I once read an article (source : http://ieeexplore.ieee.org/document/7561271/ ) about measure absorbance using photodiode The formula used is :

\$A= -\log{\frac{I}{Io}} = - \log \left(\frac{V_{sample}-{V_{zero}}}{V_{solvent}-V_{zero}}\right) \$

Where Io and I are incident light intensity and transmitted light intensity

ܸVsample and Vsolvent is voltage when the light transmits a sample and solvent respectively . Vzero is the voltage at zero light

I understand why they used Vsample - Vzero , but the substraction Vsolvent - Vzero seems to be wrong , this is true only if the solvent does not absorb the light , but in fact even water absorb light . So what did i miss here?

  • I don't understand why its only true if the solvent doesn't absorb light. You're measuring the attenuation of light transmission due to the solute. – Scott Seidman May 16 '17 at 15:24
  • Absorbance applies only at the matched wavelengths predicted, not all wavelengths. Use a Bragg diffractor – Tony EE rocketscientist May 16 '17 at 15:29
  • that cheap one in article uses RGB when RYGtGaBV different LEDs is better for measuring absorption where Gt and Ga are true and aqua green. One can also add IR and UV – Tony EE rocketscientist May 16 '17 at 15:38

Received amplitude is < 1 thus -log of ratio <1 is a positive value indicating absorption of sample(+solvent)/solvent.

PD with TIA is much more accurate than a PhotoTransistor due to huge variance of hFE. So calibration is critical for every unit and temperature stability etc.

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Most photoreceptors have a dark current, which will cause a voltage at the output of an amplifier. The equation removes this effect from the numerator and from the denominator.

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