I am trying to understand the inner workings of a photoelectric object sensor I have in my hands. (modulated with 24 kHz) Here is the very beginning of the receiver part:


simulate this circuit – Schematic created using CircuitLab

This doesn't look like other photodiode detection circuits I have seen around. This actually doesn't look like anything I saw before. It seems like there is a band pass filter in the non inverting input. Then, the filtered signal is amplified by -10, and sent to the output.

  1. Am I correct?

  2. What are the pros/cons of this configuration against other classical photodiode configurations?

  3. How can I calculate the corner frequencies for such a band pass filter? (If it is one)

  • 2
    \$\begingroup\$ You should add values for V1, C1 and C2 even if it is an estimation because V1 having a value > +0.5 V doesn't make sense (D1 would be in forward mode). If C1 is 100 nF or 100 pF makes a huge difference on circuit behavior. Also there is no DC path to ground for the + input of the opamp: that simply cannot work like this. \$\endgroup\$ Commented Jan 10, 2019 at 10:51
  • \$\begingroup\$ @Bimpelrekkie sorry for the photodiode, it should be reverse biased. Fixed \$\endgroup\$
    – C K
    Commented Jan 10, 2019 at 11:23
  • \$\begingroup\$ I think, R1 and R2 used for adjusting photodiode sensitivty. It is not about filter. Also, this circuit may designed for a special purpose, Where did you find this? It may help us to understand. also gain is not -10. it is 11. you may try to simulate only C1 R3 C2 to find cutoff. \$\endgroup\$
    – Koray
    Commented Jan 10, 2019 at 11:41
  • \$\begingroup\$ @Koray it is from a U shaped photoelectric sensor. I found the connections and drew the schematic. \$\endgroup\$
    – C K
    Commented Jan 10, 2019 at 13:05
  • \$\begingroup\$ What Bimpelrekkie said, also AC gain is +11, not -10. \$\endgroup\$ Commented Jan 10, 2019 at 14:37

1 Answer 1


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 – 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

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


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