Current direction in a multivibrator

I have been studying multivibrators lately, but I have hit a wall right now. I have attached a sketch for an astable multivibrator. I know the basic stuff. When Vout is high then Vin will rise as CT1 is charging. Since the op-amp has negative feedback, then Vref will follow Vin until the upper limit of the Schmitt trigger is hit. Then Vout will turn low.
My problem is that in the split second where Vout changes from Vcc to 0V, then CT1 will discharge through R130. Vref is also higher than 0V, so a current will go through R107 and towards Vout. So what I see is two positive currents moving towards Vout. But this current has to go somewhere. Vdd of the op-amp is connected to ground, so I can't imagine a current moving into the output of the op-amp, but it is the only solution as I see it. I hope someone can clarify how the currents are moving when the capacitor is discharging.

• Aslund, Your reasoning about the current is correct. You only need to imagine the internal structure of the op-amp output stage. It is the so-called "push-pull stage" that consists of two complementary transistors - upper and lower. When the output voltage is high, the current exits from the upper transistor; when the voltage is low, the current enters the lower transistor and goes to ground (your case). If you need more explanations, I can draw the current path (loop)... but it will be more useful for you to do it yourself. Commented Mar 12, 2021 at 21:54

However, you should note that your circuit also has positive feedback through R107. Your assumption that $$\V_{IN} = V_{REF}\$$ is not necessarily valid.