The first stage is the transimpedance amplifier, meaning an amplifier for which the output
voltage is proportional with the input
current. It turns amps into volts.
Intuitively, it works like this:
- in an ideal opamp, no current flows in/out of the input pins in+ and in- of the opamp
- an opamp with negative feedback will do its best to keep zero volts difference between its in+ and in+ pins
- the PIN diode with all the resistors and input capacitors on the leftside produces
current pulses (not voltage)
Now, if you tie in+ pin of the opamp to GND, then the opamp (here IC1a in Fig.1 from the gamma detector) will vary its output so to keep its in- pin also at zero volts. However, the current in/out of any input pin is zero, so all the current pushed by the PIN diode will go through the feedback resistor R2 (and through the capacitor C1, too).
For, let's say 1uA coming from the PIN photodiode, the voltage drop on the R2 will be 1uA*10M=10V (if we neglect C1). Because one terminal of the R2 is at zero volts (in- must be zero, because in+ is tied at GND), then the other terminal of R2 (which is also the output of the first amplifier) will show 10V.
That's how a
transimpedance amplifier works, in principle, and that's how IC1a works, too.

About the AC response of IC1a, the cutting frequency is indeed about 3.4kHz (
https://www.ti.com/lit/an/sboa268a/sboa268a.pdf), but the roll off of a first order low pass filter is only 20dB/decade, so at 34kHz it will still amplify, just that it will amplify 10 times less than at DC. However, the output voltage from the first opamp might be still enough, considering that after the first opamp there are 3 more op amps (working as voltage amplifiers) further amplifying the signal 1000 times (60 dB all 3 together).
If, say the diode produces 1uA, then the output will be 10V at DC and roughly about 1V at 34kHz, and the LTspice simulation shows about 1Vpp for 5us pulses of 1uA.
