Sensor Help

Discussion in 'The Projects Forum' started by victorment, Jul 14, 2011.

  1. victorment

    Thread Starter Member

    Jul 4, 2010
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    0
    I have an LDR and I want to detect the intensity of the light and its output varies only 2mV of two different light intensity. What should I do to make 1v difference in every intensity of the 2 lights detected. Could Someone help me. Thanks in advance.
     
  2. debjit625

    Well-Known Member

    Apr 17, 2010
    790
    186
    Do you know how to use operational amplifier ?
     
  3. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    First, it would be a good idea to show how you have your LDR connected.
    Please post a schematic of your circuit with values of components used.
     
  4. victorment

    Thread Starter Member

    Jul 4, 2010
    52
    0
    Here is my circuit diagram.

    [​IMG]
     
  5. victorment

    Thread Starter Member

    Jul 4, 2010
    52
    0
    Yes. I know. Should I amplify the voltage? would the distance between the output voltage get larger?

    Thanks
     
  6. SgtWookie

    Expert

    Jul 17, 2007
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    Try increasing the 1k resistor to ~50k Ohms for starters. That might bring you up to about 100mV difference. If the difference is still less than 1v for the two lights, then increase the resistance some more. Using a 1 MEG pot in series with a 10k resistor will help to speed up this process quite a bit.
    2mV is too small of a difference to calculate what amount of change needs to be made to the fixed resistance.

    Also, you have not shown us what else might be connected to the analog voltage output. If there is some load other than a multimeter, you really need to tell us about it.

    If there is a load, you will need to use a buffer/voltage follower between the LDR/resistor junction and the load.

    A common opamp connected as a voltage follower makes a good buffer for up to perhaps 20mA. There are methods to increase that current if necessary.
     
    Last edited: Jul 15, 2011
  7. CDRIVE

    Senior Member

    Jul 1, 2008
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    You should have also posted the LDR model. There are a host of models, all with widely varying specs. Once the specs are known you can logically calculate the proper value for fixed resistor (R).
     
  8. SgtWookie

    Expert

    Jul 17, 2007
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    Well, without having the LDR specifications available, one could just remove that 1k resistor, and measure the resistance of the LDR with the 1st light, then measure the resistance of the LDR with the 2nd light.

    Then calculate what the value of the 1k resistor would need to be in order to obtain a 1v difference between the two readings; or if it's even possible to do so without an amplifier.
     
  9. CDRIVE

    Senior Member

    Jul 1, 2008
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    Touché! A fine example of field expedience, or black box analysis.;)
     
  10. victorment

    Thread Starter Member

    Jul 4, 2010
    52
    0
    I have connected the output to a voltage doulbler. here is the circuit. I add a 10k pull down resistor on the output to discharge the capacitor so that it would vary. But I did not show it on the schematic.

    I would connect the circuit to a PIC to control 3 solenoid valve with respect on the voltage output of the voltage doubler to switch on one of the solenoid valve. Should I use an ADC circuit then connect the binary output to the PIC?
    Can you give me any ideas what kind of circuit should I build or correct me if I am wrong. Thanks in advance...



    [​IMG]
     
  11. Adjuster

    Well-Known Member

    Dec 26, 2010
    2,147
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    This does not seem to follow from your initial inquiry. What (if anything) has it got to do with your LDR?
     
  12. CDRIVE

    Senior Member

    Jul 1, 2008
    2,223
    99
    I thought it was just me. This is .... uh... schizophrenic! What's the point of asking for help if you're going to ignore the advice you get?
     
  13. debjit625

    Well-Known Member

    Apr 17, 2010
    790
    186
    First of all tell us the full story,we all are confused....:confused: .
     
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