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I am trying to measure the voltage coming out from my OpAmp after connecting it to my Photodiode. As a Photodiode I am using the BPW34S from Vishay and the Op Amp is the AD822ANZ from Analog Devices. According to the Analog Devices Photodiode Circuit Wizard the following image should be my connection:

enter image description here

The Cathode of my Photodiode is connected towards the Op Amp, while the Anode is connected to the Ground. I am using a 9V Battery to power my Op Amp as a single supply. The '-IN1' is connected as shown in the above image, '+IN1' is connected to the ground, as is 'V-' and 'V+' is connected to the positive side of the 9V Battery. As ground I am using the negative side of the battery. (I have also tried using the 5V and GND output of a RaspberryPi 4 but I thought that maybe I was not getting enough Voltage).

I have measure the voltage between the Vout and the Ground of the battery with a Voltmeter, but I always get a value of 0V as output from my OpAmp and there is no change when I put more light on my Photodiode.

My Photodiode is working when I connect it simply with a resistor and measure the Voltage passing through the resistor. But I need it to work with an OpAmp as I want to connect it to an ADC afterwards.

I am doing something wrong with the wiring or is there a problem with my Op Amp?

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    \$\begingroup\$ Double check anode and cathode as otherwise, this circuit should work and the op-amp is a good choice. \$\endgroup\$ – Andy aka Mar 2 at 15:02
  • \$\begingroup\$ Your circuit looks like a TIA (Trans Impedance Amplifier) to me which converts the current from the photodiode into a voltage. As the current from the photodiode is usually quite small, the TIA needs a large conductance. In your circuit that conductance is very small because Rf is only 178 ohms, if your photodiode really outputs 11.3 mA then that could be OK though. If you want to amplify the voltage of the photodiode, this is not the correct circuit. You might want to build a "buffer" circuit using the opamp. \$\endgroup\$ – Bimpelrekkie Mar 2 at 15:12
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    \$\begingroup\$ You will need more than 10mW of light to create 11mA of photocurrent. That is a lot. How much light are you putting on it? \$\endgroup\$ – Kevin White Mar 2 at 15:26
  • \$\begingroup\$ Yes that was the problem of my circuit. I optimized it to realise when a laser was shining into it (hence the small resistance) but was testing it with normal light and the light from my phone. When changing the resistance to around 100kOhm, the measured voltage is definitely higher and reacts more! Thanks a lot! \$\endgroup\$ – Samboff Mar 2 at 15:32
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If you are using a single supply, ground isn't in your allowable common mode input range, which would stop at 100 mV for the AD822. You need a negative supply, or a different circuit.

Also, current would drive this circuit output below zero, which won't work with a single supply.

A quick fix would be to tie the anode and noninverting input positive with the same positive voltage reference

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  • \$\begingroup\$ If I move both to the positive 9V side of the battery, I just get a reading of 8.6 V, which doesn't change with the amount of light intensity. This is also the V when I read the Voltage of the battery. \$\endgroup\$ – Samboff Mar 4 at 8:22
  • \$\begingroup\$ 9V is bad on the other side. Op amps don't work too near the rails \$\endgroup\$ – Scott Seidman Mar 4 at 13:08
  • \$\begingroup\$ You need a voltage in between \$\endgroup\$ – Scott Seidman Mar 4 at 13:08

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