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I am designing a simple photodiode light sensor circuit. The circuit will use a transimpedance amplifier to generate a usable voltage from the light level detected by the photodiode. The most basic form of the circuit looks like this:

However, many designs that I see also include a very small feedback capacitor (usually a few pF) in parallel with the feedback resistor. Most of the information online also assumes that the amplifier will be operating in AC, with a pulsed laser or some such thing to generate light. Is this feedback capacitor necessary when using the amplifier in DC?

I apologize if this question has already been asked, but I couldn't find a clear answer in my research.

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Yes, the feedback capacitor is necessary with most op-amps.

Op-amps don't like capacitance to ground on their negative input, and photodiodes generally have lots of area, which means lots of capacitance. So the feedback capacitance in parallel with the feedback resistance gives more phase margin and keeps the circuit from oscillating.

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It actually restricts the bandwidth but improves stability.

Its necessary value depends on the actual type of opamp and its input capacitance.

Without it, the transimpedance amplifier might oscillate with a time constant of around \$R_F\cdot C_{in}\$.

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Is this feedback capacitor necessary when using the amplifier in DC?

It might be necessary but, that depends on the circuits that follow. Generally not needed (and note that not fitting it doesn't make the op-amp less stable as some might have said).

For the TIA itself (and AC performance), it's used to reduce noise gain (to prevent a lot of high frequency stuff appearing at the output). It doesn't improve stability; it slightly worsens stability but not to any point where things can go wrong.

Noise gain is due to capacitance on the inverting input pin: -

enter image description here

The effect it has is this: -

enter image description here

So, to reduce that increase in noise, we use a feedback capacitor.

Images from here

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  • \$\begingroup\$ Looking at that graph in your answer. The difference in slopes between the Aol plot and the 1/beta plot is 40dB/decade and so that amplifier will oscillate. A correctly sized parallel feedback capacitor will introduce a feedback zero, flattening the 1/beta curve at the critical intersection frequency. This reduces the difference in slopes between the two plots, increasing phase margin and stabelising the amp. \$\endgroup\$
    – user173271
    Commented Aug 11, 2022 at 8:56
  • \$\begingroup\$ @James the plot Andy shows is from the grounded cap, which causes the oscillation, not from the remedying feedback cap. \$\endgroup\$
    – tobalt
    Commented Aug 11, 2022 at 9:37
  • \$\begingroup\$ @tobalt Exactly! \$\endgroup\$
    – user173271
    Commented Aug 11, 2022 at 9:56
  • \$\begingroup\$ @Andyaka I understand perfectly. What you are missing is that the grounded cap introduces a feedback pole which adds -90 degrees to the loop response (beta response), as indicated by the +20dB/decade slope of the 1/beta response in your graph. At the point where the two responses cross (on your graph) the loop gain is unity, there is -90 degrees lag from the open loop gain (-20dB/decade) and a further -90 degrees lag from the feedback (beta) response. Therefore -180 degrees lag at unity gain resulting in very likely oscillation. \$\endgroup\$
    – user173271
    Commented Aug 11, 2022 at 10:05
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    \$\begingroup\$ @James OK, point taken, you have convinced me that there is a point of potential instability when there is no feedback capacitor hence, what I said earlier, I take back completely and, thank you for the lesson I've learnt today. I mean it too; anyone who corrects something in my head is worthy of thanks. I even went so far to simulate using an AD8065 (I've used several times in low-speed TIAs) and saw exactly what you stated. But, the capacitor still reduces unwanted noise! \$\endgroup\$
    – Andy aka
    Commented Aug 11, 2022 at 12:15

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