Electronic – Why is an op-amp necessary for photo detecting circuit generally

operational-amplifierphotodiode

I am trying to build a photodetecting circuit with a photodiode Hamamatsu Si PIN photodiode S6036

As I don't have much knowledge about circuit, I think a reverse voltage supply and a resistor are enough to build the circuit:

But, when I googled for a photodetecting circuit, the most examples are using op-amp. With a further search, I found that a such circuit is called transimpedance amplifier:


(source: wikimedia.org)

With my rough calculation, in the first figure, V_OUT across R will be I * R (Suppose that I is the current by a photodiode.) And in the second figure, V_OUT at the end of the op-amp will be I_P * R_F.

Both equations have the same form. Then, what reason makes TIA circuit si the widely used photodetecting circuit? TIA circuit can have higher gain? bandwidth? or stability?

Best Answer

Both circuits are used and it depends on the application which of the two you should choose. There are two important differences between the modes the photodiode is used in those circuits:

  1. The first mode is called photoconductive. It has a relative large reverse voltage across the diode which results in
    • low junction capacitance (good for high speed application)
    • high dark current (leakage current; bad for linearity)
  2. The second mode is called photovolatic. It has zero voltage across the diode (note: the polarity of the diode doesn't matter much; if you reverse the diode it will just reverse the polarity of the output voltage; but still works)
    • high junction capacitance (bad for high speed application)
    • no dark current, very good linearity (for quantitative measurements)

Photoconductive mode would be advantageous e.g. for digital data transmission.
Photovoltaic mode would be advatageous e.g. for measuring irradiance.

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