operational-amplifier
I've wired this circuit :
simulate this circuit – Schematic created using CircuitLab
When Vin- not connected to R1 I get Vin+ – Vin- in interval [0, 0.5]V when moving potentiometer
After, if I connect Vin- to R1 I get Vin+ – Vin- in interval [0, 1.8]V when moving potentiometer
Vout is in range [1, 3]V
Why does Vin- range change when I connect it to inverting input?
lah
The current produced by the photodiode in the presence of light is a reverse current not a forward current. This means current flows from R1 thru D1 to ground.
This means that to keep both input pins of the op-amp at the same voltage (e.g. 0V), the op-amp output has to rise above 0V to feed current into R2 thus keeping -Vin and +Vin at precisely the same potential\$^1\$. As more light hits the photodiode the op-amp output has to rise higher.
If the photodiode were the other way round photo-current would be traveling into R2 and this means the op-amp output has to provide negative voltage and, of course it can do this when it has a negative supply.
I'm assuming you are using rail-to-rail op-amps in this analysis.
Same problem with solar cells - you need the left hand end of R1 to be pulled negative by the solar cell for this to work.
\$^1\$ If you don't understand why -Vin = +Vin then this is another question and is unrelated to photo-diodes or solar cells.
The datasheet is really quite clear about this:
This opamp isn't specified for only 5 V supply.
These parameters shown above are for ±15 V supplies. The common mode input range only goes to within 3 V of either supply. Nothing is said how that scales to lower supply voltages, so assume it is at least that much. That means there is no common input range left with only a 5 V supply.
Likewise, the output can't drive to within 3 V of the supply rails with a 10 kΩ load, and not to within 5 V with a 2 kΩ load.
You need to actually read the datasheet before designing a part into a circuit.
Best Answer
Because some current flows through R1, which "ruins" your voltage divider.
I.e. your voltage divider only works "properly" when negligible current flows through the wiper of the potentiometer. When R1 is disconnected, no current can flow, and the pot acts as a "proper" voltage divider. When you connect R1, a non-negligible current flows through R1, causing your voltage divider to "droop".