operational-amplifierphotodiode
In the below schematic,
Photodiode is used ( not diode)
OP-AMP OPA 380
So, Vo = If X Rf
Case1:
IPh= A , Rf= 10K, VCC pin =5V
Vout = A X Rf = A(10K)
Case2:
IPh= A , Rf= 10K, VCC pin =3V
Vout = A X Rf = A(10K)
Does it means, Vout will be irrespective of VCC voltage?
Thank you in advance.
The way you're checking your circuit is not appropriate. The dark current specified in the datasheet is for 20 V reverse voltage applied. However your circuit does not reverse bias the photodiode.
The datasheet shows typical dark current vs reverse bias:
With no reverse bias, you should expect 0 dark current, as in any other diode with 0 V applied.
You need to test your circuit with some light applied to the photodiode. An input signal on the order of 10 to 100 nW (depending on wavelength) should produce a measurable output signal. Be aware that your photodiode has a strongly directional response, and the input light must be entering from nearly "straight on" to be detected. This will limit your ability to detect diffuse reflections.
There is nothing more you need to do. Your circuit without the diode is stable all by itself. R2 provides DC feedback so that the output will be driven to the same as the positive input, which is ground.
Think of it this way. Your circuit is basically a transimpedance amplifier. By disconnecting the diode, you are feeding 0 current into the transimpedance amplifier. 0 current is a valid signal in this case, and no different from the diode being in the dark.
Also, the datasheet is being too specific about what to do with unused opamps. You want to drive them so that they don't pick up noise or oscillate, but otherwise use little current. There are various ways to achieve this. They only show some examples.
Best Answer
Yes, the output voltage of your amplifier is independent of the power supply voltage, as long as that power voltage is large enough. Of course it can't be so large as to blow up the opamp either.
Note that the direction of the photocell current of the diode is down in this instance. That means the current flows right to left thru the resistor, and the right end of the resistor will therefore be at higher voltage than the left end. Due to the closed loop feedback, the opamp will make its output whatever it takes so that its negative input matches its positive input, which is ground. The output will therefore swing from ground (0 V) at no light to some maximum a full light. The power supply range must include that, plus whatever headroom the opamp needs to operate correctly.
If this is a "rail to rail" input and output opamp, then the negative output of the supply can be connected to ground, and the positive needs to be at least whatever the maximum output voltage will be due to the brightest light.