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A common noise model of a Photodiode amplifier is the following: enter image description here

(for example in https://e2e.ti.com/cfs-file/__key/communityserver-discussions-components-files/14/noise11_2D00_Photodiode_2D00_Noise-1.pdf)

There, the contribution of the source \$i_{n,OPA}\$ is considered to only affect the output without the noise current considering the path through the photodiode shunt resistance \$R_{sh}\$. They calculate: $$E_{noI} = \sqrt{i_{n,PD}^2 + i_{n,OPA}^2} * R_F * BW$$

In my understanding, the effect of the input current noise of the OpAmp should be based on the OpAmps source impedance, which would be \$R_{sh} || R_F\$ and therefore in certain cases the output referred noise of \$i_{n,OPA}\$ would be much lower than considering only \$R_F\$. Now I trust the authors of that document more than myself, yet the question remains whether their formula is just a simplification for the case \$R_F << R_{sh}\$ or if the noise current (for some reason I have yet to understand) will always take the path through the feedback resistor?

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  • \$\begingroup\$ You've added the tag 'transimpedance amplifier'. Do you know what such an amplifier does? What's the input impedance of an ideal TI amplifier? What's the input current of an opamp? How does the opamp behave to have zero input current when there's an external current impressed upon it? \$\endgroup\$ Commented Feb 25 at 8:40
  • \$\begingroup\$ Doesn't the same argument also apply for the signal current Seppde? \$\endgroup\$ Commented Feb 25 at 8:44
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    \$\begingroup\$ @Andyaka indeed but for the signal current (or its noise component, especially the thermal noise component) I considered it (more) logical that it cannot take the path through the resistor that created it.... AHHH. Hold your horses, I'm starting to understand \$\endgroup\$ Commented Feb 25 at 9:05
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    \$\begingroup\$ Take care (I have) to distinguish between opamp, which has infinite input impedance, and TI amp, which is an opamp with feedback resistor and grounded +ve input, which has a zero input impedance. \$\endgroup\$ Commented Feb 25 at 9:09
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    \$\begingroup\$ Cj does appear once you have to do stability calculations with Cf. If you do full calculations with a TI amplifier, and compare it with just a voltage follower across a load resistor which puts the output voltage across Cj, you'll see an order of magnitude difference in bandwidth. 'Cj doesn't appear in the bandwidth calculation' is a lazy back-of-the-envelope way of expressing this difference. \$\endgroup\$ Commented Feb 25 at 10:51

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Hopefully, after the comments above you now realize that because the negative feedback produces a virtual ground at the inverting input, both signal and noise currents flow exclusively into the feedback resistor. Clearly this is the ideal situation and is, I suspect, the scenario which is discussed in the pdf file you linked.

Reality is different. For instance: -

  • As frequency rises the op-amp's open-loop gain falls and the virtual ground gets less ideal. As a result, current will gradually begin to flow in \$R_{SH}\$ but, considering how large \$R_{SH}\$ is compared to a non-ideal virtual ground impedance (maybe 10 Ω to 1000 Ω in extremes), the error will still be very small because \$R_{SH}\$ is usually in the high 100's of kΩ to MΩ.
  • Input offset voltages will mean that there is an offset seen between inverting input and non-inverting (grounded) input. This will cause a small current to flow in \$R_{SH}\$.
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