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I'm going to build photodiode amplification circuit which is then connected to ADS1115 controlled by Raspberry Pi, it has sampling 3.3k. The operational amplifier I wanted to use is TLE2072CP.

The idea is to measure positive values of voltage which increases with the illumination of the photodiode. The light range I want to measure is 400-500 nm.

I wanted to use BPX61 photodiode or BPX21-o.

I think I can use the circuit like this:

enter image description here

where $$V_{cc}=+5V$$ and $$V_{ee}=-5V$$ which are the minimum required voltages for TLE2072CP.

Should I use another circuit? I'm looking for complete solution with correct values of elements.

I also wanted to ask is this circuit better for measuring changes of intensity when the diode is illuminated all the time?

I’ve built the circuit in the figure 6 with additional non-inverting amplifier connected to the output 6. Why increasing the gain of the second amplifier reduces Vout?

I want to measure low light levels, about 1 nx.

enter image description here


I have also a question, capacitors C2, C4 shouldn’t be connected to 0V like the lower leg of C3 and C5 instead of ground?

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    \$\begingroup\$ ti.com/lit/pdf/tidu535 I do not know why want all those components hanging off the noninverting input of the first opamp. Without more specifications it is uncertain whether the second stage is even correct, let alone required. You can't omit the photodiode from the schematics like this nor can you omit the power rails being used because they both will play a role in determining things. \$\endgroup\$
    – DKNguyen
    Jun 26, 2022 at 22:08
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    \$\begingroup\$ What range of light do you wish to measure? \$\endgroup\$ Jun 26, 2022 at 22:11
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    \$\begingroup\$ What bandwidth do you want? \$\endgroup\$ Jun 26, 2022 at 22:13
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    \$\begingroup\$ As drawn the offset at the output could be +/- 1.5V with zero light. \$\endgroup\$ Jun 26, 2022 at 22:14
  • \$\begingroup\$ Did you have any specs? For gain, offset? \$\endgroup\$ Jun 26, 2022 at 22:52

2 Answers 2

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You want a complete circuit?

Then you need to specify a few things.

What is the incident power you expect on your detector? What bandwidth are you looking for? What signal-to-noise do you need?

Your first circuit probably won't work. Look at the data sheet, and find "Dark current". Apply this to your first stage and then to the second, and you'll see that you might have a dark voltage of 14 volts - which isn't going to happen with +/- 5 volt supplies.

If you are measuring very low light levels, the second circuit is the way to go. In this circuit, dark current doesn't occur, so you get very good performance at low levels. Of course, you will also have a lower bandwidth than the first circuit, but that's the tradeoff you'll have to make.

You need to get a rough estimate of the optical power which will fall on the active area of the photodiode. Use that and the efficiency at that wavelength (from the data sheet) to determine the PD current. Multiply that by the feedback resistor and you'll get your nominal output voltage. The larger the resistor, the greater the sensitivity - and the lower the bandwidth. Since you have not specified the bandwidth you need, I can't help you on that either.

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I have also a question, capacitors C2, C4 shouldn’t be connected to 0V like the lower leg of C3 and C5 instead of ground?

Those four capacitors provide bypassing for the opamp's power rails. If you're not familiar with bypassing you should look it up before building that circuit, but essentially the smaller value capacitors should be placed as close as possible to supply pins of the opamp (pins 4 and 7) and connected to ground (or 0V as you called it). The purpose of this is to hold the supply rails as stable as possible by bypassing the inductance on your power supply lines with a low impedance capacitor to ground.

The larger value capacitor does not need to be physically close to the opamp, and could be omitted. The small value capacitor could be made larger but must be close to the opamp.

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  • \$\begingroup\$ The 0V is connected to pin 4 and the ground is connected to pin 3. It looks like single power supply, so shouldn't be capacitors C2 and C4 connected to 0V (pin 4) instead of being connected to the ground which is on pin 3? \$\endgroup\$ Jul 8, 2022 at 0:37

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