How does this op-amp photodiode circuit behave?
Although you can call it a "photodiode circuit", the underlying type of circuit is more commonly called a transimpedance amplifier (TIA). It converts a current - in this case the photocurrent - into a voltage.

The circuit you have outlined is simply put a combination of such a TIA and a non-inverting amplifier. Ignoring C2 for reasons mentioned in the comments, the TIA is basically D1 + R1 + R2.







simulate this circuit &ndash; Schematic created using CircuitLab

The bandwidth of this circuit is typically limited by the photodiode capacitance \$C_{PD}\$. The maximum bandwidth will be

$$BW (Hz) = \frac{1}{2\pi\cdot R_{tia}\cdot C_{PD}}$$

and the transimpedance gain will obviously be \$R_{tia}\$.

The voltage output is then high-pass-filtered by C1 + R3 (which unfortunately also worsens the transimpedance gain and bandwidth a bit), and finally amplified by OA1 + R4 + R5 by a factor of \$1+\frac{R_4}{R_5} = 11\$.

I can't really find any performance benefits for the circuit in the question. However, I would grant that it is easier to make and use without much worry about stability and noise. Given that it has to work at only 24kHz you can probably consider it a good enough circuit. The following more traditional circuit would probably allow you to get higher speeds (assuming you still want to cancel any low-frequency background light like in your circuit).







simulate this circuit

C2 is only there in case the circuit is unstable (which is usually the case) and is usually a fraction of \$C_{PD}\$. The circuit will still work pretty well without L1. The bandwidth is then given by

$$BW(Hz) \approx \frac{1}{2\pi\cdot R_{tia}\cdot C_2}$$