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Photodiode circuit design

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stcoso:
Hi.. I'm trying to design a little photodiode amplifier with some parts that i have at hand... i attached some schematic.
 I have a problem  :-// ... the amplifier works opposite as i would expect  :o ... Basically the output voltage goes down when i shine some light on it. :wtf:


I'm using a BPV10 pin photodiode and a ada4857 op amp.



Thank you in advance.

StillTrying:
PD is in the Power Down direction.

stcoso:
Oh f**k :palm: :palm: :palm: :palm: :palm:

janoc:

--- Quote from: stcoso on March 29, 2019, 11:37:46 am ---Hi.. I'm trying to design a little photodiode amplifier with some parts that i have at hand... i attached some schematic.
 I have a problem  :-// ... the amplifier works opposite as i would expect  :o ... Basically the output voltage goes down when i shine some light on it. :wtf:

--- End quote ---

Um, and what exactly are you expecting to happen?

The photodiode will generate current into the inverting input of the amplifier (making it "positive"), so the opamp will use its output to bring that node through the feedback resistor back to matching the non-inverting input (the inverting input behaves as a virtual ground here). The output voltage will go down because that's what is needed to compensate the photodiode current out.


mikerj:

--- Quote from: janoc on March 29, 2019, 02:53:36 pm ---
--- Quote from: stcoso on March 29, 2019, 11:37:46 am ---Hi.. I'm trying to design a little photodiode amplifier with some parts that i have at hand... i attached some schematic.
 I have a problem  :-// ... the amplifier works opposite as i would expect  :o ... Basically the output voltage goes down when i shine some light on it. :wtf:

--- End quote ---

Um, and what exactly are you expecting to happen?

The photodiode will generate current into the inverting input of the amplifier (making it "positive"), so the opamp will use its output to bring that node through the feedback resistor back to matching the non-inverting input (the inverting input behaves as a virtual ground here). The output voltage will go down because that's what is needed to compensate the photodiode current out.


--- End quote ---

The photodiode in the schematic is reverse biased (since the non-inverting input is biased at +5v) and operates in photoconductive mode.  More light will cause the diode to sink more current, and the output voltage of the op-amp will increase to keep the inverting input at the same voltage as the non-inverting.

As StillTrying mentioned, the power down pin is connected to the wrong rail.

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