Transimpedance Amplifier for Photodiode with low Shunt-Resistance

I am designing a transimpedance amplifier to amplify the signal of a photodiode. I plan to use the IC ADA4350 (datasheet), it contains an Op-Amp and analog switches to select different feedback paths for different gains.

I plan to use 3 different photodiodes, one of them has a very large shunt resistance of 38GOhm but two of them have a very low shunt resistance, one is 2MOhm and the other 100kOhm. (datasheet diode 1, datasheet diode 2)

I want to use 5 different selectable gains, from 1k to 10M in decade steps. I calculate the feedback compensation capacitor for a bandwidth of 25kHz. I simulated the circuit in LTSpice, and using the diode with the large shunt resistance of 38GOhm the results are as expected. But when I simulate one of the other two diodes that have a low shunt resistance I encounter the following problem: If the feedback resistor Rf is smaller than the photodiode's shunt resistance, it works as expected. But if the feedback resistor is equal to or larger then the shunt resistor, the amplifier circuit doesn't work. The output goes into saturation.

Can somebody help me to understand what is the cause for this behaviour? How can I modify the circuit to make it work?

Kind regards,

Marcel

• The behavior you see is totally expected, you really need to study and understand this transimpedance amplifier you're working on. I suggest you start reading at: en.wikipedia.org/wiki/Transimpedance_amplifier and also electronicdesign.com/analog/… use google search to find more articles. A high value feedback resistor increases the transimpedance of the amplifier, make the transimpedance too high and the amplifier will saturate. That's totally expected. – Bimpelrekkie Oct 22 '19 at 9:25
• The input current in the simulation is 10nA. If the transimpedance gain is 10M, this would result in an output voltage of 0.1V, which would not saturate the output. – Marcel Oct 22 '19 at 10:16

$$\V_{BIAS}\$$ is DC amplified by the factor $$\1+\dfrac{R_F}{R_{SHUNT}}\$$.
So, if $$\V_{BIAS}\$$ is 1.8 volts and $$\R_F\$$ = $$\2\times R_{SHUNT}\$$ the op-amp tries to put a DC voltage at its output of 5.4 volts and this is above the 5 volt power rail.