Electronic – Why do we use an opamp to produce a voltage output above 1V in this circuit

operational-amplifierphotodiode

I wonder what advantages (regarding noise or other important factors) the opamp circuit:

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holds over a circuit consisting of only a photodiode and a resistor (the resistor has to be placed where the voltage (V) is):

enter image description here

The math should be the same:
$$ U_{out} \propto R * I_{photodiode} $$

I'm curious what your ideas are.

P.S.: I want to use a circuit to pass a voltage proportional to the photodiode current to the ADC of an ┬ÁC.

Best Answer

It's all down to speed. What your circuit doesn't show is the self-capacitance of the photodiode: -

enter image description here

Given that the signal produced by the photodiode is current (Iph shown above), if this is rapidly changing like in an optical data receiver, the junction capacitance will have a significant effect on rise and fall times.

However, with a transimpedance amplifier we are, in effect, shorting out the capacitance and now, the current signal takes the path of lowest resistance and that is into the virtual earth node of the inverting input. This vastly improves high frequency performance.