+ Post New Thread
Results 1 to 5 of 5
  1. #1
    Full Member level 2
    Points: 729, Level: 5

    Join Date
    Nov 2012
    Posts
    126
    Helped
    7 / 7
    Points
    729
    Level
    5

    Need Help: Converting LDR resistance to LUX

    I need to display the LUX on the LCD by using LDR. My problem is in calibration. Actually it can be done by comparing the real LUX meter with my LDR resistance and find the factor. But I don't have LUX meter for now, so I decide to use the following graph which is provided by the datasheet:

    Click image for larger version. 

Name:	lx.png 
Views:	44 
Size:	17.5 KB 
ID:	86416

    Now, can somebody help me to find the multiplication factor or maybe the equation that shows the relation between LDR resistance (y-axis) with LUX (x-axis). I really have a problem with this logarithmic graph. Help please....

    •   Alt31st January 2013, 12:43

      advertising

        
       

  2. #2
    Advanced Member level 4
    Points: 10,248, Level: 24

    Join Date
    Sep 2001
    Location
    Argentina
    Posts
    1,131
    Helped
    361 / 361
    Points
    10,248
    Level
    24

    Re: Need Help: Converting LDR resistance to LUX

    Hi ArdyNT,

    I[lux] = 10000 / (R[kΩ]*10)^(4/3)

    Regards

    Z


    1 members found this post helpful.

    •   Alt31st January 2013, 12:58

      advertising

        
       

  3. #3
    Full Member level 2
    Points: 729, Level: 5

    Join Date
    Nov 2012
    Posts
    126
    Helped
    7 / 7
    Points
    729
    Level
    5

    Re: Need Help: Converting LDR resistance to LUX

    Hi there, thank you.

    Can you tell me how can you convert that logarithmic graph to that equation? Is there any reference? (I really want to understand it too)



    •   Alt31st January 2013, 13:06

      advertising

        
       

  4. #4
    Advanced Member level 4
    Points: 10,248, Level: 24

    Join Date
    Sep 2001
    Location
    Argentina
    Posts
    1,131
    Helped
    361 / 361
    Points
    10,248
    Level
    24

    Re: Need Help: Converting LDR resistance to LUX

    I'll try to proceed stp-by-step.
    We convert the log-log graph into a linear one by this change of variables:

    x = log10(I[lux])
    y = log10(R[kΩ])

    Then we have the same graph but the abscissa axis is linear (ranging from -1 to 5) and the y too (ranging from -1 to 3).
    It is a straight line whose equation is

    (y-2)/x = (y+1)/(x-4)

    It simplifies to

    x = -4/3 * (y+1) + 4

    Then, reaplacing and grouping terms, etc...:

    log10(I[lux]) = -4/3 * ( log10(R[kΩ])+1 ) + log10(10000)

    log10(I[lux]) = log10(R[kΩ]*10)^(-4/3) + log10(10000)

    log10(I[lux]) = log10 [ (R[kΩ]*10)^(-4/3) * 10000 ]

    I[lux]) = (R[kΩ]*10)^(-4/3) * (10000)

    I[lux] = 10000 / (R[kΩ]*10)^(4/3)

    Regards

    Z


    2 members found this post helpful.

  5. #5
    Full Member level 2
    Points: 729, Level: 5

    Join Date
    Nov 2012
    Posts
    126
    Helped
    7 / 7
    Points
    729
    Level
    5

    Re: Need Help: Converting LDR resistance to LUX

    Ok, thank you very much.



--[[ ]]--