a freshly charged AA NiMH cell in good condition is about 1.4 volts.
What op amp are you using? With many, there are limits on the allowable input voltage range.
Post a complete circuit diagram. The battery appears to be floating, so the reference for V1 and V2 is not clear.
As mentioned, the voltage plotted that looks wierd is actually the battery reaching the charged state at 1.4 V (OP's output voltage is about 1.75 times the battery voltage.)The battery is alternating between charging and discharging at the charged voltage. (probably a few hundred mV)
That's not an issue
We have no way of knowing what the rest of the charging circuitry looks like, and how it is connected to the existing circuitry.
I don't understand how a bleed resistor could solve that. Furthermore I want to get reasonable precise capacity values by integrating the current over time. For this any unmeasured current will increase my error quite a bit.
The OP AMP output is not governing the "jump".
Also it cannot be the cell because the "jump" also occurs with other cell chemistries, following the same pattern. The start of the "jump" varies a bit (2.43V - 2.443V) but the upper value is always 2.46.
@MarkT: Decoupling sounds interesting. However I have no idea how to do this. Would you mind giving me an example or an online link for that? Also suggestions how to make the circuit more robust would be nice.
Your schematic isn't 100% clear to me. For one thing, you have two nodes named Vout but I know they're not connected to each other.From the red arrows, however, it seems that you have a mode where you charge the battery through a FET/shunt which is controlled to provide a constant current?Since your cells get ground through this FET/shunt you are putting a common mode signal at the inputs to your op amp. It sounds like this has the potential to cause the results you're seeing, although I don't have a good reason why it'd happen at the end of charge (when you'd expect the FET to be fully on).Edited to Add: wait...no...if your FET was fully on at the end of charge, then you'd have 12V across your battery. I don't understand how your FET is controlled...