If you want a positive output then you can eliminate the second stage inverter as shown below.
View attachment 277024

I want to make exactly that hand drawn schematics. on the last stage he drew positive output. maybe he did it wrong. can you explain how each one of that 3 stages works?

 

I want to make exactly that hand drawn schematics. on the last stage he drew positive output. maybe he did it wrong. can you explain how each one of that 3 stages works?

The first stage is the integrator/ inverting amplifier. In the original drawing you see the output is a negative signal, therefore the input must be positive. The second stage is another inverter amp with a gain of 1 to just invert the signal back to positive as shown. The last stage is a filter-buffer to clean up the signal without changing the polarity as the signal is connected to the positive input of the op amp.
If you go back to the circuit in post #35 and input a positive signal the output will also be positive.

 

The first stage is the integrator/ inverting amplifier. In the original drawing you see the output is a negative signal, therefore the input must be positive. The second stage is another inverter amp with a gain of 1 to just invert the signal back to positive as shown. The last stage is a filter-buffer to clean up the signal without changing the polarity as the signal is connected to the positive input of the op amp.
If you go back to the circuit in post #35 and input a positive signal the output will also be positive.

Thank you man it was very helpful. Problem was I added wrong signal to the input. Finally... but I still can't understand why he added 1/2 VCC to the photo diode.

 

You want the output to be able to swing both positive and negative about a reference point. But where is the reference point?

If the opamp is powered from a 0 to +12V supply, for example, the signal cannot go below 0V.

If you set the reference point to 1/2 Vcc then the output can swing both sides of 1/2 Vcc. This is what we mean by pseudo ground.

 

but I still can't understand why he added 1/2 VCC to the photo diode.

I don't think that was his intent. He probably meant you will need a 43 volt source for the diode and a 1/2Vcc circuit for the op amp bias and let you decide what circuit that would entail including the photo diode type, op amps and Vcc supply.
Redrawn I see the complete circuit as this:

 

I don't think that was his intent. He probably meant you will need a 43 volt source for the diode and a 1/2Vcc circuit for the op amp bias and let you decide what circuit that would entail including the photo diode type, op amps and Vcc supply.
Redrawn I see the complete circuit as this:
View attachment 277093

Your intelligence is great. Probably you are right. I can't say anything. Just I want to thank you all guys especially sghioto for his helpful circuit drawn. That circuit was for a scintillation detector which is designed for detection for nuclear radiation in gamma range. Scintillation crystal sends photons to the specially designed silicon photo diode. It can send millions of pulses in very short time.