Transimpedance Amplifier for Photodiode with low Shunt-Resistance
\$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.

There may be other factors at play but this seems the most likely one.