit yourself (which is what I would have done), then have a look at this:
http://pcbheaven.com/circuitpages/Car_Dome_Light_Off_Delay/

Do share with us what you decide on doing. Good luck!

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Thank you maersk and Davy. What I actually want is just the fading effect for aesthetic purposes :) I don't really need a delay to be added ( The thread title needs to be edited I guess). My dome light goes off when I put the key to the ignition position or when I lock the door, I want to make it fade out nicely as soon as the key is in ignition position or doors are locked. Davy's circuits serve the purpose but they add a delay too.

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Thank you maersk and Davy. What I actually want is just the fading effect for aesthetic purposes :) I don't really need a delay to be added ( The thread title needs to be edited I guess). My dome light goes off when I put the key to the ignition position or when I lock the door, I want to make it fade out nicely as soon as the key is in ignition position or doors are locked. Davy's circuits serve the purpose but they add a delay too.

If it is a plugin solution that you are seeking then you need to turn to LEDs with a switch mode controller that dumps stored energy in a capacitor (I've done a similar thing to convert parking and brake light sometime back). Still it could be a bit bulky. A better solution is http://www.circuitdb.com/?p=343 - an extension to the Davy's proposals above. You can keep the switch off delay to 0 seconds in your case.

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OK guys. I tried out the simplest method of all, which was putting capacitors in parallel to the light bulb. The experiment was a partial success. I used 2 of 4700uF 16V caps in parallel and that capacity was nowhere near enough. It only lit the dome light about 1/3 of a second and that too with low brightness. I guess this method to be succeeded I'll have to at least use total capacitance of 40-50,000uF, which is not practical due to the size of the capacitors.

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OK guys. I tried out the simplest method of all, which was putting capacitors in parallel to the light bulb. The experiment was a partial success. I used 2 of 4700uF 16V caps in parallel and that capacity was nowhere near enough. It only lit the dome light about 1/3 of a second and that too with low brightness. I guess this method to be succeeded I'll have to at least use total capacitance of 40-50,000uF, which is not practical due to the size of the capacitors.

Not only the caps are bulky but when they charge the current drawn at the switch on is very high (I=12/R where R is the wiring resistance which could be well below one ohm) before the exponential decay begins. This could result in premature wear off in door switch contacts.

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Also, having a capacitor parallel to the bulb means that when the capacitor discharges, the current doesn't flow to the bulb totally, most part of the current travels outside the circuit to where the resistance is much lower. To prevent the current from flowing outside, you have to use a diode (a rectifier diode would do the trick I suppose). This way the capacitor is guaranteed to discharge only through the bulb.

Someone correct me if I'm wrong, it has been quite a while since I got my hands on a circuit so I have forgotten most of what I have learnt in school :)

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Also, having a capacitor parallel to the bulb means that when the capacitor discharges, the current doesn't flow to the bulb totally, most part of the current travels outside the circuit to where the resistance is much lower. To prevent the current from flowing outside, you have to use a diode (a rectifier diode would do the trick I suppose). This way the capacitor is guaranteed to discharge only through the bulb.

Someone correct me if I'm wrong, it has been quite a while since I got my hands on a circuit so I have forgotten most of what I have learnt in school :)

The circuit is supposedly open when the capacitor starts to discharge machan, so there shouldn't be any current flow outside the bulb + cap combo right ?

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The circuit is supposedly open when the capacitor starts to discharge machan, so there shouldn't be any current flow outside the bulb + cap combo right ?

Yes, you're right. I probably had my wires crossed while I was thinking about it earlier. Since you need the delay after the door closes, we can safely assume that the circuit is open.

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