+ Post New Thread
Results 1 to 5 of 5
 31st January 2013, 12:43 #1
 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:
Now, can somebody help me to find the multiplication factor or maybe the equation that shows the relation between LDR resistance (yaxis) with LUX (xaxis). I really have a problem with this logarithmic graph. Help please....
 31st January 2013, 12:43
 31st January 2013, 12:58 #2
 Join Date
 Sep 2001
 Location
 Argentina
 Posts
 1,131
 Helped
 361 / 361
 Points
 10,197
 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.
 31st January 2013, 12:58
 31st January 2013, 13:06 #3
 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)
 31st January 2013, 13:06
 31st January 2013, 14:18 #4
 Join Date
 Sep 2001
 Location
 Argentina
 Posts
 1,131
 Helped
 361 / 361
 Points
 10,197
 Level
 24
Re: Need Help: Converting LDR resistance to LUX
I'll try to proceed stpbystep.
We convert the loglog 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
(y2)/x = (y+1)/(x4)
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.
 31st January 2013, 19:10 #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.
+ Post New Thread
Please login
LinkBacks (?)

Untitled document
Refback This thread12th May 2014, 17:57