i would like measure a tiny bit off reflected light from a modulated led in an ambient light environment. I'm afraid the large ambient light output wil saturate my amplifier with the high amplification i need for the modulated led output.

Is there such a thing as a selective photodiode amplifier ? So only the light at the led carrier frequency gets amplified ?

  • \$\begingroup\$ A simple High-pass Filter can help here. \$\endgroup\$ – JimmyB Jun 21 '16 at 14:51
  • \$\begingroup\$ You might also want an optical band-pass filter at your frequency (light color) of interest. \$\endgroup\$ – user2943160 Jun 21 '16 at 15:03
  • try to optically filter as much unwanted light as possible (e.g. if your LED emits IR use a filter that filters out non-IR light; if your LED emits blue light use a filter that filters out other colors especially IR)
  • use an amplifier that amplifies only signal changes (AC coupling)
  • if that is not enough use amplification and band pass filtering (requires knowledge of signal frequency)
  • if that is not enough use a lock-in amplifier (requires knowledge not only of signal frequency but also of phase)
  • \$\begingroup\$ thanks a lot, i'll try these out. In your opinion: can the output of a photodiode be ac coupled directly into the first amplifier stage ? All the reference schematics i've seen so far have the diode amplified by a transimpedance amplifier and ac coupled after this first stage. \$\endgroup\$ – JakkeFire Jun 22 '16 at 7:53
  • \$\begingroup\$ I think it can't be directly AC coupled. First use an transimpedance amplifier like you have seen (no AC couling yet) and than AC couple ist output to another amplifier. It's important that the amplification (\$V_{out}/I_{photo}\$ ratio, i.e. feedback-resistor) of the transimpedance amplifier is not too large in order to avoid saturation by unwanted light. The more unwanted light you have the smaller the amplification in the transimpedance amplifier needs to be. That's why it is good to filter out as much unwanted light as possible. \$\endgroup\$ – Curd Jun 22 '16 at 9:20

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