The capacitor gives (defines) the frequency band of the amplifier, I think.

As the photo diode itself is an intrinsic low-pass, receiver circuits try to compensate this to some extend - implementing high-passes where possible.

Be aware that the 'extreme' resistances (generally any resistance > 500k) will create problems as parasitic resistances (depending on temperature, humidity, cleanliness etc.) will 'play' with your circuit.

Hello Xianqian,

First apply the basic equation to the RC combination as f = 1/(2*pi* R * C).

This will give you the high-frequency roll-off for the amplifiers. If you apply the same equation to the parallel circuit see that you shall get the two values of the frequencies.

This should give you some idea of the signals that your circuit is handling.

Best Luck!

A useful way to look at the problem would be to simply lay out the complex equations and see where the maths take you. For a more intuitive approach what I like to do is look at the extremes:

The 2nd stage amplifier has RC filters both at its input and at the feedback and taking example no2. the input RC constant corresponds to 40Hz and the output at 14,5MHz. At very low frequency (

I'm sorry to reply so late. I am engaged in reviewing the final exam. This paper  will give an answer:《How to enlarge the bandwidth without increasing the noise in op-amp-based transimpedance amplifier》.

Good question although I understand it.

The preamplifier has to be integrating = very low bandwidth to keep the noise down and have high gain. However, we can partially recover the bandwidth for input signal by so called zero-pole compensation technique (which is used). It has to be understood via Laplace transform domain:

We are given some transfer function (integrating preamp)

f(s) = 1/(1 + s*Tau0)

We want to replace it with another Tau1, so will be Tau1