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I want the LED light(12V) to turn on like normal when you apply power to it but I want it to fade out over 3-5 seconds when you turn it off. HOW CAN I DO THIS? Capacitor,resistor,relay,a combo of several?!Can you please tell me some specific items I need to gather for this project?I have been given MANY generalized answers regarding this(such as...Well maybe some sort of capacitor might work.Or maybe a resistor of some type.) But I'm really needing some specifics(SIMPLE IS BETTER in this case)I'm not an engineering buff and I'm no electronics junkie either.I just need some answers to anyone who has some valid info to offer. Thanks in advance! I guess more specifically an LED Lightbar.I'm unsure exactly how many LED's are inside it cause the bar is a sealed type and package does not say(but suppose we assume their is between 10 to 15 20ma bulbs).If you do have an accurate diagram of your proposed answer,please submit a link or an email address so I may review it or contact you personally. Thanks
And got the following answer:
You need: 1 RED LED (1.25v Forward @ 30ma) 1 resistor (Value see text) 1 capacitor (Value see text) First lets start by defining the LED and its series resistance Rs Assume 12 VDC supply (Vs) and a LED with 1.25V forward Voltage(the voltage at which the LED turns on) means we have to drop 10.75 volts across Rs. A typical LED will glow with at least 5ma current and max current of 20 to 30 ma. (Brightly) Lets use 20ma current. (Since you want fading you have to start with a larger current. However if you just want the LED on for three seconds you may redesing using a smaller current like 10ma. Smaller current will result in a smaller and cheaper capacitor) So by Ohms Law Rs = 10.75 / .020 = 537Ohms Rs By connecting our LED and RS to Vs the LED will turn on and off as Vs is turned on and off. You need to install a capacitor in parallel with the Vs supply but which remains connected to the LED and Rs. (Put a cap in parallel with Rs an LED) The capacitor will supply power to the LED when the Vs is turned off. The key here is to size the capacitor so that it can supply enough current to the LED for at least 3 seconds. A capacitor hold a charge measured in Coulombs. A coulomb is 1 amp flowing for 1 second. C= Amp*Sec Since we know we know the LED will be drawing 20ma for three seconds we can compute charge. C = .020*3 = .06 Coulombs The number of coulombs which may be stored on a capacitor is based on the Voltage AND Capacitance in Farads: C = F*V Since the LED won't light below 1.25V we will use the Voltage dropped across the series resistor (10.75V) .06C = F * 10.75V F = .06 / 10.75 = 0.0055 Farads or 5.5mF
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