Consider this circuit that filters the input signal and amplifies it:

If I am not mistaken, it is preferable to set a high gain in the first stage because, by amplifying the second stage instead, I would also amplify the noise of the first stage
All correct so far?

Then I wonder:
1) If the gain of the 1st stage (Rf*I) is simply set by the resistor Rf .. why would I also need the 2nd stage to amplify?
.. the only answer I can think of is that: high gain --> high Rf --> low freq. cutoff .. so I have the gain-bandwidth trade-off that does not allow me to set Rf to my liking
Is this the "correct" answer?

2) What is the direction of the input current on the photodiode? Toward GND? Or does it depend from case to case?

 

The first stage is not a high gain filter. It is a transimpedance amplifier. It is used to convert current to voltage.

https://en.wikipedia.org/wiki/Transimpedance_amplifier

Another way of looking at the first stage, it is a charge integrator.

Charge accumulates on the capacitor Cf. As an integrator, the voltage output would continue to rise until it cannot go any further (only limited by the supply voltage). Rf (across Cf) is there to discharge the capacitor over time. In high speed pulse detection systems, active discharge is sometimes applied, i.e. a transistor is used to discharge the capacitor very rapidly.

 

In addition to what M C says, gain-Bandwidth is always a consideration. And the value of Cf will affect the frequency response. The photo-diode current would be away from the "ground" connection.

What is interesting is that some LEDs will produce a significant voltage with light shining on them, while others do not . I have seen this with red green, and even amber LEDs. I have not experimented with IR LEDs in that aspect.

 

What will be the purpose of the Circuit ?

Are You trying to generate a "Pulse" at some particular threshold of Light-Intensity,
or a continuously varying Analog-Voltage ?
.
.
.

 

A serious consideration with "transimpedance amplifier circuits, especially those boosting the signals from polymer capacitive sensors, is that the material's electrical properties change with temperature. That sort of change affects a whole lot of how the circuit works.

 

The first stage is not a high gain filter. It is a transimpedance amplifier. It is used to convert current to voltage.

https://en.wikipedia.org/wiki/Transimpedance_amplifier

Another way of looking at the first stage, it is a charge integrator.
View attachment 323057
Charge accumulates on the capacitor Cf. As an integrator, the voltage output would continue to rise until it cannot go any further (only limited by the supply voltage). Rf (across Cf) is there to discharge the capacitor over time. In high speed pulse detection systems, active discharge is sometimes applied, i.e. a transistor is used to discharge the capacitor very rapidly.

In addition to what M C says, gain-Bandwidth is always a consideration. And the value of Cf will affect the frequency response. The photo-diode current would be away from the "ground" connection.

What is interesting is that some LEDs will produce a significant voltage with light shining on them, while others do not . I have seen this with red green, and even amber LEDs. I have not experimented with IR LEDs in that aspect.

Thank you for the clarification.
This configuration, when is it called a transimpedance amplifier and when is it called an integrator? Are they exactly the same thing and can it be called either way? Or are there conditions such that it can be called one way or the other?

 

when is it called a transimpedance amplifier and when is it called an integrator?

You are trying to compare apples and oranges.

A transimpedance amplifier converts current to voltage.
An integrator integrates either current of voltage, depending upon its configuration.

 

Thank you for the clarification.
This configuration, when is it called a transimpedance amplifier and when is it called an integrator? Are they exactly the same thing and can it be called either way? Or are there conditions such that it can be called one way or the other?

Transimpedance amplifier
Charge coupled amplifier
Integrator
Low pass filter

These are different names for different applications.

If you look at an op amp integrator configuration it looks the same as a transimpedance amplifier.



In this integrator configuration, a constant input voltage is integrated (summed) over time producing a linear ramp. The direction of the ramp reverses on the opposite input voltage polarity. This circuit suffers from the fact that there is no DC voltage reference. Hence the output triangular wave can be any average DC voltage. A resistor in parallel with C1 would solve that problem.

For the same reason, the charge coupled amplifier below will accumulate (integrate) charge on capacitor CF. The charge will continue to accumulate until the amplifier output voltage reaches the power supply rail. The parallel resistor RF is there to discharge the capacitor.




Notice that this op amp low-pass filter circuit below looks exactly the same as the two before. That is because a low-pass filter is an integrator. Rather than using the other terms, one would call it a low-pass filter when the purpose of the circuit is to remove high frequencies from the input signal, for example, in audio applications.




Below is a simple RC low pass filter. It is also an integrator circuit. Capacitor C accumulates charge and tends to smooth out any high frequency fluctuations. Hence high frequency signals are attenuated. The two graphs show the difference between a low-pass and high-pass filter. Interchanging R and C in the circuit will create a high-pass filter. The X-axis is frequency while the Y-axis is amplitude or voltage. These are called frequency response curves or Bode plots.

 

Transimpedance amplifier
Charge coupled amplifier
Integrator
Low pass filter

These are different names for different applications.

If you look at an op amp integrator configuration it looks the same as a transimpedance amplifier.

View attachment 323197

In this integrator configuration, a constant input voltage is integrated (summed) over time producing a linear ramp. The direction of the ramp reverses on the opposite input voltage polarity. This circuit suffers from the fact that there is no DC voltage reference. Hence the output triangular wave can be any average DC voltage. A resistor in parallel with C1 would solve that problem.

For the same reason, the charge coupled amplifier below will accumulate (integrate) charge on capacitor CF. The charge will continue to accumulate until the amplifier output voltage reaches the power supply rail. The parallel resistor RF is there to discharge the capacitor.

View attachment 323198


Notice that this op amp low-pass filter circuit below looks exactly the same as the two before. That is because a low-pass filter is an integrator. Rather than using the other terms, one would call it a low-pass filter when the purpose of the circuit is to remove high frequencies from the input signal, for example, in audio applications.

View attachment 323199


Below is a simple RC low pass filter. It is also an integrator circuit. Capacitor C accumulates charge and tends to smooth out any high frequency fluctuations. Hence high frequency signals are attenuated. The two graphs show the difference between a low-pass and high-pass filter. The X-axis is frequency while the Y-axis is amplitude or voltage. These are called frequency response curves or Bode plots.
View attachment 323200

Thank you very much for the clarification!