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I was trying to work with a photodiode that is supposed to detect very tiny amount of light. Using different kind of lenses I succeeded amplifying the light amount that will fall on the photodiode. Now I have to work on the photodiode’s ability to convert the light into electrons / increase the pd's sensibility. Hence I had experienced only little success getting the photodiode to receive the small amount of light and convert it into electrons.

Now I learned that an Op amp will significantly help to amplify the signal provided by the photodiode. Moreover I learned that a lock-in amplifier might be a solution in my case. Being a rookie in electronics I first want to work with an op amp (lm358n) and then try my luck with the more complicated lock in amplifier. Working on a circuit with an op amp I came up with this circuit (see figure) considering the suggestion given in this post as well How to Use SFH235 IR Photodiode Correctly?.

However the photodiode does not really react to the provided light. I guess there is something wrong with my circuit. enter image description here

Can anyone give me a hint what I am doing wrong or what I have to do to get it right?

Datasheet OP Amp http://www.ti.com/lit/ds/symlink/lm158-n.pdf

Datasheet Photodiode http://www.farnell.com/datasheets/57158.pdf

As Ignacio Vazquez-Abrams mentionied I have just test the circuits provided in the http://www.ti.com/ww/en/bobpease/assets/AN-31.pdf document. enter image description here

Indeed the sensitivity of the photodiode has increased. One question left: The circuit on the right side makes use of an LM 108 and a capacitor on pin 8 ( due to the datasheet pin 8 of the op amp stands for Comp). Not having this pin on L358 what can I do for that?

Morever the arduino analog pin normally indicates a value btw 0-1023. Now using the op amp I get a maximum value of 767 ( this means approx. 3/4 of 1023). Is there any particular reason for that or am I doing any mistake?

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    \$\begingroup\$ Have you tried application note 31 yet? \$\endgroup\$ Commented Jun 2, 2015 at 0:31
  • \$\begingroup\$ Do you talk about this one? ti.com/ww/en/bobpease/assets/AN-31.pdf ....i just see it;there are some interesting circuits \$\endgroup\$
    – Sathees
    Commented Jun 2, 2015 at 0:33
  • \$\begingroup\$ What is the power supply voltage provided to the op-amp? \$\endgroup\$
    – The Photon
    Commented Jun 2, 2015 at 4:15
  • \$\begingroup\$ And is the Arduino input actually connected to the negative terminal of the 5 V supply? \$\endgroup\$
    – The Photon
    Commented Jun 2, 2015 at 4:17
  • \$\begingroup\$ So far the postiive was connected to 5 V and the negative one to GND...so far as the first the arduino is connected to a USB port ...if I get it work the way I want it I will start using a 6V battery (4xAA) \$\endgroup\$
    – Sathees
    Commented Jun 2, 2015 at 7:52

2 Answers 2

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The circuit on the right side makes use of an LM 108 and a capacitor on pin 8 ( due to the datasheet pin 8 of the op amp stands for Comp). Not having this pin on L358 what can I do for that?

The LM108 has pins for external compensation. As you've already figured out you will not find these pins on the LM358 because the LM358 is internally compensated (e.g. the capacitor is within the chip). So if you use the LM358 you don't need that capacitor.

Morever the arduino analog pin normally indicates a value btw 0-1023. Now using the op amp I get a maximum value of 767 ( this means approx. 3/4 of 1023). Is there any particular reason for that or am I doing any mistake?

The output of your amplifier does not generate a voltage that the arduino would interpret as full scale. You can increase the amplification by choosing a bigger resistor between the output and the negative input. If you want your amplification 25% higher just make the resistor 25% bigger.

Note that the voltage at the LM358 output can't go as high as the supply voltage. It will always have a maximum output voltage that is 1.5V to 2V below the supply. That means, if your arduino analog input would give you a full scale at - lets say - 5V, then you need to supply your OpAmp with 7V at least. Otherwise the output will clamp and never reach the full scale.

You'll also likely don't want to connect your OpAmp output directly to the analog input. Analog inputs have a maximum input voltage, and you should never exceed this. If you do current will flow through a protection diode. If the current is to large that could blow up your arduino. Adding a resistor between the OpAmp output and the analog input will limit the current without influencing the reading much. 2.2k is what I usually use as a rule of thumb value.

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  • \$\begingroup\$ Thx for your enlightening answer!! Actually I wanna add a digital pin instead of an analog pin. But I test it with the analog pin so that I can see changes in value adding little light to the photodiode. I wanna the digital pin to turn into 1 if I add the small amount of light! Given the fact that the Arduino has 5v I will need to cross the analog value arrnd. 512 to turn the digital value into one... Now I have to work on noise cancelling since the daylight literally distrubs my light source... \$\endgroup\$
    – Sathees
    Commented Jun 4, 2015 at 11:01
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Your circuit uses a forward biased photodiode.

Under these conditions, the junction electric field is unable to effectively separate generated EHPs as it is weakened by the applied bias.

To get a noticeable current, you must reverse bias the photodiode, thereby reinforcing the electric field and allowing for effective separation of EHPs generated in the space charge region. See if that gets you better results. You may not need an Op Amp.

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  • \$\begingroup\$ Thx for your reply. I did a mistake on my circuit. I have just updated the picture above. I was actually using a reverse biased photodiode. However the photodiode's sensibilty due to the very low amount of light was limited so that I am conisderung an op amp. \$\endgroup\$
    – Sathees
    Commented Jun 2, 2015 at 11:03
  • \$\begingroup\$ What is an ehp? \$\endgroup\$ Commented Jun 2, 2015 at 12:03
  • \$\begingroup\$ Sorry, should have been clearer. EHP stands for "electron hole pair" \$\endgroup\$
    – Dr Coconut
    Commented Jun 2, 2015 at 12:04

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