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

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Discussion Overview

The discussion revolves around designing a light circuit that remains powered briefly after the main power is cut. Participants explore the use of a diode and capacitor to achieve this effect within the context of an existing project involving multiple light circuits.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests using a small signal diode (e.g., 1n4148) in conjunction with a capacitor to keep the LED on for a short duration after power loss.
  • The time constant (\tau) for the circuit can be calculated using the formula R*C, where R is the resistance and C is the capacitance.
  • Another participant questions whether the ballast resistor refers to the resistance of the flasher circuit or if an additional resistor is needed in line with the capacitor.
  • There is a discussion about the forward voltage drop of the diode (0.7V) and its implications for the required input voltage to maintain 5VDC for the flasher circuit.
  • A later reply clarifies that if the intention is to power the flasher circuit, an additional ballast resistor may not be necessary.

Areas of Agreement / Disagreement

Participants express varying interpretations of the circuit design, particularly regarding the role of the ballast resistor and the voltage requirements. No consensus is reached on the optimal configuration or the necessity of additional components.

Contextual Notes

Participants acknowledge the need for specific measurements, such as the current required by the circuit, to determine appropriate component values. There are also references to external resources for further understanding of LED resistor requirements.

Who May Find This Useful

This discussion may be useful for hobbyists and engineers interested in circuit design, particularly those working with LED circuits and power management techniques.

nafifics
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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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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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