So i am in the middle of a physical chemistry laboratory and i have built a sort

Question

So i am in the middle of a physical chemistry laboratory and i have built a sort a sort of self-made spectrometer with a photodiode and LED to detect the voltage drop as a solution is increasingly concentrated with solute/analyte. The multimeter measures the voltage drop across the photodiode as more solute is added. How does one convert voltage into units of light intensity which is what i need in order to find the absorbance? It would also be helpful to know how to calculate current into light intensity since the specs of the LED incident intensity are in amps, although i understand how to convert the voltage into current using ohm's law.

Explanation / Answer

A photo-diode's voltage is not proportional to light intensity, but its current is. The amount of light from an LED is low however, so the current is also small. The best way to measure this very small current is to use an op-amp as a current to voltage converter. This will need a low bias current op-amp like the LT1012, and a dual power supply. Use 2 x 6V or 9V batteries.. The circuit is so simple you can easily build it using the dead bug technique or a piece of matrix board (look these things up).

LEDs range from a few milli candela to a few candela luminous intensity in the beam, depending on the particular LED and its driving current and colour. Find the data sheet for your led and you will know its specified output at rated current.

Power in the light from a red led..
One candela is 1/683W per steradian for 555nm green. If we consider a red led colour of 630nm, then more power is needed to equal 1 candela. Correcting from a photopic response curve (the spectral weighting for candela, lumens, lux etc) the power of a red led to equal one candela is 1/683W * 1.38888 = ~2mW/steradian. Other colour leds will need a different factor to correct for the colour. You could look up a photopic table.

Power on the photodiode at 100mm..
A candela is the unit of luminous intensity, and is the equivalent of lumens per steradian. We can determine the power from the LED per square meter at a given distance because the area in 1 steradian is r^2 (talking about a sphere, so radius = distance). Using a distance of 100mm from the source, the equivalent area of a steradian is 100 x 100 = 10,000 square mm. Let's say the photo-diode has an area of 1mm^2, so we get 1/10,000 of a stereo radian on the photo-diode at 100mm, so in this case, a candela from a red led gives 2mW/10,000 = 0.2uW on the photo-diode active area. This is different for other distances and other size photo-diodes..

Current from the photo-diode..
As the sensitivity of a silicon photo-diode is about 0.43A/W, with 0.2uW we have 0.2uW x 0.43A = 0.086uA per candela. This is 86 nano-amps per candela of red led at a distance of 100mm.

At half the distance the current is 4 times, at twice the distance the current is 1/4 the current. With other colours the correction for spectrum (colour) will be different. If the photo-diode is more than 1mm^2 it will be a proportionally higher current.

That is why you need the amplifier. If you calculate the current from the photo-diode you can calculate the resistor needed for the amplifier, and then know its output in volts per micro-amp etc, and so per watt or micro-watt of red led light on the cross sectional area of the photo-diode (like 1mm^2).

The link below shows how this is done. If you use a 10 megohm resistor (a value you can buy), the amplifier will be 100nA per volt, so 4.3uW/V for a red LED. Knowing the distance and the colour and the area of thge photodiode from its spec, you can then work out the candela as lumens per steradian (the reverse of the above procedure). If the led is not monochromatic (e.g. a white led), you need to integrate the spectrum to find the photopic correction to the value at 555nm. Similarly for the photo-diode spectral response, which is usually shown in data sheets

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