Krdytkyu

So I have bought a photodiode, which I intend to hook up to the analog port of my Arduino, so that I can measure differences in the light that the photodiode picks up.

I realized, that when I don't ground the photodiode, I get a lot of mV spanning from 2600mV to 5400mV when I measure with my multimeter - Obviously, however, I cant supply my analog ports with 5.4v or possibly more, as I would destroy the Arduino.

I then tried to ground my anode with the output just before the resistor as follows:

^{simulate this circuit – Schematic created using CircuitLab}

Unfortunately now, I get close to nothing with a 10kΩ resistor. even with 40kΩ, the output is nothing to brag about - I would at least expect something in 1000mV range, but I barely get 200mV.

What's going on here - Am I doing anything wrong?

You have a datasheet that references two different versions with different responses, nevertheless back of envelope calculations are...

Your photodiode (datasheet references two) has a 6.3uA or 13uA typical Isc at 100Lux. 6.3uA*40K= 252mV, 13uA*40K=520mV.

So no major surprise there...

This doesn't take into account your light source spectrum which will modify these values (see the spectral response curve).

If you want higher voltages, you will need some amplification (plenty of info on "photodiode amplifier") to get you started.

A specification for accuracy is missing.

As others have pointed out, leakage current of your Arduino input pin likely limits the accuracy, especially for low-light measurements. Leakage current of the photodiode is probably less than that of Arduino, but also impacts low-light accuracy. Leakage currents are affected greatly by temperature.

If you're using analog-to-digital converter, increase the current-sensing resistor to a large value to improve sensitivity, but be aware that large value resistors, while fairly temperature-stable themselves, make temperature-variable leakage currents impact accuracy to a greater degree.

If you're only interested in light level changes, then to a first approximation you can calibrate out the offsets caused by leakage by taking a zero-light reference reading as a stored reference, and measure light-level changes from that point. A change in temperature might require a new reference reading.

Consider too the maximum light level you anticipate. This will set the maximum value of current sense resistor...as the voltage across the resistor approaches +5V, the diode's voltage decreases - your detected voltage is no longer proportional to light intensity.

A requirement for very wide-range detection (both low-light and high-light situations) might suggest switching in various values of current sense resistors. An extra I/O pin can be programmed for a logic low, or else a high-impedance state where its attached resistor is disconnected. But now leakage is from two I/O pins:

^{simulate this circuit – Schematic created using CircuitLab}
In this circuit, high-sensitivity is achieved by programming the I/O pin to be high-impedance, so that R2 (100k) is not connected. R1 (10Meg ohm) is the active current-sense resistor. However the leakage current of the R2 I/O pin still impacts accuracy.

To decrease sensitivity in high-light situations, the I/O pin is set to be active logic low. Now you have 10Meg in parallel with 100K as the current-sense resistor.

This is pretty basic.

Your DMM is 10 MΩ input which shunts the low current into a higher voltage instead of 10 kΩ. Since 10 MΩ is 1k bigger than 10 kΩ, so is the voltage yet the same power P = VI.

Depending on your experiment and desired dynamic range of voltage, change the value of R1 to obtain the desired conversion of PD from I to V.
(conversion factor is just Ohm's Law V=IR). Perhaps 100k to 1M is more suitable.

Panasonic makes an inexpensive 5mm radial "Light sensor" that compresses the range with a log scale so >4 decades of light input from near dark to bright sun gives an output to 5V output with a selected R-value to choose your optimum input light range.

The sensitivity, S for Silicon, is 0.6 A/W at λ=λp with a declining sensitivity from IR towards blue. This is the same as 0.6 uA/uW. The output current depends on the surface area of ~ 5 x 5 mm like a tiny PV power source.

Consider 1uA times R1 = 1Mohm * 1uA = 1V output. This simply uses the shunt resistor to convert the photon microamps * R = microvolts.

For overvoltage protection, the analog inputs are usually ESD protected with Diodes but limited to 5mA or so RMS so in bright sunlight at 100 klux, you might expect a short circuit current of 13 uA/100 lux * 100 klux = 13 mA.

To prevent this from flowing into the internal Analog port, they use an ESD-protection shunt diode to Vdd, but it is rated for only 5mA steady or so.

You can solve this in many ways to prevent Vin > 5V by PD cathode voltage drops (0.7V) from 5V or add an input series R to the Analog port from 1M shunt to 10k series.

"Shunt" means parallel or bypass.

Can you figure out my suggestions?

As pointed out in another answer, the configuration you have will saturate (no voltage across the diode) and cause non-linearity at full scale.

You don't describe your application at all (light value range) so it's hard to divine an answer.

For example, Let's assume you are going to be sensing in the range of 1-10k Lux (See here for ranges/scenarios). At maximum Lux level you'd expect the S1223 to pass about 6.3mA at full scale. You want that full scale to be either 5V or 3.3V depending on the Arduino you are using. You need to be able to adjust your full scale setting (calibration) but are limited to the resolution of the A/D for the low end of the scale (about 6uA/LSB). You also here run into leakage currents for the A/D input that may dominate a reading on the very low end.

You also need to ensure that there is a reasonable voltage across the photo diode to prevent no-linearity, so driving the diode from the same supply as your MCU is not going to work well.

Assuming you are driving the Arduino from Vin and not from the MCU VCC level, you could do the following:

^{simulate this circuit – Schematic created using CircuitLab}

Assuming a 5V Arduino and a Vin of 9V or greater, the above will ensure that you don't exceed the A/D voltage yet always have at least 4V across the photo diode.