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Reversing a Photoresistor?

  1. Feb 13, 2010 #1
    I'm a total newbie at this, but I have a photoresistor set up so that it controls the shorting on an audio circuit. When light is on, it shorts and cuts the sound. When it is dark, it closes and allows the circuit to run normally. How can I reverse this so that light will make the sensor close the short, thus allowing sound to run through?

    More simply, I just want the photoresistor to open with darkness, and cause more resistance with increasing light. Sorry for being so layman about it, but like I said, I'm very new at this. Thanks for any help.
  2. jcsd
  3. Feb 13, 2010 #2


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    http://dl.dropbox.com/u/4222062/photo%20R.PNG [Broken]

    It sounds like you have something like the first arrangement above.
    The photoresistor becomes low resistance compared with the series resistor, so less of the incoming signal appears across it.

    If so, you can reverse the two components and get the opposite effect where most signal is passed if the lighting is bright and little if it is dark.

    R could be 10 times the value of the photoresistor when it is in bright light. You could measure this with a multimeter on "ohms".
    Last edited by a moderator: May 4, 2017
  4. Feb 13, 2010 #3
    I tried putting a resistor before and after the photocell, but it seems to have the same effect either way. Are there any other ways to do this?
  5. Feb 14, 2010 #4


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    Maybe you should draw a schematic of what you have tried?

    If you find the program MSPaint in Windows you can easily draw up the circuit with that. Or, you could just modify my drawings.
    Use the "attach thing" at the top of the eidt screen to attach your drawing. It looks like a paper clip.

    The important thing is that the photo resistor is either in series with the signal or across from the signal path to ground.
    Last edited: Feb 14, 2010
  6. Feb 14, 2010 #5
    I've got the cell setup between two points that short the whole circuit. I want the cell to "close" when lit, so that it controls whether or not the circuit is shorting. I realized that a night light does the same thing. In light, it closes the circuit. Do you know of a way I could just break open a night light and connect it through that?
    I really appreciate all the help, thank you.
  7. Feb 14, 2010 #6


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    A photoresistor gets lower in resistance as light falls on it and will reduce the signal volts when connected as the LH diagram above. You need to get the values of the resistance on the diagram and the load (the following amplifier input) for best results.

    In the RH diagram, you also have a potentially working system.

    The precise effect of the photoresistor in the circuit depends upon its actual values in light and dark and also the value of the other R (and the output and input resistances of the audio feed and following amplifier).

    If it really does cut out the audio signal, in the first case, when lit, then it must be going to a very low value c/w R. Can you ascertain the actual change in levels?
    As vk6kro says, it would be a good idea to measure some actual values if you want a good answer to this question.
    There will probably be more in a nightlight than just a photoresistor, btw.
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