photodiode
The current in photo diode because of generation rate G and depletion width W is given as qGW (neglecting diffusion currents).
When the light is incident on the diode excess carriers are generated in depletion region in pairs and these are swept to respective P and N regions.
If the above is true, then why we are not taking current because of both holes and electrons i.e. current should be 2qGW.
Just an idea - you could try using the propagation delay through a logic gate to form very narrow pulses. By adding more gates you can increase the width of the pulse.
A simple (variable) pulse generator (e.g. a 555 astable) could be used to alter pulse rate. The circuit works on the rising edge only.
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
We agree that current is constant in a circuit.
Take an arbitrary point in your voltage source - diode circuit. Make a cut. How many charges per second flow through that? That's current.
You'll see that for every recombination, exactly one electron (or one hole) flows. I don't see how that would lead to a factor of 2.