Your circuit is apparently working fine, you just don't realize it.
With the drain left floating, the MOSFET cannot pull any current, and therefor the integrator keeps providing a lower and lower gate drive, trying to draw current.
With the drain grounded and the reference at 0 volts, the integrator goes just low enough to start pulling a tiny amount of current and satisfies the loop.
What you now need to do is get a set of power resistors, such as 10 ohm / 25W, 20 ohm / 25 watt, etc. and use them as loads. From the voltage across your load resistors you can determine the current through them.
However, before you do that, you need desperately to provide a heatsink for your MOSFET. Its' peak power dissipation at a current of 2A will be nearly 30 watts, worst case (for very low load resistances).
You also need to make sure that your shunt resistor is of high enough power. If you're following the app note religiously and using a .2 ohm resistor, it needs to be a 1-watt or better unit, since at 2 amps it will dissipate (2 x 2 x .2) = .8 watts.
You would also do well to reconsider your power wiring. The contacts on a solderless breadboard may not handle 2 amps well.
Finally, you need to make damn sure you can't provide a reference voltage greater than 2 volts.
And finally finally, I suggest you do a good deal of testing with resistors, in order to convince yourself that the circuit really is working. I suspect that you may find it doesn't work nearly as well when you give it a highly inductive load like a coil. It may work, and it may not, since the effects of the inductance will tend to counteract the effects of the loop integrating capacitor, and the result may be instability.
Otherwise, congratulations.
There is nothing is wrong with your circuit. The wide pulse width is because you're feeding your diff_amp_out to a comparator.
It can only supply you with a pulse of full voltage if the input (in this case) is even slightly above zero volts. I like how your scope shows the tiny delays from stage to stage.
You could adjust the 'trip' level of the comparator, but the output will always be +5 V.
If you just want to amplify diff_amp_out, add another AD8003 op-amp and set your gain to fit your needs. At least you keep your true signal pulse width. Admittedly, working with these nS pulses is tough.
CFA amps are designed for current feed back, which lowers the input impedance, but greatly increases its bandwidth and high frequency response compared to VFA.
All video and RF amplifiers are the CFA type. The VFAs are best for audio and precision DC measurements.
Best Answer
I've found this (below) useful. It's a site offered by Analog Devices and appears to give useful information on the design of TIAs for photodiodes: -
The wizard produces several output specifications such as ENOB, SNR, pulse response, bode plot, spectral noise etc.. All good stuff. It also recommends the best/most suitable op-amp. I think it's a good start to any design and allows you to play around with values.
If you are not using high capacitance photodiodes and speed isn't that important the JFET amplifier (bootstrap circuit) is probably not needed in your design.