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I'm about to build a Luxmeter. I want it to give me a respons of 1mV/lux and be able to detect 10.000lux.

Here is part of the solution I'm about to use (4. Luxmeter): http://sales.hamamatsu.com/assets/applications/SSD/si_pd_circuit_kspd1043e07.pdf

The problem comes when I want to record the output signal with my CR1000. If I use the single ended input on the ADC I'll have the range of 0-5V and accuracy of 1.333mV (this is not good enough and also, we would only be able to "see" up to 5.000lux). But if I use the differential input of -5V to +5V I'll have the full range and accuracy of 0.667mV witch will work.

But for this to work, I'll use the construction example in the attached pdf and then what? A circuit who can give me negative response added to the coupling point above the "Volt meter" in the example circuit.

Could anyone give me an solution of how this could work?

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Your approach isn't a great idea for this application. You almost never see it that way in instrumentation design.

What you want to do is implement a series of sensing scales. That is you have a system that can work over different light intensity ranges and you switch between them to effectively increase your dynamic range (range of light input you can handle) without increasing your voltage range. As a bonus you can get higher sensitivity from the decreased power supply noise (greater CMRR) if you're careful.

Here is an architecture that I helped develop:

enter image description here

The module looked like this:

enter image description here

It could go from complete darkness to well beyond 10k lux with accuracy better than the commercial meters of its day on a single 5V rail. You can find a number of papers by me and others on the project here.

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  • \$\begingroup\$ That is an impressive project, thanks for sharing! So what I acctually could do is slightly alter the circuit and measure over all three resistors with the ADC or Muxing the load resistors to one ADC port, right? \$\endgroup\$
    – Christian
    Commented Jan 21, 2013 at 15:27
  • \$\begingroup\$ @Christian -- You got it! That's the basic idea. \$\endgroup\$ Commented Jan 21, 2013 at 21:48

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