How Can I Keep a Light Circuit Powered Briefly After Cutting Power?

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To keep a light circuit powered for one to two seconds after cutting power, a small signal diode (like a 1N4148) and a capacitor in parallel with the LED are recommended. This setup allows current to flow when power is on while blocking reverse current when power is off. The time the LED remains lit can be adjusted by selecting an appropriate capacitor value based on the circuit's resistance. It is important to consider the forward voltage drop of the diode, which may require boosting the input power to 5.7V if the circuit needs 5V. Properly calculating the resistance based on the flasher circuit's average current consumption is also essential for optimal performance.
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Hi folks! First post here, I have a question regarding what should be a simple electric circuit and I hope you can help.

I want to add a light circuit to a project that has 2 other light circuits already. I want this 3rd light circuit to remain on very briefly (one or 2 seconds would be enough) when power is cut but not the other light circuits. I believe a diode and a capacitor will do the trick (an in-line diode and a capacitor across power (5V). I don't know for sure so I'd like some confirmation that this will work and is there an online program that will allow me to caculate to cap and diode specs? Other than 5V power I don't know the current required yet, but I can measure it if needed. It's a circuit with 4 leds powered from a flasher board.

Thanks very much in advance.
Jeff
 
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Welcome to PhysicsForums!

If you put a small signal diode (e.g. a 1n4148) going to a capacitor in parallel with your LED and ballast resistor, this should do the trick. Obviously, you want the diode to allow current through when power is on, but block reverse current when power is off.

The combination of the ballast resistor and capacitor will have a time constant (\tau) equal to R*C:
http://en.wikipedia.org/wiki/RC_time_constant

By choosing a sufficiently large value for C, you can keep your LED on for a little while after you cut power to the circuit.

Good luck!
 
Thanks MATLABdude! This more or less confirms what I was thinking. By ballast resistor I assume you are talking about the resistance of the flasher circuit? Or do I need to add a resistor in line with the cap somewhere as shown in the pic. I have also been told the diode forward voltage is 0.7V, so if the flasher circuit needs 5VDC, I will have to boost input power to 5.7VDC. Is this correct?

powerhold.jpg
 
Sorry, I was under the impression that you wanted to drive the LED directly and with a current limiting resistor, not through your flasher circuit. The ballast resistor (a.k.a. current-limiting resistor) is used to control current to the LED and drop voltage to what your LED needs:
http://led.linear1.org/why-do-i-need-a-resistor-with-an-led/1/

If you want to power your flasher circuit, you don't need a(n additional?) ballast resistor. While not remotely correct, you can come up with a rough resistance ("R") based on the average current consumption of the flasher circuit (V=I*"R")
 
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