So I have been working on this circuit that basically will activate an LED when the lux levels drop low enough. The time that the LED will remain active for is determined by VR4. I'm having two issues that I want to overcome, however. The first is that I want to make it so that LED "D6" will turn off after the time that I have chosen has passed (say 10 seconds, etc) and not turn back on again until the lux levels have risen above the threshold and then dropped below again. The second is that I want to remove LED "D5" from the circuit without altering functionality, as it is only on when lux levels are high, but the point of the circuit is that no LED is on when the lux levels get to a certain level or after the set time has passed. enter image description here

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    \$\begingroup\$ Simply remove R12 and D5. I think the circuit already is set for one shot mode so the first part is done, if not, look up one-shot 555. \$\endgroup\$
    – Passerby
    May 20 '17 at 21:01
  • \$\begingroup\$ What does your circuit do now? \$\endgroup\$ May 20 '17 at 21:06
  • \$\begingroup\$ @Passerby thank you for the diode part. With the one-shot 555, it works for the most part, however the LED won't turn off until I turn up the lux levels and I want the LED to turn off once the timer is up. \$\endgroup\$
    – RyKen
    May 20 '17 at 21:12
  • \$\begingroup\$ You might need two 555 or a 556. One to trigger on the lux change, and another to handle the timer part. Someone else may have an answer to that. \$\endgroup\$
    – Passerby
    May 20 '17 at 21:19
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    \$\begingroup\$ perhaps you could draw a graph showing various lux inputs vs time and what you wish the LED to do. At present if the lux level is low D6 will be on. If the lux level is low for just a fraction of a second D6 will be on for the time set by vr4 & c2 \$\endgroup\$
    – sstobbe
    May 20 '17 at 23:56

For the D5 question, you can simply remove it. It will not have any effect on the balance of the circuit.

The monostable mode of the 555 timer normally requires that the trigger pulse be shorter than the timer length. If not, the 555 stays triggered prolonging the length of the output pulse beyond the programmed time.

A simply fix for this is to change your circuit as follows:

enter image description here

(Image credit to tcrosley on StackExchange 180786)

Remove R9 and R10 in your circuit, add the components shown, and place the emitter and collector of Q2 in place of the SW1 switch shown.

Your timer will then run for the programmed length and not retrigger until your LUX sensor turns off and again turns on the transistor.

  • \$\begingroup\$ I don't think this will work because it relies on a fast transition of SW1 from open to closed in order to generate a pulse. Unless the OP's application involves a fast transition from light to dark (and even if it does, but the LDR doesn't respond fast enough) then their Q2 will turn on too slowly to generate a pulse - some sort of Schmitt trigger will be needed. \$\endgroup\$
    – nekomatic
    Oct 1 '19 at 13:04

If you can't design any LUX specs , then try this circuit which works with SHarp/Vishay/etc Photo Diodes PD ( not photo transistors) With suitable sunlight and aperture You can choose current for LED OFF and set R as PD current sink with 1.1V /250K = 4.4uA for twilight to turn on LED.

The threshold is typically 1/3 and 2/3 of Vdd for hysteresis

  • e.g. 1.1V, 2.2V for Vdd = 3.3V

I chose a 74ALVC14 because it operates down to 3V easily and consumes almost no current when LED is OFF and can easily drive 20mA with its internal RdsOn shown in simulation as an added R.

I would suggest 2 or 3 AA with added R or better 1 LiPo cell for more power with more LEDs if you want. enter image description here tinyurl.com/lxov7ft

  • paste above into browser addressbar for JAVA simulator.
  • Pot is just to simulate twilight

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