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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:

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?

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  • \$\begingroup\$ It isn't you can also roll your own discrete component amplifier. \$\endgroup\$ – PlasmaHH Jul 13 '15 at 16:07
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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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  • \$\begingroup\$ Good, you can also put reverse bias on the opamp circuit. (connect diode to negative supply vs ground.) The main advantage of the opamp is you can get more speed. It's limited by PD capacitance and value of resistance in the top circuit. \$\endgroup\$ – George Herold Jul 13 '15 at 19:31
  • \$\begingroup\$ Yes...but if you reverse bias the photodiode you loose the low dark current (high linearity) feature . \$\endgroup\$ – Curd Jul 13 '15 at 20:39
  • \$\begingroup\$ I'm a bit confused since you are saying photovoltaic with op-amp is good vs bad for high speed. With some googling, photoconductive mode seems to be good for high speed. I am curious that which configuration can make photovoltaic have high frequency response. I am not opposing you, I have very little knowledge about circuit and I am really curious about it. \$\endgroup\$ – Jeon Jul 14 '15 at 2:23
  • \$\begingroup\$ Yes, a variation of the second circuit can also be good for high speed (that's what George Herold was pointing out). What matters is not whether there is an OpAmp or not (of course also the 1st circuit could use an OpAmp) but whether the PD is biased or not. Biasing causes a lower junction capacitance (which is good for high speed) but it also causes higher dark current. \$\endgroup\$ – Curd Jul 14 '15 at 7:07
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An opamp is used in these situations because it can reduce the apparent impedance of the photodiode. A photodiode has very high impedance, so it's easy for measurements - such as with an ADC - to affect the result.

The opamp provides a buffer, which allows you to achieve fast response times and accurate measurements.

If you add a load resistance to both circuits, you should be able to calculate the difference in the results.

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  • \$\begingroup\$ In a TIA configuration the input impedance of the op-amp is immaterial because all the photodiode sees is a virtual earth. \$\endgroup\$ – Andy aka Jul 13 '15 at 20:26
  • \$\begingroup\$ @Andyaka Oops, good point. \$\endgroup\$ – Nick Johnson Jul 14 '15 at 9:02

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