How to modify photocell to turn on later

Discussion in 'The Projects Forum' started by Patrick Clark, May 19, 2016.

  1. Patrick Clark

    Thread Starter New Member

    Apr 10, 2016
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    Hello All,

    To activate a floodlight at my house I have a simple AC photocell that takes AC in (BLACK wire and WHITE wire) and uses a LDR to connect the Live wire (BLACK) to the Load wire (RED) when the light level falls below a certain level.

    My issue is that it comes on too early, about 30-40 minutes before it's actually dark enough for my floodlight to be needed.

    I would like to modify the circuit to delay turn-on. I opened up the photocell housing and I see that the LDR is connected to a set of contacts and it appears as though current heats one of the contact blades and is it warms up, the tension on it causes it to make contact with the contact blade next to it, and that allows current to flow out the Load wire.

    The LDR measures roughly 2K Ohms in full light to 400K Ohms in dark.

    Would adding a resistor in series with the LDR bring the resistance up and therefore delay the turn-on?

    Thanks!



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  2. #12

    Expert

    Nov 30, 2010
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    It looks to me like it's firing an electromagnet to make the main contacts close.
    Anyway, look at the logic. The photocell is increasing resistance as darkness happens. That means the current is decreasing so the contacts must be normally closed and the electromagnet is holding them open. More current seems necessary, even if it isn't an electromagnet. You would place a resistor in parallel with the photocell. Somewhere in the range of 4K to 10k should work.
     
  3. Patrick Clark

    Thread Starter New Member

    Apr 10, 2016
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    A 10K resistor in parallel with the LDR would give me a resistance of around 1.67K to 9.67K instead of 2K to 400K. That seems to be quite a drastic change to make. Thoughts?
     
  4. SLK001

    Well-Known Member

    Nov 29, 2011
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    If you don't like 10k, try 2MEG. It is the 400k that it is reaching too fast, so experimenting with high parallel values may get you where you want to be. 2Meg in parallel with 400k is about 333k.
     
  5. #12

    Expert

    Nov 30, 2010
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    The switching point is NOT 400K ohms. It is a lot lower. It is when the light just starts fading, not complete darkness where you find 200k. The last one I built switched at 680 ohms, which is not even in the range where yours is working. My photocell eventually arrived at 2 million ohms in complete darkness, but that is NOT the switching point.
    I'm thinking yours probably switches somewhere in the 10k to 20k range.
    If the light stays off all the time with 10k in parallel, try 22k, then 47k then 100k. You will find it eventually.
    Or, you could measure the resistance of the actual switching point by taking the photocell out and using some resistors, then measure the photocell at the time of twilight when you want it to switch and calculate the resistance needed.

    ps, this means I am disagreeing with SLK001. He's calculating to find something in parallel with 200k which only happens in dead darkness with no moon. The truth is, you're working in the range where the light is fading, but it's still a lot brighter than the inside of a black sack at the bottom of a coal mine.
     
  6. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    If you can measure the LDR at full light and full dark, measure it when the output is activated, and later at the darkness you want as the transition point. These two resistances plus Ohm's Law should be enough to calculate a solution.

    ak
     
  7. SLK001

    Well-Known Member

    Nov 29, 2011
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    Not really disagreeing... It's just a matter of degree. I agree that it might switch earlier than 400k (someone will have to measure it), but it would depend on the LDR. A 1 or 2MEG resistor will give a small change - maybe too small at the switch point, but it will help the OP to figure out the value he needs.

    I also wonder if that screw in the original post's pic is for adjusting something (like maybe, when to switch?).
     
  8. #12

    Expert

    Nov 30, 2010
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    Me too.
     
  9. AlbertHall

    Well-Known Member

    Jun 4, 2014
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    Agree with SLK001 that that screw is there for making exactly the required adjustment.
    I suspect that the working method is that the LDR is in series with a heating element which heats a bimetal strip and so activates the switch. That would fit well with that adjusting screw.
     
  10. Alec_t

    AAC Fanatic!

    Sep 17, 2013
    5,813
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    Give the adjusting screw a tweak and see what effect it has.
     
    GopherT likes this.
  11. Patrick Clark

    Thread Starter New Member

    Apr 10, 2016
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    Thanks for all the input, guys! Before I do the testing with the adjustment screw and different values of resistors, let me ask this.

    I normally use cheap Arduino based board for my light sensors which allows me to dial in exactly what light levels the switching happens.

    I came across 2 of these thermal type photocells and thought I'd use them in my current project to keep from having to have a wall wart powering the Arduino board 24/7, thinking I'd cut down on power consumption during daylight hours.

    However, if the photocell in this thermal type sensor is low resistance during daylight, and allowing current to flow in order to keep the bimetal strip heated, that's using power too.

    So in your opinions, which would use the most power during daylight hours, the thermal sensor or a wall wart powering a Arduino?
     
  12. GopherT

    AAC Fanatic!

    Nov 23, 2012
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    Thermal sensor. A decently efficient wall wart and an Arduino with almost nothing to drive takes almost nothing until the ADC senses dark on the LDR (or the day/date looks up when it should turn on based on the sunset times you entered into the arduino code for your location.
     
  13. #12

    Expert

    Nov 30, 2010
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    If it's a thermal type, it's probably very inefficient, but you have it in your hand (and we don't). Why don't you just measure the current instead of asking a bunch of strangers to guess?
     
  14. Patrick Clark

    Thread Starter New Member

    Apr 10, 2016
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    On the off chance that someone here has extensive experience with them and knows the answer to which would be more efficient. If you don't know you can just refrain from chiming in. That's how I function, anyway.
     
  15. #12

    Expert

    Nov 30, 2010
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    If anybody here had extensive experience, they would have told you. Meanwhile, you prefer to spend over 2 hours on, "an off chance" that a stranger can guess the actual values instead of achieving an indisputable measurement which you can do in less than 2 minutes? Your choice not to help yourself when you have everything you need disqualifies you for my prime criteria: I help people that want to help themselves. You can continue to waste your time as long as you want without hearing from me again.
     
  16. Patrick Clark

    Thread Starter New Member

    Apr 10, 2016
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    You've assumed a few things incorrectly, Sir. I do NOT have a way to measure current. The cheap-o meter I have does resistance and DC voltage and that's all not current.

    This post hasn't been up that long so you don't know WHO may have weighed in later, and with what level of experience with these photocells. You shouldn't make broad statements like that, it makes you look foolish.

    Also, I haven't spent over 2 hours on anything. I post a question then I go about my day. Maybe you post things and sit by your computer waiting on replies, but I don't have that luxury.

    If you don't want to help, don't. But YOU wasted a few seconds to type in how lame I am and that you don't want to help. Now please just bugger off and go "help" someone else.
     
  17. DUFFER

    New Member

    May 3, 2013
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    Mr Clark

    I guess the others are just too polite to point it out, but if you can measure the resistance of a resistor and the voltage drop across it, you can too measure current. You should get a fairly accurate reading because even the crappiest meters today are fairly accurate in the voltage and resistance modes.(E=IR)
     
  18. GrahamRounce

    New Member

    May 6, 2009
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  19. GrahamRounce

    New Member

    May 6, 2009
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    Why not just partly cover up the photocell?
     
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