I don't understand the nature of the input signal. There's no rail to rail TIA, so the V- is created to avoid that issue? The signal is primarily positive going? The negative supply is just to handle the undershoot?
No "undershoot" is allowed. In my circuit, a 45pA pulse, with a 820K TIA gain and the FET in place makes a 39uV signal, negative going.
The following stage lays on another 41dB of gain, bringing it up to about 5.2mV positive going.
If we decided that is as big as the pulse gets, we put in another x 120 gain, to finally get a 650mV pulse copy of the original current, still positive going.
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In another scenario, where I assume a "big" diode pulse is 2.85nA, The first stage gain with the FET + Rf =330K delivers a negative going 840uV.
Then, a +20dB inverting stage delivers a positive going about 86mV
A final +27dB non-inverting gain stage brings it up to +2.0V.
The internal reference is 2.048 volts
What final gain we use all depends on how big a current pulse happens from the most energetic photon we ever care to display
This is why I am after being able to have switched gain ranges.
In theory, a Am241 source could provoke a pile of 59.3KeV pulses from tungsten, along with some 8.4KeV and 9.7KeV. We only find out when we show it a carbide end mill
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The op-amp active driven 0V supply circuit
It is a circuit I have used before, and I completely understand how it can confuse. Imagine temporarily that the output is not connected to 0V, but instead is called Vout. Then imagine the negative end is indeed connected to the "0V". We then have a simple circuit where "0V" is the negative supply for the opamp, 3.3V is the positive supply. and the voltage divider delivers +0.8V, which is buffered by the opamp unity gain follower.
Now if instead we "disconnect" the 0V from the bottom of the 3.3V supply, and connect the output of the opamp to that 0V. The circuit still operates in exactly the same way, except instead of asserting that the opamp output is +0.8V, it has to assert that V+ is 2.50 volts, and that V- is -0.8V, up to the limit of the current the opamp output is capable of giving. It becomes a dual regulator for the TIA low current circuits, limited to +2.5V, and offering the possibility of -0.8V for the op-amps, so never requiring them to go all the way to the rail.
The key thing is that the 3.3V supply negative end, if it had a 0V from anywhere else, is not the same 0VA as we are using for the TIA.
The scheme works well, but has it's inconveniences. You get the same result using two separate regulators, but the need is to contrive a no-noise negative supply. The really important thing is that no signal drives or powers the ADC above +2.5V. It's absolute maximum is 2.7V, and not recommended ever.
You may be right that the good way is just to use batteries.