First: thank a lot for the kind reply. Now I understand.
On a keysight/agilent 33210A for example, https://mse.gist.ac.kr/~flexible/data/file/equipment/873648049_T7A1xeQO_2.Agilent_33210A_2810_MHz_FunctionArbitrary_Waveform_Generator29.jpg on the right of the output there is a marking "42Vpk" indicating you can float the shield of the output BNC upto 42V above earth ground. If the symbol is just a earth ground symbol, it is directly grounded, and you have to take care where you hook it, much like a scope ground lead. If your scope and sig gen are both plugged into the wall, measure resistance between the BNCs on both and you can see if the sig-gen is floating or earth referenced. The equipment manual should specify too.
EDIT: looked up the SDG1010, the symbol indicates its earthed directly, so no floating the SG output.
at the waveform-generator's output, besides 50 Ohm, there is also the GND-symbol. So it's directly grounded (as you have also found it in the manual) ... hmmm ... this bring me to the following assumption: for sure, I should not connect it to my 0.7V. Because the waveform-generators GND would pull those 0.7V down to GND. So, I assume, I have to connect that waveform-generator-GND with NOTHING (perhaps only with the housing of my circuit for shielding???)
As far as the first comment, a resistor makes a good voltage to current converter
ahhh ... now I think I got it. I know the waveform-generator's voltage at the output, I also know the voltage at the inverting-input of the TIA-OpAmp (=0.07V). I also have the waveform-generator's output resistance. Hence, based on U
genertor - 0.07V = R
sum * I
target I should be able to controll my I
target by tuning U
generator and R
sum. But wouldn't the voltage from the waveform-generator bias my photodiode? Respectively, how could I avoid biasing my photodiode .... hmmm ... I need to draw some schematic to get a better understanding of this. As soon as I have something reasonable, will post it.
Also, in your schematic, you have a bias, but both ends of your photodiode are hooked up to op-amp inputs, meaning they will be at the same potential, no bias applied ( By the rules of ideal op-amps) If youre trying to reverse bias your photodiode, then the anode of the photodiode should be connected to your 0v, so 0.7v of reverse bias would be applied. Changing this would mean your sig gen output - current pump in and out -, and scope reference lead, will all be attached to 0V, and everything should work ok.
No, I do not want to reverse-bias the photodiode. To avoid a bias, the TIA's non-inverting input and the photodiode's anode have the same potential. That this common potential is 0.7V is on purpose, since I want to have a DC-offset. But I see your point, that my Current Pump will generator a voltage different than 0.07V at the photodiode's cathode which will bias it. But I can't follow you, how to avoid this bias ("but both ends of your photodiode are hooked [...]"
However, ConKbat, thanks a million time for your kind effort so far!